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Full-length undergraduate textbook, Dover 1985 reprint of the 1963 Harper & Row edition, by Tenenbaum (Cornell) and Pollard (Purdue). Organized as numbered lessons covering first-order types, applications, linear equations, operators and Laplace transforms, second-order problems, systems, and series methods. This is a published book by others, kept in Phil's math downloads.

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ORDINARY DIFFERENTIAI. AnEleme MothemotiEQPATIQNSryTextbook grneerrng onudenrs of eScrences Morris Tenenboum Horry Pollard ORDINARY DIFFERENTIAL EQUATIONS AnElementary Tartbookfor Students ofMathematics, Engineering, andtheSciences Morris TenenbamnCornell University Harry PollardPurdue University DOVER PUBLICATIONS, INC., NEW YORK Copyright ©1963byMorris Tenenbaum andHarry Pollard Allrights reserved under PanAmerican andInternational Copyright Con- ventions. ThisDover edition, ﬁrstpublished in1985, isanunabridged andcorrected republication ofthework ﬁrstpublished byHarper &Row, Publishers, Inc., NewYork, in1963. Manufactured intheUnited States ofAmerica Dover Publications, Inc.,31East2ndStreet, Mineola, N.Y. 11501 Lilrrary ofCongress Cataloging inPublication Data Tenenbaum, Morris. Ordinary differential equations. Reprint. Originally published: NewYork :Harper &Row, 1963. Bibliography: p. Includes index. 1.Differential equations. I.Pollard, Harry, 1919- II.Title. QA372.T4 1985 515.3'5 85-12983 ISBN 0486-64940-7 Contents PREFACE FOR THE TEACHER XV mnrncn ronTHEsrunmrr xvii l. 2.BASIC CONCEPTS 1 Lesson 1.How Differential Equations Originate. 1 Lesson 2.The Meaning oftheTerms Setand Function. Im- plicit Functions. Elementary Functions. A.The Meaning oftheTerm Set.6B.The Meaning oftheTerm Function ofOne Independent Variable. 6'C.Function ofTwo Inde- pendent Variables. l1 D.Implicit Function. 14 E.The Elemen- tary Functions. 17'5 Lesson 3.The Dilferential Equation. A.Deﬁnition ofanOrdinary Diﬁerential Equation. Order ofaDiﬁer- ential Equation. 20 B.Solution ofaDiﬂerential Equation. Explicit Solution. 21C.Implicit Solution ofaDiﬁerential Equation. 2420 Lesson 4.The General Solution ofaDifferential Equation. A.Multiplicity ofSolutions ofaDifferential Equation. 28B.Method ofFinding aDifferential Equation ifItsn-Parameter Family ofSolu- tions IsKnown. 81C.General Solution. Particular Solution. Initial Conditions. 3828 Lesson 5.Direction Field. A.Construction ofaDirection Field. The Isoclines ofaDirection Field. 88 B.The Ordinary andSingular Points oftheFirst Order Equation (5.11). 4138 SPECIAL TYPES OF DIFFERENTIAL EQUATIONS OF THE FIRST ORDER 4-6 Lesson 6.Meaning oftheDifferential ofaFunction. Separable Differential Equations. A.Differential ofaFunction ofOneIndependent Variable. 47B.Dif- ferential ofaFunction ofTwo Independent Variables. 50 C.Differ- ential Equations with Separable Variables. 51 V4-7 viConranrs Lesson 7.First Order Diﬂerential Equation with Homogeneous Coefficients. 57 A.Deﬁnition ofaHomogeneous Function. 67 B.Solution ofaDif- ferential Equation inWhich theCoefficients ofdz:anddyAreEach Homogeneous Functions oftheSame Order. 58 Lesson 8.Differential Equations with Linear Coefficients. 62 A.AReview ofSome Plane Analytic Geometry. 6'2 B.Solution of aDifferential Equation inWhich theCoefficients ofdzand dyare Linear, Nonhomogeneous, andWhen Equated toZeroRepresent Non- parallel Lines. 63 C.ASecond Method ofSolving theDifferential Equation (8.2) with Nonhomogeneous Coefficients. 66 D.Solution ofaDifferential Equation inWhich theCoefficients ofda:anddyDeﬁne Parallel orCoincident Lines. 67 Lesson 9.Exact Differential Equations. 70 A.Deﬁnition ofanExact Differential and ofanExact Differential Equation. 72 B.Necessary andSufficient Condition forExactness andMethod ofSolving anExact Differential Equation. 73 Lesson 10. Recognizable Exact Differential Equations. Integrating Factors. 80 A.Recognizable Exact Diﬁerential Equations. 80 B.Integrating Factors. 82 C.Finding anIntegrating Factor. 84 Lesson ll. The Linear _Diﬂ'erential Equation oftheFirst Order. Bernoulli Equation. 91 A.Deﬁnition ofaLinear Diﬁerential Equation oftheFirst Order. 91 B.Method ofSolution ofaLinear Differential Equation oftheFirst Order. 92 C.Determination oftheIntegrating Factor e-lP(”)"”. 94 D.Bernoulli Equation. 95 Lesson 12. Miscellaneous Methods ofSolving aFirst Order Differential Equation. 99 A.Equations Permitting aChoice ofMethod. 99 B.Solution by Substitution andOther Means. 101 3.PROBLEMS LEADING TO DIFFERENTIAL EQUATIONS OF THE FIRST ORDER 107 Lesson 13. Geometric Problems. 107 Lesson I4. Trajectories. 115 A.Isogonal Trajectories. 115 B.Orthogonal Trajectories. 117 C.Orthogonal Trajectory Formula inPolar Coordinates. 118 CONTENTS Lesson 15. Dilution and Accretion Problems. Interest Prob- lems. Temperature Problems. Decomposition and Growth Problems. Second Order Processes. A.Dilution andAccretion Problems. 122 B.Interest Problems. 126 C.Temperature Problems. 129 D.Decomposition and Growth Problems. 131 E.Second Order Processes. 184 Lesson 16. Motion ofaParticle Along aStraight Line- Vertical, Horizontal, Inclined. A.Vertical Motion. 13.9 B.Horizontal Motion. 160 C.Inclined Motion. 164 Lesson 17. Pursuit Curves. Relative Pursuit Curves. A.Pursuit Curves. 168 B.Relative Pursuit Curve. 177 Lesson 17M. Miscellaneous Types ofProblems Leading to Equations oftheFirst Order A.Flow ofWater Through anOriﬁce. 183 B.First Order Linear Electric Circuit. 184 C.Steady State Flow ofHeat. 185 D.Pres- sure—Atmospheric andOceanic. 186' E.Rope orChain Around a Cylinder. 188 F.Motion ofaComplex System. 189 G.Variable Mass. Rocket Motion. 191 H.Rotation oftheLiquid inaCylin- der. 1.93 LINEAR DIFFERENTIAL EQUATIONS OF ORDER GREATER THAN ONE Lesson 18. Complex Numbers and Complex Functions. A.Complex Numbers. 197 B.Algebra ofComplex Numbers. 200 C.Exponential, Trigonometric, andHyperbolic Functions ofComplex Numbers. 201 Lesson 19. Linear Independence ofFunctions. The Linear Differential Equation ofOrder n. A.Linear Independence ofFunctions. 205 B.The Linear Differ- ential Equation ofOrder n.207 Lesson 20. Solution ofthe Homogeneous Linear Differential Equation ofOrder nwith Constant Coefﬁcients. A.General Form ofItsSolutions. 211 B.Roots oftheCharacteristic Equation (20.14) Real andDistinct. 213 C.Roots ofCharacteristic Equation (20.14) Real butSome Multiple. 214 D.Some orAllRoots oftheCharacteristic Equation (20.14) Imaginary. 217vii 122 138 168 I83 196 197 205 211 vm CONTENTS Lesson 21. Solution oftheNonhomogeneous Linear Differential Equation ofOrder nwith Constant Coefficients. A.Solution bytheMethod ofUndetermined Coeﬂicients. 221 B.So- lution bytheUseofComplex Variables. 230 Lesson 22. Solution oftheNonhomogeneous Linear Differential Equation bytheMethod ofVariation ofParameters A.Introductory Remarks. 233 B.The Method ofVariation of Parameters. 233 Lesson 23. Solution ofthe Linear Differential Equation with Nonconstant Coefficients. Reduction ofOrder Method. A.Introductory Remarks. 241 B.Solution oftheLinear Differential Equation with Nonconstant Coefficients bytheReduction ofOrder Method. 242 OPERATORS AND LAPLACE TRANSFORMS Lesson 24. Differential and Polynomial Operators. A.Deﬁnition ofanOperator. Linear Property ofPolynomial Opera- tors. 26l B.Algebraic Properties ofPolynomial Operators. 265 C.Exponential Shift Theorem forPolynomial Operators. 260 D.So- lution ofaLinear Differential Equation with Constant Coeflicients by Means ofPolynomial Operators. 262 Lesson 25. Inverse Operators. A.Meaning ofanInverse Operator. 269 B.Solution of(25.1) by Means ofInverse Operators. 272 Lesson 26. Solution ofaLinear Differential Equation by Means ofthe Partial Fraction Expansion ofInverse Operators. A.Partial Fraction Expansion Theorem. 283 B.First Method of Solving aLinear Equation byMeans ofthePartial Fraction Expansion ofInverse Operators. 288 C.ASecond Method ofSolving aLinear Equation byMeans ofthePartial Fraction Expansion ofInvere Operators. 290 Lesson 27. The Laplace Transform. Gamma Function. A.Improper Integral. Deﬁnition ofaLaplace Transform. 292 B.Prop- erties oftheLaplace Transform. 295 C.Solution ofaLinear Equa- tionwith Constant Coeﬂicients byMeans ofaLaplace Transform. 296 D.Construction ofaTable ofLaplace Transforms. 302 E.The Gamma Function. 306 6. 7. 8.Comnnrs PROBLEMS LEADING TO LINEAR DIFFERENTIAL EQUATIONS OF ORDER TWO Lesson 28. Undamped Motion. A.Free Undamped Motion. (Simple Harmonic Motion.) 313 B.Deﬁ- nitions inConnection with Simple Harmonic Motion. 317 C.Exam- plesofParticles Executing Simple Harmonic Motion. Harmonic Oscil- lators. 328 D.Forced Undamped Motion. 388 Lesson 29. Damped Motion. A.Free Damped Motion. (Damped Harmonic Motion.) 34?’ B.Forced Motion with Damping. 359 Lesson 30. Electric Circuits. Analog Computation. A.Simple Electric Circuit. 36.9 B.Analog Computation. 375 Lesson 30M. Miscellaneous Types ofProblems Leading to Linear Equations oftheSecond Order A.Problems Involving aCentrifugal Force. 380 B.Rolling Bodies. 381 C.Twisting Bodies. 383 D.Bending ofBeams. 888 SYSTEMS OF DIFFERENTIAL EQUATIONS. LINEARIZATION OF FIRST ORDER SYSTEMS Lesson 31. Solution ofaSystem ofDifferential Equations. A.Meaning ofaSolution ofaSystem ofDiﬁerential Equations. 398 B.Deﬁnition andSolution ofaSystem ofFirst Order Equations. 894 C.Deﬁnition andSolution ofaSystem ofLinear First Order Equa- tions. 396 D.Solution ofaSystem ofLinear Equations with Con- stant Coeﬁicients bytheUseofOperators. Nondegenerate Case. 898 E.An Equivalent Triangular System. 405 F.Degenerate Case. f|(D)g2(D) —ﬂ1(D)fg(D) =0.413 G.Systems ofThree Linear Equations. 415 H.Solution ofaSystem ofLinear Diﬁerential Equa- tions with Constant Coefﬁcients byMeans ofLaplace Transforms. 418 Lesson 32. Linearization ofFirst Order‘Systems. PROBLEMS GIVING RISE TO SYSTEMS OF EQUATIONS. SPECIAL TYPES OF SECOND ORDER LINEAR AND NON- LINEAR EQUATIONS SOLVABLE BYREDUCING TOSYSTEMS Lesson 33. Mechanical, Biological, Electrical Problems Giving Rise toSystems ofEquations. A.AMechanical Problem—Coupled Springs. 440 B.ABiological Problem. 447 C.AnElectrical Problem. More Complex Circuits. 451 Lesson 38. Lesson 39.xConrnurs Lesson 34. Plane Motions Giving Rise toSystems ofEquations A.Derivation ofVelocity andAcceleration Formulas. 459 B.The Plane Motion ofaProjectile. 468 C.Deﬁnition ofaCentral Force. Properties oftheMotion ofaParticle Subject toaCentral Force. 470 D.Deﬁnitions ofForce Field, Potential, Conservative Field. Conser- vation ofEnergy inaConservative Field. 473 E.Path ofaParticle inMotion Subject toaCentral Force Whose Magnitude IsProportional toItsDistance from aFixed Point O.476‘ F.Path ofaParticle in Motion Subject toaCentral Force Whose Magnitude IsInversely Pro- portional totheSquare ofItsDistance from aFixed Point O.481 G.Planetary Motion. 491 H.Kepler's (1571-1630) Laws ofPlane- tary Motion. Proof ofNewton's Inverse Square Law. 492 Lesson 35. Special Types ofSecond Order Linear and Nonlinear Differential Equations Solvable byReduction toa System ofTwo First Order Equations. A.Solution ofaSecond Order Nonlinear Diﬁerential Equation in Which y’andtheIndependent Variable :cAreAbsent. 500 B.Solu- tionofaSecond Order Nonlinear Differential Equation inWhich the Dependent Variable yIsAbsent. 502 C.Solution ofaSecond Order Nonlinear Equation inWhich theIndependent Variable 2:IsAbsent. 508 Lesson 36. Problems Giving Rise toSpecial Types ofSecond Order Nonlinear Equations. A.The Suspension Cable. 506 B.ASpecial Central Force Prob- lem. 521 C.APursuit Problem Leading toaSecond Order Nonlinear Differential Equation. 528 D.Geometric Problems. 528 SERIES METHODS Lesson 37. Power Series Solutions ofLinear Differential Equations. A.Review ofTaylor Series andRelated Matters. 531 B.Solution ofLinear Diﬁerential Equations bySeries Methods. 537 Series Solution ofy’=f(x,y). Series Solution ofaNonlinear Differential Equation ofOrder Greater Than One and ofaSystem ofFirst Order Differential Equations. A.Series Solution ofaSystem ofFirst Order Differential Equations. 555 B.Series Solution ofaSystem ofLinear First Order Equations. 559 C.Series Solution ofaNonlinear Diﬁerential Equation ofOrder Greater Than One. 562459 500 506 531 531 548 555 Lesson 40. Ordinary Points and Singularities ofaCom-rm-s xi Linear Differential Equation. Method ofFrobenius. 570 A.Ordinary Points and Singularities ofaLinear Diﬁerential Equa- tion. 570 B.Solution ofaHomogeneous Linear Differential Equation About aRegular Singularity. Method ofFrobenius. 572 Lesson 4-1. The Legendre Differential Equation. Legendre Functions. Legendre Polynomials Pg(x). Properties ofLegendre Polynomials P;,(x) A.The Legendre Differential Equation. 591 B.Comments. 591 onthe Solution (41.18) oftheLegendre Equation (41.1). Legendre Functions. Legendre Polynomials P;,(:c). 593 C.Properties ofLegendre Poly- nomials P;,(:c). 598 Lesson 42. The Bessel Differential Equation. Bessel Function ofthe First Kind J;,(x). Differential Equations Leading toaBessel Equation. Properties of];,(x). 609 A.TheBessel Differential Equation. 609 B.Bessel Functions ofthe First Kind J;,(:c). 611 C.Differential Equations Which Lead toa Bessel Equation. 615 D.Properties ofBessel Functions oftheFirst Kind J;,(:c). 619 Lesson 43. The Laguerre Differential Equation. Laguerre Polynomials L;,(x). Properties ofL;,(x). 624- A.TheLaguerre Differential Equation andItsSolution. 624 B.The Laguerre Polynomial L;,(z). 625 C.Some Properties ofLaguerre Polynomials L;,(:c). 627 10.NUMERICAL METHODS Lesson 44-. Starting Method. Polygonal Approximation. Lesson 4-5. AnImprovement ofthePolygonal Starting Method. Lesson 46. Starting Method—Taylor Series.631 632 64-1 64-5 A.Numerical Solution ofy’=f(:c,y) byDirect Substitution inaTaylor Series. 646 B.Numerical Solution ofy’=f(a:,y) bythe“Creeping Up” Process. 646 Lesson 4-7. Starting Method—Runge-Kutta Formulas. Lesson 4-8. Finite Differences. Interpolation. A.Finite Differences. 659 B.Polynomial Interpolation. 661653 659 xii ll.CONTENTS Lesson 49. Newton’s Interpolation Formulas. A.Newton’s (Forward) Interpolation Formula. 663 B.Newton’s (Backward) Interpolation Formula. 668 C.The Error inPolyno- mial Interpolation. 670 Lesson 50. Approximation Formulas Including Simpson’s and Weddle’s Rule. Lesson 51. Milne’s Method ofFinding anApproximate Numerical Solution ofy'=_f(x,y). Lesson 52. General Comments. Selecting h. Reducing h. Summary and anExample. A.Comment onErrors. 690 B.Choosing theSizeofh.691 C.Re- ducing andIncreasing h.692 D.Summary andanIllustrative Exam- ple.694 Lesson 53. Numerical Methods Applied toaSystem ofTwo First Order Equations. Lesson 54. Numerical Solution ofaSecond Order Diﬁerential Equation. Lesson 55. Perturbation Method. First Order Equation. Lesson 56. Perturbation Method. Second Order Equation. EXISTENCE AND UNIQUENESS THEOREM FOR THE FIRST ORDER DIFFERENTIAL EQUATION y'=_f(x,y). PICARD’S METHOD. ENVELOPES. CLAIRAUT EQUATION. Lesson 57. Picard’s Method ofSuccessive Approximations. Lesson 58. AnExistence and Uniqueness Theorem fortheFirst Order Differential Equation y'=f(x,y) Satisfying y(¥o) =yo- A.Convergence andUniform Convergence ofaSequence ofFunctions. Deﬁnition ofaContinuous Function. 728 B.Lipschitz Condition. Theorems from Analysis. 731 C.Proof oftheExistence andUnique- nessTheorem fortheFirst Order Diﬁerential Equation y’=f(a:,y). 733 Lesson 59. The Ordinary and Singular Points ofaFirst Order Differential Equation y'==f(x,y). Com-mzrs Lesson 60. Envelopes. A.Envelopes ofaFamily ofCurves. 748 B.Envelopes ofa1-Param- eterFamily ofSolutions. 754 Lesson 61. The Clairaut Equation. I2.EXISTENCE AND UNIQUENESS THEOREMS FOR ASYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS AND FOR LINEAR AND NONLINEAR DIFFERENTIAL EQUATIONS OF ORDER GREATER THAN ONE.WRONSKIANS. Lesson 62. AnExistence and Uniqueness Theorem foraSystem ofnFirst Order Differential Equations and fora Nonlinear Differential Equation ofOrder Greater Than One. A.The Existence andUniqueness Theorem foraSystem ofnFirst Order Diﬁerential Equations. 763 B.Existence andUniqueness The- orem foraNonlinear Differential Equation ofOrder n.765 C.Exist- ence and Uniquen Theorem foraSystem ofnLinear First Order Equations. 768 Lesson 63. Determinants. Wronskians. A.ABrief Introduction totheTheory ofDeterminants. 770 B.Wronskians. 774 Lesson 64. Theorems About Wronskians and the Linear Independence ofaSetofSolutions ofa Homogeneous Linear Differential Equation. Lesson 65. Existence and Uniqueness Theorem forthe Linear Differential Equation ofOrder n. Bibliography Indexxiii 74-7 757 763 763 770 778 783 791 793 Preface fortheTeacher INwarrmo THIS BOOK, ithasbeen ouraimtomake itreadable forthe student, toinclude topics ofincreasing importance (such astransforms, numerical analysis, theperturbation concept) and toavoid theerrors traditionally transmitted inanelementary text. Inthislastconnection, wehave abandoned theuseoftheterminology “general solution” ofa differential equation unless thesolution isinfact general, i.e.,unless the solution actually contains every solution ofthedifferential equation. We have also avoided theterm “singular solution.” Wehave exercised great care indeﬁning function, differentials andsolutions; inparticular wehave tried tomake itclear that functions have domains. Ontheother hand, thisaccuracy hasbeen secondary toourmain pur- pose: toteach thestudent how tousedifferential equations. Wehope and believe that wehave notoverlooked any ofthemajor applications which canbemade comprehensible atthis elementary level. You will ﬁnd inthis text anextensive listofworked examples and homework problems with answers. Weacknowledge ourindebtedness tothepublishers fortheir coopera- tion and willingness toletususenew pedagogical devices and toProf. C.A.Hutchinson forhisthorough editing. M.T. H.P. Ithaca, New York West Lafayette, Indiana XV Preface fortheStudent Tms BOOK HASBEEN WRITTEN primarily foryou, thestudent. Wehave tried tomake iteasy toread andeasy tofollow. Wedonotwish toimply, however, that youwillbeable toread this text asifitwere anovel. Ifyouwish toderive anybeneﬁt from it,you must study each page slowly and carefully. You must have pencil and plenty ofpaper beside yousothat youyourself canreproduce each step andequation inanargument. When wesay“verify astatement, ”“make asubstitution, ”“add two equations,” “multiply twofactors,” etc., you yourself must actually perform these operations. Ifyoucarry outthe explicit and detailed instructions wehave given you, wecan almost guarantee that youwill, with relative ease, reach theconclusion. One ﬁnal suggestion—as you come across formulas, record them and their equation numbers onaseparate sheet ofpaper foreasy reference. You may also ﬁnd itadvantageous todothesame forDeﬁnitions and Theorems. M.T. H.P. Ithaca, New York West Lafayette, Indiana xvii Chapter 1 Basic Concepts LESSON 1.How Diﬁerential Equations Originate. Weliveinaworld ofinterrelated changing entities. The position of theearth changes with time, thevelocity ofafalling body changes with distance, thebending ofabeam changes with theweight oftheload placed onit,thearea ofacircle changes with thesizeoftheradius, the path ofaprojectile changes with thevelocity andangle atwhich itisﬁred. Inthelanguage ofmathematics, changing entities arecalled variables andtherate ofchange ofonevariable with respect toanother aderiva- tive. Equations which express arelationship among these variables and their derivatives arecalled differential equations. Inboth thenatural andsocial sciences many oftheproblems with which they areconcerned give risetosuch differential equations. But what weareinterested in knowing isnothow thevariables and their derivatives arerelated but only how thevariables themselves arerelated. Forexample, from certain facts about thevariable position ofaparticle anditsrate ofchange with respect totime, wewish todetermine how theposition oftheparticle is related tothetime sothat wecanknow where theparticle was, is,orwill beatanytime t.Differential equations thus originate whenever auni- versal lawisexpressed bymeans ofvariables and their derivatives. A course indifferential equations isthen concerned with theproblem of determining arelationship among thevariables from theinformation given tousabout themselves andtheir derivatives. Weshall useanactual historical event toillustrate how adifferential equation arose, how arelationship wasthen established between thetwo variables involved, andﬁnally hpw from therelationship, theanswer toa very interesting problem was determined. Intheyear 1940, agroup of boys was hiking inthevicinity ofatown inFrance named Lascaux. They suddenly became aware that their dog had disappeared. Inthe ensuing search hewasfound inadeep hole from which hewasunable to climb out. When oneoftheboys lowered himself into thehole tohelp extricate thedog, hemade astartling discovery. The hole was once a l 2BASIC CONCEPTS Chapter 1 part oftheroof ofanancient cave that hadbecome covered with brush. Onthewalls ofthecave there were marvellous paintings ofstags, wild horses, cattle, and ofaﬁerce-looking black beast which resembled our bull.* This accidental discovery, asyoumay guess, created asensation. Inaddition tothewall paintings and other articles ofarchaeological interest, there were alsofound thecharcoal remains ofaﬁre. Theproblem wewish tosolve isthefollowing: determine from thecharcoal remains how long agothecave dwellers lived. Itiswellknown thatcharcoal isburnt wood andthatwith time cer- tainchanges take place inalldead organic matter. Itisalsoknown that allliving organisms contain twoisotopes ofcarbon, namely C12andC“. The ﬁrst element isstable; thesecond isradioactive. Furthermore the ratio oftheamounts ofeach present inany macroscopic piece ofliving organism remains constant. However from themoment theorganism dies, theC“that islostbecause ofradiation, isnolonger replaced. Hence theamount oftheunstable C14present inadead organism, aswell asits ratio tothestable C12, changes with time. The changing entities inthis problem aretherefore theelement C“and time. Ifthelawwhich tells ushow oneofthese changing entities isrelated totheother cannot be expressed without involving their derivative, then adifferential equation willresult. Lettrepresent theelapsed time since thetreefrom which thecharcoal came, died, and letacrepresent theamount ofC“present inthedead treeatanytime t.Then theinstantaneous rateatwhich theelement C“ decomposes isexpressed inmathematical symbols as .1(1.1) 315- Wenow make theassumption that this rate ofdecomposition ofC14 varies astheﬁrst power of2:(remember asistheamount ofC“present atanytime t).Then theequation which expresses thisassumption is d(1.11) 3’;=—kz, where Ic>0isaproportionality constant, andthenegative sign isused toindicate that re,thequantity ofC“present, isdecreasing. Equation (1.11) isadifferential equation. Itstates that theinstantaneous rate of decomposition ofC14islctimes theamount ofC“present atamoment of time. Forexample, ifk=0.01 and tismeasured inyears, then when 2:=200units atamoment intime, (1.11) tells usthat therateofdecom- position ofC14atthat moment is1/100 of200orattherateof2units per ‘You canseesome ofthese pictures inPrimitive ArtbyErwin O.Christensen, Viking Press, 1955, andinThePicture History ofPainting byH.W.andD.J.Jansen, Harry N.Abrams, 1957. Lesson 1 How DIFFERENTIAL EQUATIONS ORIGINATE 3 year. If,atanother moment oftime, 2:=50units, then (1.11) tellsusthat therate ofdecomposition ofC14atthat moment is1/100 of50oratthe rate of1-unit peryear. Ournexttaskistotrytodetermine from (1.11) alawthatwillexpress therelationship between thevariable :1:(which, remember, istheamount ofC“present atanytime t)andthetime t.Todothis, wemultiply (1.11) bydt/:1: andobtain (1.12) ‘if=—kdt. Integration of(1.12) gives (1.13) logx=—kt +c, where cisanarbitrary constant. Bythedeﬁnition ofthelogarithm, we canwrite (1.13) as (1.14) :4:=e"°‘+° =e°e"°' =Ae'“, where wehave replaced theconstant e°byanew constant A. Although (1.14) isanequation which expresses therelationship between thevariable :2:andthevariable t,itwillnot.give ustheanswer weseek until weknow thevalues ofAandlc.Forthispurpose, wefallback on other available information which asyetwehave notused. Since time is being measured from themoment thetree died, i.e.,t=0atdeath, we learn from (1.14) bysubstituting t=0init,that :1:=A.Hence wenow know, since :1:istheamount ofC“present atanytime t,that Aunits of C“ were present when thetree, from which thecharcoal came, died. From thechemist welearn that approximately 99.876 percent* ofC“ present atdeath will remain indead wood after 10years and that the assumption made after (1.1) iscorrect. Mathematically thismeans that when t=10,:1:=0.99876A. Substituting these values of2:and tin (1.14), weobtain (1.15) 0.9987611 =Ae"°", 0.99876 =e-1°”. Wecannow ﬁnd thevalue ofItineither oftwoways. There aretables which tellusforwhat value of—l0k, e'1°" =0.99876. Division ofthis value by-10 willthen give usthevalue ofk.Orifwetake thenatural logarithm ofboth sides of(1.15) there results (1.2) log0.99876 =—-10k. ‘There issome diﬁerence among chemists inregard tothisﬁgure. The oneused above isbased onahalf-life ofC14of5600 years, i.e.,halfofC14present atdeath will decompose in5600 years. Itisanapproximate average of5100 years, thelowest half- lifeﬁgure, and6200 years, thelargest half-life ﬁgure. 4BASIC Concnrrs Chapter 1 From atable ofnatural logarithms, weﬁnd (1.21) —0.00124 =-1010, It=0.000124 approximately. Equation (1.14) now becomes xZ Ae—O-0001241, where Aistheamount ofC14present atthemoment thetreedied. Equation (1.22) expresses therelationship between thevariable quan- tity:0andthevariable time t.Wearetherefore atlastinaposition to answer theoriginal question: How long agodidthecave dwellers live? Byachemical analysis ofthecharcoal, thechemist wasable todetermine theratio oftheamounts ofC14toC12present atthetime ofthedis- covery ofthecave. Acomparison ofthis ratio with theﬁxed ratio of these twocarbons inliving trees disclosed that 85.5 percent oftheamount ofC“ present atldeath had decomposed. Hence 0.145A units ofC“ remained. Substituting thisvalue for:1:in(1.22), weobtain (1.23) 0.145.-1 =Ae-°~°°°‘“‘1 e-11.000124! log0.145=-0.000124: ~1.9310 =—0.0001241 1=15573. Hence thecave dwellers lived approximately 15,500 years ago. Comment 1.3. Differential equation (1.11) originated from theas- sumption that therate ofdecomposition ofC“varied astheﬁrst power oftheamount ofC“present atany time t.The resulting relationship between thevariables was then veriﬁed byindependent experiment. Assumptions ofthiskind arecontinually being made byscientists. From theassumption adifferential equation originates. From thedifferential equation arelationship between variables isdetermined, usually inthe form ofanequation. From theequation certain predictions canbemade. Experiments must then bedevised totestthese predictions. Ifthepre- dictions arevalidated, weaccept theequation asexpressing atrue law. Ithashappened inthehistory ofscience, because experiments performed were notsensitive enough, that laws which were considered asvalid for many years were found tobeinvalid when new andmore reﬁned experi- ments were devised. Aclassical example isthelaws ofNewton. These were accepted asvalid forafewhundred years. Aslong astheexperi- ments concerned bodies which were macroscopic andspeeds which were reasonable, thelaws were valid. Ifthebodies were ofthesizeofatoms or thespeeds near that oflight, then newassumptions hadtobemade, new Lesson 2A Tm: MEANING orTHETERM Set 5 equations born, new predictions foretold, andnew experiments devised to testthevalidity ofthese predictions. Comment 1.4. The method wehave described fordetermining the ageofanorganic archaeological remain isknown asthecarbon-14 test.* EXERCISE I l.Theradium inapiece ofleaddecomposes ataratewhich isproportional to theamount present. If10percent oftheradium decomposes in200years, what percent oftheoriginal amount ofradium willbepresent inapiece of leadafter 1000 years? 2.Assume thatthehalflifeoftheradium inapiece ofleadis1600 years. How much radium willbelostin100years? 3.The following item appeared inanewspaper. “The expedition used the carbon-14 testtomeasure theamount ofradioactivity stillpresent inthe organic material found intheruins, thereby determining thatatown existed there aslong agoas7000 B.c.” Using thehalf-life ﬁgure ofC14asgiven in thetext, determine theapproximate percentage ofC14stillpresent inthe organic material atthetime ofthediscovery. ANSWERS 1 1.59.05 percent. 2.4.2percent. 3.Between 32percent and33percent. LESSON 2.The Meaning oftheTerms Setand Function. Implicit Functions. Elementary Functions. Before wecanhope tosolve problems indifferential equations, wemust ﬁrst learn certain rules, methods and laws which must beobserved. In thelessons that follow, weshall therefore concentrate onexplaining the meaning ofcertain terms which weshall useandondevising methods by which certain types ofdifferential equations canbesolved. Weshall then apply these methods tosolving awide variety ofproblems ofwhich theoneinLesson 1wasanexample. Webegin ourstudy ofdifferential equations byclarifying foryoutwo ofthebasic notions underlying thecalculus andoneswhich weshall use repeatedly. These arethenotions ofsetandfunction. LESSON 2A. The Meaning oftheTerm Set. Each ofyouisfamiliar with theword collection. Some ofyouinfactmay have ormay have had collections—such ascollections ofstamps, ofseashells, ofcoins, ofbutter- ﬂies. Inmathematics wecallacollection ofobjects aset, andtheindi- ‘Dr. Willard F.Libby wasawarded the1960 Nobel Physics Prize fordeveloping this method ofascertaining theageofancient objects. HisC14half-life ﬁgure is5600 years, thesame astheoneweused. According toDr.Libby, themeasurable agespan by thistestisfrom 1000 to30,000 years. 6BASIC CONCEPTS Chapter 1 vidual members ofthesetelements. Asettherefore may bedescribed byspecifying what property anobject must have inorder tobelong toit orbygiving alistoftheelements oftheset. Examples ofSets. 1.The collection ofpositive integers lessthan 10 isaset. Itselements are1,2,3,4,5,6,7,8,9. 2.The collection ofindividuals whose surnames areSmith isaset. 3.The collection ofallnegative integers isaset. Itselements are ---,-4, -3, -2, ——1. Since toeach point onaline,there corresponds oneandonly onereal number, called thecoordinate ofthepoint, and toeach real number there corresponds oneandonly onepoint ontheline, wefrequently refer toapoint onalinebyitscorresponding number andvice versa. Deﬁnition 2.1. The setofallnumbers between anytwopoints ona lineiscalled aninterval andisusually denoted bytheletter I. Ifthetwopoints onalinearedesignated byaandb,then thenotation (2.11) I:a<:0<b willmean thesetofallrealnumbers :0(orofallrealvalues of1:)which lie between thepoints aandb,butnotincluding aandb.Forconvenience, weshall frequently omit theIandwrite only (2.111) a<x<b torepresent thissetofnumbers. Similarly, (2.12) I:—oo<2:<oowillmean thesetofallrealvalues of1:. I:a§:1:§bwillmean thesetofallreal values ofas between aandb,including thetwoendpoints. I:a§:0<bwill mean thesetofallreal values of:0 between aandb,including abutnotb. I:-1<1:<3,2:=10,willmean thesetofallnum- bersbetween -1and3plusthenumber 10. I:2:g0willmean thesetofallpositive realvalues of:0 plus zero. I::1:=awillmean thesetconsisting ofthesingle num- bera. LESSON 2B. The Meaning oftheTerm Function ofOne Inde- pendent Variable. Iftwovariables areconnected insome waysothat thevalue ofoneisuniquely determined when avalue isgiven totheother, wesaythatoneisafunction oftheother. (This concept willbegiven amore precise meaning inDeﬁnitions 2.3and2.31below.) Lesson 2B THE MEANING orTHETERM Function 7 Weshall show byexamples below thatthemanner inwhich therela- tionship between thevariables isexpressed isunimportant. Itmay be byanequation, ofthekind with which youarefamiliar, orbyother means. Itisonly important forthedeﬁnition ofafunction that there bethis unambiguous relationship between thevariables sothat, when avalue is given toone, acorresponding value totheother isthereby uniquely determined. Example 2.2. Letlbethelength oftheside ofasquare and Aits area. Itisthen customary tosaythattheareaAdepends onthelength l,sothat lisgiven anindependent status andAadependent one. How- ever, there isnovalid reason why lcould notbeconsidered asbeing dependent onA.The decision astowhich variable inaproblem istobe considered asdependent and which independent liesentirely within the discretion oftheindividual. The choice will usually bedetermined by convenience. Itiscustomary towrite, whenever itispossible todoso, ﬁrst thedependent variable, then anequals sign, then theindependent variable inamanner which expresses mathematically therelationship be- tween thetwovariables. Ifinthisexample, therefore, weexpress therelationship between our twovariables Aandlbywriting , (a) A=Z2; wethereby give toAadependent status and tolanindependent one. Equation (a)now deﬁnes Aasafunction oflsince foreach l,itdetermines Auniquely. Therelationship between thetwovariables, expressed mathe- matically byequation (a),is,however, notrigidly correct. Itsays that foreach value ofthelength l,A,thearea, isthesquare ofl.Butwhat if weletl=-3? The square of—3is9;yetnoarea exists ifthesideofa square haslength lessthan zero. Hence wemust place arestriction onl andsaythat (a)deﬁnes thearea Aasafunction ofthelength lonly for asetofpositive values oflandforl=0.Wemust therefore write (b) A=l2, lg0. Example 2.21. The relationship between twovariables xandyisthe following. If2:isbetween 0and 1,yistoequal 2.If:1:isbetween 2and 3,yistoequal Theequations which express therelationship between thetwovariables are,with theendpoints oftheinterval included, (9') y=2: y=\/5, 25:053. These twoequations now deﬁne yasafunction of2:.Foreach value ofx inthespeciﬁed intervals, avalue ofyisdetermined uniquely. The graph 8Bxsrc Concnrrs Chapter 1 ofthisfunction isshown inFig. 2.211. Note that these equations donot deﬁne yasafunction ofxforvalues ofxoutside thetwostated intervals. Y 2 1 i (0,0) 1 2 3 X Figure 2.211 The reason isobvious. Wehave notbeen told what thisrelationship is. Forvalues of:1:therefore, equal tosay3-or——1or4,etc., wesaythat y isundeﬁned orthat thefunction isundeﬁned. Example 2.22. InFig.2.221, wehave shown thetemperature Tofa body, recorded byanautomatic device, inaperiod of24consecutive hours. The horizontal axisrepresents thetime inhours; thevertical axis thetemperature Tatany time tg0.Even though wecannot express therelationship between thevariables land Tbyanequation, there can benodoubt that aprecise, unique relationship between thetwovariables T 10' 55 n v I | OYQ 3 6 9 12 15 18 21 24 I() _ , Figure 2.221 exists. Foreach value ofthetime t,thegraph willgive aunique value of thetemperature T.Hence thegraph inthiscasedeﬁnes Tasafunction oft. The main feature wewished toemphasize intheabove differing exam- pleswasthat forthedeﬁnition ofafunction itwasnotessential tobeable tosetuptherelationship between thetwovariables byasingle equation. (Most ofthefunctions that youhave encountered thus farwere ofthis type.) Aswementioned attheoutset, what isessential forthedeﬁnition Lesson 2B Tun MEANING orTHETERM Fumxtum 9 ofafunction isthat therelationship between thetwovariables bespeciﬁc and unambiguous sothat foreach value taken onbyanindependent variable onaspeciﬁed set,there should correspond oneandonlyonevalue ofadependent variable. Asyou canverify, alltheexamples wegave above hadthisonecommon important property. Weincorporate allthe essential features ofafunction inthefollowing deﬁnition. Deﬁnition 2.3. Iftoeach value ofanindependent variable :1:ona setE(thesetmust bespeciﬁed) there corresponds oneand only onereal value ofthedependent variable y,wesaythatthedependent variable y isafunction* oftheindependent variable 2:onthesetE. Itiscustomary tocallthespeciﬁed setEofvalues oftheindependent variable, thedomain ofdeﬁnition oftheindependent variable and to callthesetofresulting values ofthedependent variable, therange ofthe dependent variable ortherange ofthefunction. Using thisterminology, wemay deﬁne afunction alternately, asfollows. Deﬁnition 2.31. Afunction isacorrespondence between adomain setDandasetRthat assigns toeach element ofDaunique element ofR. Comment 2.32. Afunction isthus equivalent toarulewhich tells us howtodetermine theunique element yoftherange which istobeassigned toanelement a:ofthedomain. When wesaytherefore thattheformula (a) y=\/x—l, xgl, deﬁnes yasafunction ofx,wemean thattheformula hasgiven usarule bywhich, foreach value oftheindependent variable asinitsdomain D: atQ1,wecandetermine theunique assigned value ofyintherange R. (Here Rg0.)The ruleisasfollows: foreach :1:inD,subtract oneand take thepositive square root oftheresult. Forexample, totheelement 2:=5ofD,thefunction orruledeﬁned by(a)assigns theunique value 2ofR.Conversely, wecanﬁrst give therule, andthen express therule, ifpossible, byaformula. Comment 2.33. Wemay attimes forconvenience refer toanequa- tionorformula asifitwere afunction. Forexample, wemayfrequently refer totheequation (b) y=x2, -—oo<x<oo, asafunction. What weactually mean isthat theequation y=2:2deﬁnes afunction; that isy=:02gives usarulebywhich toeach :1:inDwecan ‘Inadvanced mathematics, thefunction iscalled arealfunction. Since weshall for themost partbeconsidering only realfunctions, weshall omit theword realwhenever arealfunction ismeant. 10Bxsrc Concarrs Chapter 1 assign ayinR.Here theruleis:foreach xinD,thenumber yinRassigned toitisobtained bysquaring x. Inview ofDeﬁnition 2.3,suchfrequently encountered formulas as y=w2. ll: V1—x2: _a:2+5a: y_ :c—3 aremeaningless because they donotspecify thedomain, i.e.,thesetof values ofx,forwhich theformulas apply. Inpractice, however, weinter- pretsuchformulas todeﬁne functions forallvalues of2:forwhich theymake sense. The ﬁrst equation, therefore, deﬁnes afunction forallvalues ofx, thesecond forvalues of:1:intheinterval -1§:0§1,andthelastfor allvalues ofztexcept :0=3. Because oftheabove comment, thefunction deﬁned by y=\/1—-x2 isnotthesame asthefunction deﬁned by y=\/1-1:’, 0§a:§1. (The ﬁrst isdeﬁned forallxin—1§x§1;thesecond only for:1:in 0§:1:§1.) Itiscustomary and convenient torecord thefact that thedependent variable yisafunction oftheindependent variable as(these aretheletters most commonly used forthedependent andindependent variable respec- tively), bymeans ofthesymbolic expression (2.34) y=f(:v). Itisread as“yequals fofx"or“yisafunction of2:.” ByDeﬁnition 2.3and(2.34), wecould therefore write forthetempera- ture Example 2.22, (2.35) T=fa). Wethen sayTisafunction oftandrefer tothegraph itself asthedeﬁni- tion orrulewhich tells uswhich Ttoassign toeach value oft. Comment 2.36. Weshall attimes write anexpression in1:,say (a) f(x)=:z:2+e", —oo <:v<oo andrefer tof(:c)asafunction. What wemean isthatf(:c), here (:02+e’), gives arulebywhich foreach x,wecanassign aunique value tof(x). Lesson 2C FUNCTION orTwo INDEPENDENT Vxrnxnnas ll Deﬁnition 2.4. Iff(x) isafunction ofxdeﬁned onasetE,then the symbol f(a), foranyainE,means theunique value assigned tof(:c) ob- tained bysubstituting aforre. Example 2.5. If (a) f(:t)=:z:2+2:c—|-1, O§:v§l, ﬁr1df(0).f(1).f(%),f(2),f(—1)- Solution. ByDeﬁnition 2.4wehave (b) f<0)=0”+2-0+1=1, f(1)=1’+2-1+1 =4, f(i)=(%)’+2'%+ 1=2%. f(2) isundeﬁned since 2isnotinoursetE:0§as§1, f(—1) isundeﬁned since —1isnotinoursetE. Ifseveral functions appear inasingle context sothat theuseofthesame letter foreach would beconfusing, itispermissible toreplace fbyother letters. Those most frequently used areg,h,G,H,F,etc. Similarly, we may useother letters inplace ofasandy.Those usually used aretheones attheendofthealphabet, namely u,v,w,z,s,t. LESSON 2C. Function ofTwo Independent Variables. InLesson 2B,wedeﬁned afunction ofoneindependent variable. Inananalogous manner, wedeﬁne afunction oftwo independent variables :1:and yas follows. Deﬁnition 2.6. Iftoeach element (:c,y) ofasetEintheplane (the setmust bespeciﬁed) there corresponds oneandonly onerealvalue ofz, then zissaid tobeafunction of:1:andyforthesetE.Inthisevent, :c,y arecalled independent variables and2adependent variable. Asinthecase ofoneindependent variable, thesetEissometimes called thedomain ofdeﬁnition ofthefunction and thesetofresulting values ofz,therange ofthefunction. i Inview ofDeﬁnition 2.6,aformula such as z=y\/1—-2:2 ismeaningless since itdoes notspecify thedomain ofdeﬁnition. Here again asinthecase ofonevariable, weinterpret such formulas todeﬁne functions forallvalues of2:and yforwhich they make sense. Inthis example therefore theelements ofthedomain Darethepoints (:z:,y) in theplane where —-1§:1:§1,——oo <y<oo.The domain, therefore, forwhich theformula deﬁnes zasafunction of2:andyconsists ofallpoints 12Bxsrc CONCEPTS Chapter 1 intheplane between andincluding thelines :1:=1and as=-1. Itis theshaded areainFig.2.61. z x=—1 /.0» . J’ x=1 x \ Figure 2.61 Example 2.62. Determine thedomain Dforwhich each ofthefol- lowing formulas deﬁne zasafunction ofxandy. 1_z=_._ﬂ. \/15 2.z=x+y. 3.z= . 4.z= . 5.z.= - Solutions. 1.The elements ofthedomain Dforwhich theformula deﬁnes 2asafunction ofatand y,consists ofthose points (:v,y) where —1<at<1,--1§y§1.Thedomain Distheshaded square shown Z Z x=—1 7 N _,/(cor /,~. y yﬂl x=1 x x ((1) (bl Figure 2.63 inFig.2.63(a). Itisbounded bythelines rt==l=1andy==i=l. It includes thelines y==l:1butnotthelines rt=:1:1. 2.The domain istheentire plane. Lesson 2C FUNCTION orTwo INDEPENDENT Vxmxntas 13 3.The domain consists ofthose points (:r,y) forwhich 1:2—|—y’525. Itistheshaded areainFig.2.63(b), i.e.,itisthearea outside thecircle x2+y2=25plus thepoints onitscircumference. 4.The domain consists ofthesingle point (0,0). 5.This formula does notdeﬁne 2asafunction ofasandy.There does notexist adomain which willdetermine avalue ofz. Itisevident from theabove examples, that atwo-dimensional domain may cover thewhole plane orpart oftheplane; itmay cover thewhole plane with theexception ofaholeinitsinterior; itsboundaries may be circular orstraight lines, oritmay consist ofonly aﬁnite number ofpoints. Inshort, incontrast toaone-dimensional domain, atwo-dimensional domain may assume agreat variety ofshapes andﬁgures. Itiscustomary andconvenient torecord thefact that 2isafunction ofxandybymeans ofthesymbolic expression (2-64) Z=f($.y)- Itisread as“zequals fofx,y” or“zisafunction of:c,y.” Deﬁnition 2.65. Iff(:c,y) isafunction oftwoindependent variables :e,y,deﬁned over adomain D,then thesymbol f(:t,a) forany element (x,a) inD,means thefunction ofa:obtained byreplacing ybya. Example 2.66. If (a)f(r.y) =w’+ry’+5y+3, —w<iv<w.—<><><y<w. ﬁndf(16.2).f(W1).f[$.9(1=)l- Solution. ByDeﬁnition 2.65 (b) f(:r,2)=:c2—|-4:c+10+3=:t2+4:c+13, -—oo<:z:<oo, f(:e,a)=:c2+a2:z:+5a+3, —-oo<:c<oo. And, ifg(:c) isdeﬁned forall1:, f[1..<J(Iv)] =I2+1[9(%)]2 +59(1) +3,—<><> <w<0°- Example 2.67. If (*1) f(w.y)=w+y. —1§x§1,0§y§2. ﬁndf(:c,1}), f(:z:,3). Solution. ByDeﬁnition 2.65 (b) f(x,§) =:1:+1},—l§xé1. Butf(:c,3) isundeﬁned since thedomain ofyistheinterval 0§y§2. Aspecial type ofset,called aregion, isdeﬁned asfollows. 14-BASIC CONCEPTS Chapter 1 Deﬁnition 2.68. Asetintheplane iscalled aregion ifitsatisﬁes thefollowing twoconditions: 1.Each point ofthesetisthecenter ofacircle whose entire interior con- sists ofpoints oftheset. 2.Every twopoints ofthesetcanbejoined byacurve which consists entirely ofpoints oftheset. InExample 2.62, thedomain deﬁned in2isaregion. Iftheboimdary points areexcluded from each setdeﬁned in1and3,then each resulting domain isalso aregion. Each point ofeach setsatisﬁes requirement 1, andevery twopoints ofeach setsatisﬁes requirement 2.Ontheother hand, thesetconsisting ofthepoints onalineisnotaregion. The set satisﬁes requirement 2butnot1.Alsothesetconsisting ofisolated points isnotaregion———the points inthesetdonotsatisfy either ofthetwore- quirements. Deﬁnition 2.69. Aregion issaidtobebounded ifthere isacircle which willenclose it. Comment 2.691. Inamanner analogous toDeﬁnition 2.6,wecan deﬁne afunction ofthree ormore independent variables. LESSON 2D. Implicit Function. Consider arelationship between twovariables a:,ygiven bytheformula (2.7) $2+y2-25=0. Does itdeﬁne afunction? If:0>5or:1:<——5, then theformula will notdetermine avalue ofy.Forexample, if:1:=7,there isnovalue ofy which willmake theleftsideof(2.7) equal tozero. (Why?) However, if 2:liesbetween -5and5inclusive, then there isavalue ofywhich will make theleftsideof(2.7) equal tozero. Toﬁndit,wesolve (2.7) fory andobtain (2.71) y=:l=\/25 —2:2, --5§:1:§5. When therelation between xandyiswritten inthisform, however, we seethattheformula does notdeﬁne yuniquely foravalue ofx.Hence, byourDeﬁnition 2.3,itdoes notdeﬁne afunction. Wecancorrect this defect byspecifying which value ofyistobechosen. Forexample, we canchoose anyoneofthefollowing three formulas todetermine y. (2.72) y=V25 —:02, —5§x§5. (2.73) y=-—\/25 -—.772, --5 §it§5. (2.74) y=\/25 —-x2, -5§:0§0; =——\/25-—:c2, 0<:c<5. Lesson 2D Imrmcrr FUNcrroN 15 ByDeﬁnition 2.3,each ofthese formulas nowdeﬁnes afunction. Itgives arulewhich assigns aunique ytoeach :0onthespeciﬁed interval. Now consider theformula (2.75) $2+y2+1=0, which also connects twovariables 2:andy,andaskofitthesame ques- tion. Does itdeﬁne afunction? Ifitdoes, then there must bevalues of2: forwhich itwilldetermine uniquely values ofy.Itshould beevident to youthat there arenovalues ofyforany2:.(Write theequation as 1:2+y2=-1.) Hence thisequation does notdeﬁne afunction byour Deﬁnition 2.3. Asaﬁnal example, weconsider theformula (2.76) 2:3+ya——3:cy=0, andagain askthequestion. Does itdeﬁne afunction? And ifitdoes, for what values of:1:willitdetermine uniquely avalue ofy?The answer to both questions, unlike theanswer totheprevious formula (2.7), isnot easy togive. Forunlike it,(2.76) cannot besolved easily foryinterms ofx.Hence wemust resort toother means. Thegraph ofequation (2.76) isshown inFig.2.77. Y (iii) O 2’/3 X Figure 2.77 From thegraph, weseethat forac§0andx>22/3, yisuniquely determined. Hence formula (2.76) does deﬁne yasafl1IlCl}I0l'l ofavin these twointervals, butnotintheinterval 0<:0§22/3. Itispossible, however, tomake formula (2.76) deﬁne yasafunction of:1:forallacifwe 16BASIC CONCEPTS Chapter 1 choose oneofthethree possible values ofyforeach :1:intheinterval 0<2:<22'3andchoose oneofthetwopossible values ofyfor:0=22/3. With these restrictions, wethen would beable toassert that (2.76) deﬁnes yasafunction ofasforall2:. Whenever arelationship which exists between twovariables asandyis expressed intheform (2.7) or(2.76), wewrite itsymbolically as (2-3) f(1.1/)=0- Itisreadas“fofx,yequals zero,”oras“afunction of:c,yequals zero.” Iftherelation which deﬁnes yasafunction of:cisexpressed inthe form f(a:,y) =0,itiscustomary tocallyanimplicit function ofx.When wesaytherefore that yisanimplicit function of1:,wemean, asthename suggests, thatthefunctional relationship between thetwovariables isnot explicitly visible aswhen wewrite y=f(a:), butthat itnevertheless im- plicitly exists. That is,there isafunction, letuscallitg(a:), which is implicitly deﬁned bytherelation f(am/) =0andwhich determines uniquely avalue ofyforeach 2:onasetE.Hence: Deﬁnition 2.31. The relation (232) f(fey)=0 deﬁnes yasanimplicit function of:1:onaninterval I:a<:1:<b,if there exists afunction g(x) deﬁned onIsuch that (2-33) f116.9(1)] =0 forevery atinI. Example 2.34. Show that (a) f(r.y) =rv’+y’—25=0 deﬁnes yasanimplicit function of2:ontheinterval I:—-5§at§5. Solution. Choose forg(x)anyoneofthefunctions (2.72), (2.73), or (2.74). If,forexample, wechoose (2.72), then g(x)=\/25 —x2.Itis deﬁned onI,andby(a)andDeﬁnition 2.65, (b) f[w.g(w)] =1*+l\/25—$21’—25=0. Hence Deﬁnition 2.81 issatisﬁed. Example 2.85. Show that (=1) f(r.v) =rs+ya—3x21=0 deﬁnes yasanimplicit function ofxforall:0. Lesson 2E THE ELEMENTARY FUNc'rroNs 17 Solution. Here, aspointed outearlier, itisnoteasytosolve foryin terms of2:,sothatitisnoteasytoﬁndtherequired function g(2:). How- ever, ifweselect from Fig.2.77anyoneofthegraphs shown inFig.2.86 Y Y Y X X _X 0 2’/3 0 2”" 0 2’/3 \ \Figure 2.86 torepresent thefunction g(x), then f[:t,g(:e)] =0forevery ac.Hence Deﬁnition 2.81 issatisﬁed. (Note that g(2:)_may beselected ininﬁnitely many more ways.) LESSON 2E. The Elementary Functions. Inaddition totheterms function andimplicit function, weshall refer attimes toaspecial class of functions called theelementary functions. These aretheconstants andthefollowing fimctions ofavariable 2:: 1.Powers of2::2:,2:2,2:3,etc. 2.Roots of2::\/20-,\'/at,etc. 3.Exponentials: e‘. 4.Logarithms: log2:. 5.Trigonometric functions: sin2:,cos2:,tan2:,etc. 6.Inverse trigonometric functions: Arccos2:,etc. 7.Allfunctions obtained byreplacing 2:anynumber oftimes byanyof the:other functions 1to6.Examples are:logsin2:,sin(sin2:),e'i"", e“,etc. 8.Allfunctions obtained byadding, subtracting, multiplying, anddivid- inganyoftheabove seven types aﬁnite number oftimes. Examples esin 2 e2:’+2z+1 are:22:—log2:+T» (Arc cos2:)2+7?;-;— —log(log42:). Inthecalculus course, youlearned how todifferentiate elementary functions andhowtointegrate theresulting derivatives. Ifyouhave for- gotten how, itwould beanexcellent ideaatthispoint toopen your cal- culus book andreview thismaterial. 18BASIC CoNcE1>'rs Chapter 1 1 2 3. 4. 5 6 7. 8EXERCISE 2 Describe, inwords, each ofthefollowing sets: (a)2:<0. (b)x§0. (c)a<2:§b. (d)—w<2:<5,2:=7. (e)—3<Z<—2,I>0. (f)\/§<1<1. (g)21r§22<31r. Deﬁne thearea Aofacircle asafunction ofitsradius r.Which isthede- pendent variable andwhich istheindependent variable? Draw arough graph which willshow howAdepends onr,when risgiven values between 0and5. Under certain circumstances thepressure pofagasanditsvolume Vare related bytheformula pV3/2 =l.Express each variable asafimction ofthe other. ‘ Explain thedifference between thefunction y=\/5, 2§2:§3, andthefunction y=\/E, 2:>0. =2 =7 =13Let F(2:) if2:<0, if0§2:§1, 22:if2:>4. Find(11)F(—1). (b)F(0). (<1)F(0-7). (d)F(4).(e)F(3).(f)F(2)- If 2 g(2:) = » 2:sé1, ﬁnd(a)0(2). (b)0(—5). -(0)0(1). (d)9(u). (6)9(¢2). (f)9(2=—1)- Why isthefunction deﬁned in6notthesame asthefimction deﬁned by g(:r) =2:—-1? Determine thedomain Dforwhich each ofthefollowing formulas deﬁnes zasafunction of2:andy. (a)z= - (b)z <c>z=~/F-m. <d>z= <e>.=./$-Ti». <r>:=~F-‘(Ir-1-..;2+s>. <.>.=1%j--—y-I-2. \/:t2+y2——9.=2: 9.Which ofthedomains inproblem 8areregions? 10.Iff(x,y) isafunction oftwoindependent variables 2:,ydeﬁned overadomain D,then thesymbol f(b,y) foranyelement (b,y) inD,means thefunction of yobtained byreplacing 2:byb;seeDeﬁnition 2.65. Let f(¢.;1/) =2’+21v+ 101;(Iv). we>0-Find: (3)f(-1.1).f($.b)»f(1.0).f(1»9>2)- (b)f(1.y),f(a.y),f(0.1/),f(12|ll)- (C) .f(arb)! .f(u:'))' Lesson 2—Exe1-cise 19 ll. 12 13 14 15. 16. 17 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.Draw thegraphs ofthree different functions deﬁned by 2:3+y3 —3:ty =0. SeeFig.2.77. Isthere onewhich iscontinuous forall2:? Thefollowing isastandard type ofexercise inthecalculus. lf13-1-y3—32:y=0,then 32:2+3y2%—323%—3y=0.Therefore d_1l_y 12 2;‘ d2:_y2-—2:’y I Explain bytheuseofFig.2.77what thismeans geometrically. Explain whytheprocedure followed inproblem 12,applied totherelation 2:2-1-y2+1=0andyielding theresult n__2d2: y ismeaningless. Find thefunction g(x)thatisimplicitly deﬁned bytherelation \/2:2 ~—y2+Arcc0s-Z =0,y960, Explain why.2:\/2:2 -—y2—1-Arcsing!-= 0 does notdeﬁne yasanimplicit function of2:1 Canyouapply themethod ofimplicit differentiation astaught inthecalculus tothefunction ofproblem 14,ofproblem 15? Deﬁne afunction ofthree independent variables 2:1,2:2,2:3;ofnindependent variables 2:1---,2:...Hint. SeeDeﬁnition 2.6. ANSWERS 2 (a)The setofallnegative values of2:. (b)The set(a)plus zero. (c)Thesetofvalues of2:between aandb,including bbutnota. (d)The setofallvalues of2:lessthan ﬁve,plusthenumber 7. (e)Thesetofall values of2:between -3and-2,plusallpositive values ofx. A=‘KT2, rg0. p=V"3/2, V>0;V=p'2'3, p>0. Theﬁrstfunction isdeﬁned only forallrcbetween andincluding twoand three; thesecond function isdeﬁned forall2:greater than zero. (a)2. (b)7. (c)7. (d)Undeﬁned. (e)496. (f)Undeﬁned. (a)1. (b)-6. (c)Meaningless. (d)(u2-2u+ 1)/(u —1). u941. (c)(14—222-1-1)/(2:2 —1),2:29*1. (f)(2:2—42:-1- 4)/ (2:-2),2:aé2. Function deﬁned in6ismeaningless when 2:=1. (a)—1§x§1,-1<y<1. (b)Entire plane. (c)Entire plane. (d)Area outside circle 2:2+yz=9pluspoints onthecircumference ofthe circle. (e)Line2:-1-y=0. (f)Nonexistent. (g)y#1. (b),(c),(g).Also (a)and(d)iftheir boundary points areexcluded. (a)2:2-1-22:-1-log2:;2:2-1-2b2:+log(bx); undeﬁned; 2:2-1-22:3-1-log2:3. (b)1-1-2y-1-logy;oz-1-2ay-1-logay;undeﬁned; 2:4+2x2y -1-log(ray). (c)3;undeﬁned; a2+2ab-1-log(ab); u2-1-2uv-1-log(uv). 20BASIC CoNc1~:P'rs Chapter l 13.12-I-y2+1=0does notdeﬁne afunction. 14.y=g(1) =1. 15.Theﬁrstterm requires that I2:I2IyI;thesecond term that IatI§IyI. 16.No,both examples. LESSON 3.The Differential Equation. LESSON 3A. Deﬁnition ofanOrdinary Differential Equation. Order ofaDifferential Equation. Inthecalculus, youstudied various methods bywhich youcould differentiate theelementary functions. For example, thesuccessive derivatives ofy=log1are (8.) y'='-is y”=-2%: 11'” =gs 0130. And if2=2:3—31y+2y2,itspartial derivatives with respect to2:and with respect toyarerespectively a 62 622 a’z(b)5-:=312—-3y, 5-1;=-32: +4y, §=61, 1?=4,etc. Equations such as(a)and(b)which involve variables andtheir deriva- tives arecalled differential equations. The ﬁrst involves only oneinde- pendent variable 1;thesecond twoindependent variables 2:andy.Equa- tions ofthetype (a)arecalled ordinary differential equations; ofthetype (b)partial differential equations. Hence, Deﬁnition 3.1. Letf(2:) deﬁne afunction of2:onaninterval I: a<1<b.Byanordinary differential equation wemean anequa- tioninvolving te,thefunction f(:c)andoneormore ofitsderivatives. Note. Itistheusual custom inwriting differential equations toreplace .d .f(2:)byy.Hence thedifferential equation -ggi -1-2:[f(:e)]2 =0ISusually d written asE:-1-xyz=0;thedifferential equation D,2[f(1)] +1D,,f(x) = e“asD,2y -1-:cD,y =e"orasy”+xy’=e”. Examples ofordinary differential equations are: dnu) ﬁ+y=Q (3.12) y’=e”. f 1 <3-13> i=11?" (3-14) f'(1)=f"($)- (3.15) xy’=2y. Lesson 3B EXPLICIT SOLUTION orADIFFERENTIAL EQUATION 21 (3.16) y"+(31/)3+2x=7. (8-17) (u'”)2+(u”)‘+1/’=w- (3.18> wy‘*’+2y"+(11/>”=x“- Note. Since only ordinary differential equations willbeconsidered in thistext, weshall hereafter omit theword ordinary. Deﬁnition 3.2. The order ofadifferential equation istheorder ofthehighest derivative involved intheequation. Forthedifferential equations listed above, verify that (3.11), (3.12), and(3.15) areoftheﬁrstorder; (3.13), (3.14), and(3.16) areofthesecond order; (3.17) isofthethird order; (3.18) isofthefourth order. Aworm orCAUTION. Youmight betempted toassert, ifyouwere not careful, thaty//_y//+y/__y=0 isasecond order differential equationbecause ofthepresence ofy".How- ever, y”isnotreally involved intheequation since itisremovable. Hence theequation isoforder 1. LESSON 3B. Solution ofaDiﬁerential Equation. Explicit Solu- tion. Consider thealgebraic equation (3.3) 2:2—2x-—3=0. When wesayas=3isasolution of(3.3), wemean that as=3satisﬁes it, i.e.,if:0isreplaced by3in(3.3), theequality willhold. Similarly, when wesaythefunction f(:z:) deﬁned by (3.31) y=f(x) =loga: —|—as,x>0, isasolution of (3.32) xzy” +2:cy' +y=log:1:—|—3:1:+1,2:>O, wemean that (3.31) satisﬁes (3.32), i.e.,ifin(3.32) wesubstitute the function f(:c) =log2:+:1:fory,and theﬁrst andsecond derivatives of thefunction fory’andy",respectively, theequality willhold. [Besure toverify theassertion that(3.31) doesinfactsatisfy (3.32).] Wewant youtonote twothings. First inaccordance with Deﬁnition 2.3,Wespeciﬁed in(3.31) thevalues ofxforwhich thefunction isdeﬁned. But even ifwehad not, theinterval :0>0would have been tacitly assumed since log:0isundeﬁned fora:§0.Second, wealsospeciﬁed in (3.32) theinterval forwhich thedifferential equation makes sense. Since ittoocontains theterm log2:,ittooismeaningless when :0§0. 22BASIC CONCEPTS Chapter 1 Deﬁnition 3.4. Lety=f(2:) deﬁne yasafunction of2:onaninterval I:a<2:<b.Wesaythat thefunction f(2:) isanexplicit solution or simply asolution ofanordinary differential equation involving 2:,f(2:), anditsderivatives, ifitsatisﬁes theequation forevery 2:inI,i.e.,ifwe replaee ybyf(r),y’byf'(w). 1/”byf”(r), -~'.y‘")byf(")($). thedifferen- tialequation reduces toanidentity in2:.Inmathematical symbols the deﬁnition says: thefunction f(2:) isasolution ofthedifferential equation (3-41) F(x:yry'r '''rf/(M) =01 if (3-42) F[x.f<w>.f'(w>. ---.1‘"’<x>1 =0 forevery 2:inI. Comment 3.43. Weshall frequently usetheexpression, “solve adif- ferential equation,” or“ﬁnd asolution ofadifferential equation.” Both aretobeinterpreted tomean, ﬁndafunction which isasolution ofthe differential equation inaccordance with Deﬁnition 3.4. Analogously when werefer toacertain equation asthesolution ofadifferential equation, we mean that thefunction deﬁned bytheequation isthesolution. Iftheequa- tiondoes notdeﬁne afunction, then itisnotasolution ofanydifferential equation, even though byfollowing aformal procedure, youcanshow that theequation satisﬁes thedifferential equation. Forexample, theequation y=\/—(1 —|—2:2)does notdeﬁne afunction. Tosay,therefore, that itis asolution ofthedifferential equation 2:+yy’=0ismeaningless even though theformal substitution initofy=\/——(1 —|—2:2)and y’= —:c/\/ -—(1 +2:2)yields anidentity. (Verify it.) Example 3.5. Verify that thefunction deﬁned by (a) y=2:2, —oo<2:<oo, isasolution ofthedifferential equation (b) (y”)3+(y’)”—1/—322—8==0- Solution. By(a),thefunction f(2)=22.Therefore f'(:v) =22:, f"(x) =2.Substituting these values in(b)fory,y’,y”,weobtain (c) 8+42:2—2:2——3x2—8=0. Since theleftsideof(c)iszero, (a),byDeﬁnition 3.4,isanexplicit solu- tionorsimply asolution of(b). Note that (b)isalsodeﬁned forall2:. Remark. Itistheusual practice, when testing whether thefunction deﬁned bytherelation y=_f(2:) onaninterval Iisasolution ofagiven differential equation, tosubstitute inthegiven equation thevalues ofy and itsderivatives. Intheprevious Example 3.5,therefore, ifwehad followed thispractice, wewould have substituted in(b):y=2:2,y’=2x, y"=2.Ifanidentity resulted, wewould then saythat (a)isasolution Lesson 3B Exrmcrr Sourrron 0FADIFFERENTIAL EQUATION 23 of(b). Weshall hereafter, forconvenience also follow thispractice, but youshould always remember thatitistheflmction _f(2:)anditsderivatives which must besubstituted inthegiven differential equation foryandits corresponding derivatives. And ify=f(2)does notdeﬁne afunction then yorf(2:) cannot bethesolution ofanydifferential equation. Example 3.51. Verify that thefunction deﬁned by (a) y=log:c+c, :c>0 isasolution of <1» 2'= Solution. Note ﬁrst that (b)isalso deﬁned forall2:>0.By(a), y’=1/2:. Substituting thisvalue ofy’in(b)gives anidentity. Hence (a)isasolution of(b)forallat>0. Example 3.52. Verify that thefunction deﬁned by (a) y=tanx—-2;, 2:¢(2n+1)g, n=0,=l=1,=!=2,---, isasolution of (b) 1/’=(rv+2/)’- Solution. Here y=tan2:—2:,y’=sec”2:—1=tan’ 2:.Substi- tution ofthese values in(b)foryandy’gives theidentity (c) tan22: =(2:—|—tan2:—2:)”=tan” 2:. Hence (a)isasolution of(b)ineach oftheintervals speciﬁed in(a). Comment 3.521. Note by(b)that thedifferential equation isdeﬁned forall2:.Itssolution, however, asgiven in(a),isnotdeﬁned forall2:. Hence, theinterval, forwhich thefunction deﬁned in(a)may beasolu- tion of(b),isthesmaller setofintervals given in(a). Comment 3.53. Itisalso pos- _ Y sible forafunction tobedeﬁned over aninterval andbethesolution ofadifferential equation inonly part ofthis interval. Forexample y= isdeﬁned forall2:.Itsgraph isshown inFig.3.54. Ithasnode- (0,0) X rivative when 2:=0.Itsatisﬁes thedifferential equation y’=1in Figure 3-54 theinterval 2:>0,and thediffer- ential equation y’=—-1intheinterval 2:<0.But itdoes notsatisfy any differential equation inaninterval which includes thepoint x=0. 24Bxsrc Concsrrs Chapter 1 LESSON 3C. Implicit Solution ofaDifferential Equation. T0 testwhether animplicit function deﬁned bytherelation f(2:,y) =0isa solution ofa.given differential equation, involves a.much more compli- cated procedure than thetesting ofoneexplicitly expressed byy=f(2:). The trouble arises because itisusually noteasy orpossible tosolve the equation f(2:,y) =Oforyinterms of2:inorder toobtain theneeded function g(a:)demanded byDeﬁnition 2.81. However, whenever itcanbe shown that animplicit function does satisfy agiven differential equation onaninterval I:a<2:<b,then therelation f(2:,y) =0iscalled (by anunfortunate usage)* animplicit solution ofthedifferential equation. Deﬁnition 3.6. Arelation f(2:,y) =0willbecalled animplicit solu- tion ofthedifferential equation F(xaf/If/Ii '''Yif/(M) =0 onaninterval I:a<2:<b,if 1.itdeﬁnes yasanimplicit function of2:onI,i.e.,ifthere exists afunc- tiong(2:)deﬁned onIsuch thatf[:e,g(x)] =Oforevery 2:inI,andif 2.g(:e)satisﬁes (3.61), i.e.,if (3-62) F[$.9(w).9’(¢). '''.g("’(w)l =0 forevery 2:inI. Example 3.63. Test whether (*1) f(w.y) #2’+v2—25=0 isanimplicit solution ofthedifferential equation (b) F(r,y.y’) =ye’+2=0 ontheinterval I:-5<2:<5. Solution. Wehave already shown that (a)deﬁnes yasanimplicit function of2:onI,ifwechoose forg(2:) anyoneofthefunctions (2.72), (2.73), (2.74). Ifwechoose (2.72), then (c) g(2:)=\/25—2:2, g'(2:)= -— ; -5<2: <5. Substituting in(b),g(:e)fory,g'(x) fory’,there results (<1) F[w.g(r). g'<x>1=is-w*(- +x=0. ‘Actually f(a:,y) =0isanequation andanequation isnever asolution ofadifferential equation. Only afurwtion canbeasolution. What wereally mean when wesayf(2:,y) =0 isasolution ofadifferential equation isthat thefunction g(2:)deﬁned bytherelation j'(2:,y) =0isthesolution. SeeDeﬁnition 3.6,alsoComment 3.43. Lesson 3C Im>I.IcI'r SoI.U'rIoN orADIFFERENTIAL EqUA'rIon 25 Since theleftside of(d)iszero, theequation isanidentity in2:.There- fore, both requirements ofDeﬁnition 3.6aresatisﬁed, and(a)istherefore animplicit solution of(b)onI. Example 3.64. Test whether (a) f(2:,y) =2:3+y3—-3xy=0,—-oo <:0<oo, isanimplicit solution of (b) F(x,y,y') =(y3—x)y’—y+2:2=0,-—oo <2:<oo. Solution. Unlike theprevious example, itisnoteasy tosolve fory toﬁndtherequired function g(2:). However, wehave already shown that (a)deﬁnes yasanimplicit function ofac,ifwechoose forg(2:) anyoneof thecurves inFig. 2.86. (Besure torefer tothegraphs shown inthis ﬁgure.) Ifwechoose theﬁrst, then g’(2:) does notexist when 2:=23/3. Hence, (a)cannot beasolution of(b)forall2:,butweshall show that (a)isasolution of(b)inanyinterval which excludes thispoint 2:=23/3 Since wedonothave anexplicit expression forg(2:), wecannot substitute in(b),g(2:) fory,g’(2:) fory’todetermine whether F[x,g(2c), g’(2:)] =0. What wedoistodifferentiate (a)implicitly toobtain (e)31:’+3v”;/’—312’—3y=0. (23—1):!’-v+2’=0- Since (c)now agrees with (b),weknow that theslope ofthefunction g(2:) implicitly deﬁned by(a)andexplicitly deﬁned bythegraph, satisﬁes (b)atevery point 2:inanyinterval excluding 2:=23/3.Hence both re- quirements ofDeﬁnition 3.6aresatisﬁed inanyinterval which does not include thepoint 2:=23/3.Therefore thefunction g(:e) deﬁned bythe ﬁrst graph inFig. 2.86 isanimplicit solution of(a)inanyinterval not containing thepoint 2:=23/3. Ifwechoose forg(2:) thesecond curve inFig. 2.86, then g'(:e) does not exist when 2:=0and:0=22/3.Forthisg(:e), (a)willbeasolution of (b)inanyinterval which excludes these twopoints. Ifwechoose forg(2:)thethird curve inFig.2.86, then g’(2:) does not exist when 2:=0.Forthisg(2:), (a)willbeasolution of(b)inanyinter- valwhich excludes thispoint. Comment 3.65. Theexample above demonstrates thepossibility of animplicit function being deﬁned over aninterval andbeing thesolution ofadifferential equation inonly part oftheinterval. Comment 3.651. Thestandard procedure incalculus texts toprove that (a)isasolution of(b)isthefollowing. Differentiate (a)implicitly. Ifityields (b),then (a)issaid tobeanimplicit solution of(b). Ifyou operate blindly inthismanner, then youarelikely toassert that 2:3+ 26BAsIc CONCEPTS Chapter 1 y3=0isanimplicit solution of2:+yy’=0,since differentiation ofthe ﬁrstgives thesecond. But203+y3=0doesnotdeﬁne yimplicitly asa function of2:onaninterval. Only thepoint (0,0) satisﬁes thisformula. Toassert, therefore, that22+y2=0isanimplicit solution of2:+yy’=0 because itsatisﬁes thedifferential equation ismeaningless. Example 3.66. Test whether (a) 2:y3—e_'/—1=0 isanimplicit solution ofthedifferential equation (b) (W2+22:y-1)y’+y’=0. Solution. Ifweworked blindly andused themethod ofimplicit dif- ferentiation astaught inthecalculus, then from (a),wewould obtain by differentiation (e) Zrvv’+I/3+F”!/’=0. (Zwy+e"")v' +1/’=0- Although (c)isnotidentical with (b),itcanbemade soifwereplace e_" inthesecond equation of(c)byitsvalue 2:y2-—1asdetermined from (a).Working blindly then, wewould assert that (a)isanimplicit solu- tionof(b).Butisit?Well, letussee.ByDeﬁnition 3.6,wemust ﬁrst show that (a)deﬁnes yasanimplicit function of2:onaninterval. Ifwe write (a)as (<1) y.=2 Y x=2.07 _ X y=-2.22 Figure 3.67 wesee, since e_”isalways positive, that yisdeﬁned only for2:>0. Hence theinterval forwhich (a)may beasolution of(b)must exclude Lesson 3—Exercise 27 values of2:§0.Here again asintheprevious example, wecannot easily solve foryexplicitly interms of2,sothatwemust resort toagraph to determine g(2:). Itisgiven inFig. 3.67. From thegraph, weseethat there arethree choices ofg(2:). Ifg(2:) istheupper branch, then (a)isan implicit solution of(b)forallan>0.Andifg(2:)iseither ofthetwolower branches, oneabove theliney=-2.22, theother below thisline,then (a)isanimplicit solution of(b)forI:2:>2.07 approximately. EXERCISE 3 . 1.Determine theorder ofthefollowing differential equations. (11)dy+(ry-—wew)<11=0- (b)11"+1:11”+21/(:1/)3 +11/=0-23 I” <oQ%)-4wr+w=c (ow+w"+y=o 2.Prove thatthefunctions intheright-hand column below aresolutions ofthe differential equations intheleft-hand columns. (Besuretostate thecommon interval forwhich solution anddifferential equation make sense.) (=1);/'+y=0 :I1=e""- (b)1/’=6‘ v=2’- d3 1 .(c)$= y=:cArcs1n:c+\/1-22. (<1)f'(¢) =f”(=v) 11=e‘+2- (e)my’=2y y=2:2. <cu+ﬂr=o u=VHw% (g)cos0%-2rsin0=0 r=asec20. (h)y"—y=0 y=ae‘+ be"‘. (i)f'($) =tf($) f(1) =26*”- (i)rv’+v=I/2 11%- (k)2:—|-yy'=0 y=\/16-22. 3.Show thatthedifferential equation J;-lrll/l+1—0 hasnosolutions. 4.Determine Wl16tl'l6l‘- theequations ontheright deﬁne implicit flmctions of2:. Forthose which do,determine whether they areimplicit solutions ofthe differential equations ontheleft. (of-1—m+ww=0 f—1=e+W- (b)e""—|—e"'3%=0 e2”-1-e2’=1. d<o%=—§ f+f+1=c 28BAsIc CoNcEP'rs Chapter 1 ANSWERS 3 1.(a)1. (b)2. (c)3. (d)3. 2.(a)—-e=><2:<ee. (b)—=e<a:<w. (e)——1<2:<1. (d)—ee<2<<=0. (e)2:¢0. (f)-<=e<x<ee. 3 (s)0¢=|=%,=|=§,=|=---. (h)—w<2:<ee. (i) ——°°<22<°°. (j)I?50,—2. (k)-—4<27<4. 4.(a)Yes,ifonemakes ysingle valued. Implicit solution. (b)Yes. Implicit solution, 2:aéO. (c)Function undeﬁned. LESSON 4.The General Solution ofaDifferential Equation. LESSON 4-A. Multiplicity ofSolutions ofaDifferential Equation. Weassume attheoutset that youhave understood clearly thematerial of theprevious lesson sothat when wesay“solve adifferential equation" or “ﬁnd asolution ofadifferential equation, "or“the solution ofadifferen- tialequation is,”youwillknow what ismeant (seeComment 3.43). Or ifweomit intervals forwhich afunction oradifferential equation is deﬁned, weexpect that youwillbeable toﬁllinthisomission yourself. When youstudied thetheory ofintegration inthecalculus, yousolved some simple differential equations oftheform y’=f(2:). Forexample, youlearned that, if (4-1) -y’=6’. then itssolution, obtained byasimple integration, is (4.11) y=e”—|—c, where ccantake onanynumerical value. And if (4.12) y"=e”, thenitssolution, obtained byintegrating (4.12) twice, is (4.13) y=e”+c12:—|—C2, where now clandC2cantake onarbitrary values. Finally, if (4.14) y”’=e‘, then itssolution, obtained byintegrating (4.14) three times, is (4.15) y=e’+c1223 +c2:e-1-c3, where cl,c2,c3cantake onanynumerical values. Lesson 4-A MULTIPLICITY oFSOLUTIONS oFADIFFERENTIAL EqUAT1oN 29 Two conclusions seem tostem from these examples. First, ifadifferen- tialequation hasasolution, ithasinﬁnitely many solutions (remember the c’scanhave inﬁnitely many values). Second, ifthedifferential equation isoftheﬁrstorder, itssolution contains onearbitrary constant; ifofthe second order, itssolution contains twoarbitrary constants; ifofthenth order, itssolution contains narbitrary constants. That both conjectures areinfactfalse canbeseenfrom thefollowing examples. Example 4.2. Theﬁrst order differential equation (a) (v’)"+2/’=0. alsothesecond order differential equation (b) (2/")3+23=0. each hasonly theonesolution y=0. Example 4.21. Theﬁrstorder differential equation (e) |y'|+1=0. alsothesecond order differential equation (b) ly"|+1=0, hasnosolution. Example 4.22. Theﬁrst order differential equation (a) 2:y’=1 hasnosolution iftheinterval Iis--1<rt<1.Formally onecansolve (a)toobtain (b) v=lesIel+e. butthisfunction isdiscontinuous at2:=0.ByDeﬁnition 3.4,asolution must satisfy thedifferential equation forevery 2:inI. Remark. If2:<0,then by(b)above (c) y=log(—x) +c1,2<0, isavalid solution of(a). And if2:>0,then by(b) (d) y=log2:+c2, 2:>0, isavalid solution of(a). Thelinex=0,therefore, divides theplane into two regions; inone (c)isvalid, intheother (d)isvalid. There isno solution, however, iftheregion includes theline2:=0. 30BAsIc CoNcE1>'rs Chapter 1 Example 4.23. Theﬁrst order differential equation (a) (v’—y)(y'—2y)=0 hasthesolution (b) (v-e1e”)(v —62¢“) =0. which hastwoarbitrary constants instead oftheusual one. These examples should warn younottojump immediately tothecon- clusion that every differential equation hasasolution, orifitdoes have a solution that thissolution willcontain arbitrary constants equal innum- bertotheorder ofthedifferential equation. Itshould comfort you to know, however, that there arelarge classes ofdifferential equations for which theabove conjectures aretrue, andthat these classes include most oftheequations which youarelikely toencounter. Forthese classes only, then, wecanassert: thesolution ofadiﬂerential equation oforder ncontains narbitrary constants cl,cg,---,c,,. Itiscustomary tocallasolution which contains nconstants c1,cg, ---,c,,ann-parameter family ofsolutions, andtorefer tothecon- stants cltoc,,asparameters. Inthisnewnotation, wewould say(4.11) isa1-parameter family ofsolutions of(4.1); (4.13) isa2-parameter family ofsolutions of(4.12), etc. Deﬁnition 4.3. Thefunctions deﬁned by y f(xIc1!c2! '''1ct!) ofthen—|—1variables, 2:,cl,cg,---,c,,willbecalled ann-parameter family ofsolutions ofthenthorder differential equation F(x;'J1l/,1 '''2I/(M) =01 ifforeach choice ofasetofvalues cl,cg,---,c,,,theresulting function f(:c) deﬁned by(4.31) (itwillnow deﬁne afunction of2:alone) satisﬁes (4.32), i.e.,if (4-33) F(w.f.f', ''-.f(")) =0- Fortheclasses ofdifferential equations weshall consider, wecannow assert: adiﬂerential equation ofthenthorder hasann-parameter family of solutions. Example 4.34. Show that thefunctions deﬁned by (fl) y=f($,¢1'.¢2) =2%+3-1-61¢‘+6263‘ ofthethree variables 2:,cl,cg,area2-parameter family ofsolutions ofthe Lesson 4B FINDING EQUATION FROM FAMILY 0FSoLIITIoNs 31 second order differential equation (b) F(w.v.y’,y") =v”-—32/’+221-4e=0- Solution. Leta,bbeanytwovalues ofC1,C2respectively. Then, by(a), (c) y=f(:e) =22:+3+aef+be”. [Note that (c)nowdeﬁnes afunction only of2:.]Theﬁrstandsecond derivatives of(c)are (d) y’=f’(2:) =2-1-ae’+2be2’, y”=f”(2:) =ae‘—|—4be3". Substituting in(b)thevalues off,f’andf",asfound in(c)and(d),for y,y’,y”,weobtain (e) F(2:,f,f’,f”) =ae”+4be3‘ —-6—-3ae" —-6be2” +42:+6+2ae‘+2be3‘—4:e=0. Youcanverify thattheleftsideof(e)reduces tozero. Hence byDeﬁni- tion 4.3,(a)isa2-parameter family ofsolutions of(b). LESSON 4B. Method ofFinding aDifferential Equation ifIts n-Parameter Family ofSolutions IsKnown. Weshall now show youhow toﬁnd thedifferential equation when itsn-parameter family of solutions isknown. You must bear inmind that although thefamily willcontain therequisite number ofnarbitrary constants, thenthorder differential equation whose solution itis,contains nosuch constants. In solving problems ofthis type, therefore, these constants must beelimi- nated. Unfortunately astandard method ofeliminating these constants isnotalways theeasiest touse. There arefrequently simpler methods which cannot bestandardized and which will depend onyour own in- genuity. Example 4.4. Find adifferential equation whose 1-parameter family ofsolutions is (a) y=ccos2:+:e. Solution. Inview ofwhat wehave already said, weassume that since (a)contains oneconstant, itisthesolution ofaﬁrstorder differential equation. Differentiating (a),weobtain (b) y’=—-csin2: +1. This differential equation cannot betheoneweseek since itcontains the parameter c.Toeliminate it,Wemultiply (a)bysin2:,(b)bycos2:and 32BAsIc CoNcE1>'rs Chapter 1 addtheequations. There results (c) ysin2:+y'cos:v=2:sin2:+cos:c, (y’—1)cosx-1" (1!-2:)sinz =—-0, 3y’=(x—y)tan:e+1,2:;-é:|=g,=l;%-,---I which istherequired differential equation. [We could also have solved (b)forcandsubstituted itsvalue in(a).] Note that theinterval for which (a)isasolution of(c)must exclude certain points even though the function (a)isdeﬁned forthese points. Example 4.5. Find adifferential equation whose 2-parameter family ofsolutions is (a) y=ole‘+cge"’. Solution. Since (a)contains two parameters, weassume itisthe solution ofasecond order differential equation. Wetherefore differentiate (a)twice, andobtain (b) y’=ole’ —cge"‘, (c) y”=cle’+cge"‘. Because ofthepresence oftheconstants clandcgin(c),itcannot bethe differential equation weseek. Anumber ofchoices areavailable forelimi- nating c1andcg.Wecould,- forexample, solve (a)and(b)simultaneously forclandcgandthen substitute these values in(c). This method isa standard onewhich isalways available toyou, provided youknow how tosolve thepair ofequations. Aneasier method istoobserve that the right sideof(c)isthesame astheright sideof(a). Hence, byequating their leftsides, wehave (<1) v"—v=0. which istherequired differential equation. Example 4.51. Find adifferential equation whose 2-parameter family ofsolutions is (a) y= c1sin:e+cgcos2:—l-x2. Solution. Since (a)contains twoconstants, weassume itisthesolu- tion ofasecond order differential equation. Hence wedifferentiate (a) twice, andobtain (b) y’=clcos2:—cgsin2:—|—22:, (c) y"=-01sin2:~—cgcosan+2 Lesson 4-C GENERAL ANDPARTIcULAn SoI.oTIoNs 33 Here again youcould usethestandard method ofﬁnding clandcgby solving (a)and (b)simultaneously, and then substituting these values in(c). [Oryoucould solve (b)and (c)simultaneously forclandcgand substitute these values in(a)]. Aneasier method istoobserve from (c)that (d) clsinx +cgcosx =2—-y” Substitution of(d)in(a)gives (e) y=2-y”—l-23 ory"=2:2—y+2, which istherequired differential equation. Example 4.52. Find adifferential equation whose 1-parameter family ofsolutions represents afamily ofcircles with centers attheorigin. Solution. Here thefamily ofsolutions isnotgiven tousintheform ofamathematical equation. However, thefamily ofcircles with center attheorigin is (a) 2:3—l—y3=r3, r>0. Since (a)hasonly 1-parameter r,weassume itisthesolution ofaﬁrst order differential equation. Hence wedifferentiate (a)once, andobtain (b) 2+yy’=0. which istherequired differential equation. Note that inthisexample the parameter rwaseliminated indifferentiating (a),andwewere thus able toobtain therequired differential equation immediately. LESSON 4-C. General Solution. Particular Solution. Initial Con- ditions. Ann-parameter family ofsolutions ofannthorder differential equation hasbeen called traditionally a“general” solution ofthedifferen- tialequation. And thefunction which results when wegive adeﬁnite set ofvalues totheconstants cl,cg,---,eninthefamily hasbeen called a “particular solution” ofthedifferential equation. Traditionally then, forexample, y=ce‘which isa1-parameter family ofsolutions ofy’--y=0,would becalled itsgeneral solution. And if weletc=-2, then y=-—2e’ would becalled aparticular solution of theequation. Itisevident that aninﬁnite number ofparticular solutions canbeobtained from ageneral solution: oneforeach value ofc. Ageneral solution, ifitistobeworthy ofitsname, should contain all solutions ofthedifferential equation, i.e.,itshould bepossible toobtain every particular solution bygiving proper values totheconstants cl,cg, ---,c,,.Unfortunately, there aredifferential equations which have solu- tions notobtainable from then-parameter family nomatter what values 34BAs1c CoNcE1>'rs Chapter 1 aregiven totheconstants. Forexample, theﬁrstorder differential equation (4-6) v=ev’+(1/)3 hasforasolution the1-parameter family (4.61) y=c2:+C2. Traditionally, thissolution, since itcontains therequired oneparameter, would becalled thegeneral solution of(4.6). However, itisnotthegen- eral solution inthereal meaning ofthis term since itdoes notinclude every particular solution. Thefunction 213 (4.62) y-——Z‘ isalsoasolution of(4.6). (Verify it.)And youcannot obtain thisfunc- tion from (4.61) nomatter what value youassign toc.[(4.61) isaﬁrst degree equation; (4.62) isasecond degree equation.] Unusual solutions ofthetype (4.62), i.e.,those which cannot beob- tained from ann-parameter family ortheso-called general solution, have traditionally been called “singular solutions.” Weshall show below by examples thattheuseofthese terms—general solution andsingular solu- tion—-in their traditional meanings isundesirable. Rather than being helpful inthestudy ofdifferential equations, their useleads only to confusion. Consider forexample the-ﬁrst order differential equation (4.63) y’=—2y3/3. Itssolution is (4.64) y= (Verify it.)But (4.63) hasanother solution (4.65) y=0, which cannot beobtained from (4.64) byassigning anyvalue toc.Bythe traditional deﬁnition, therefore, y=0would becalled asingular solution of(4.63). However, wecanalsowrite thesolution of(4.63) as _@’~ [Now verify that (4.651) isasolution of(4.63).] Inthisform, y=0is notasingular solution atall.Itcanbeobtained from (4.651) bysetting C=0.Hence useofthetraditional deﬁnitions forgeneral solution and Lesson 4C GENERAL ANDPARTIcULAR SoLoTIoNs 35 singular solution inthisexample leads ustotheuncomfortable contradic- tionthatasolution canbeboth singular andnonsingular, depending on thechoice ofrepresentation ofthe1-parameter family. Here isanother example. The ﬁrstorder differential equation (4-652) (y’—y)(y' —2y)=0 hasthefollowing twodistinct 1-parameter family ofsolutions (4.653) y=clef, (4.654) y=cge”. [Verify that each ofthese families satisﬁes (4.652).] Ifwecall(4.653) thegeneral solution of(4.652), asitshould becalled traditionally since itcontains therequired oneparameter, then theentire family offunctions (4.654) is,inthetraditional sense, singular solutions. They cannot beobtained from (4.653) bygiving anyvalues whatever to cl.Ifwecall(4.654) thegeneral solution of(4.652), aswell wemay in thetraditional sense, since ittoocontains therequisite oneparameter, then allthefunctions (4.653) aresingular solutions. Hence, useofthe traditional deﬁnitions forgeneral solution andsingular solution again leads us,inthisexample, totheuncomfortable contradiction that afamily of solutions canbeboth general andsingular. Inthistext, therefore, weshall notcallann-parameter family ofsolu- tions ageneral solution, unless wecanprove that itactually contains every particular solution without exception. Ifwecannot, weshall use theterm n-parameter family ofsolutions. Insuch cases, weshall make noattempt toassert that wehave obtained allpossible solutions, butshall claim only tohaving found ann-parameter family. Every solution ofthe given differential equation, inwhich noarbitrary constants arepresent, whether obtained from thefamily bygiving values tothearbitrary con- stants initorbyanyother means, willbecalled, inthistext, aparticular solution ofadifferential equation. Inourmeaning oftheterm, therefore, (4.62) isaparticular solution of(4.6), notasingular solution. Deﬁnition 4.66. Asolution ofadifferential equation willbecalled a particular solution ifitsatisﬁes theequation anddoes notcontain arbi- trary constants. Deﬁnition 4.7. Ann-parameter family ofsolutions ofadifferential equation willbecalled ageneral solution ifitcontains every particular solution oftheequation. Since there isaninﬁnite number ofways ofchoosing thenarbitrary constants cl,cg,---,c,,inann-parameter family, onemay well wonder how they aredetermined. What weusually want istheonesolution of theinﬁnitely many that willsatisfy certain conditions. Forinstance, we 36BAsIc CoNcE1>Ts Chapter 1 may observe inanexperiment, thatattime t=0(i.e., atthestart of anexperiment) a.body is10feetfrom anorigin andismoving with a velocity of20ft/sec. Theconstants then must besochosen thatwhen t=0,thesolution willgivethevalue 10feetforitsposition and20ft/sec foritsvelocity. Forexample, assume themotion ofthebody isgiven by the2-parameter family (a) 2:=16t2+c1t+ cg, where 2:isthedistance oftheparticle from anorigin attimet.Itsvelocity, obtained bydifferentiating (a),is 1)= +C1. Hence wemust choose theconstants clandcgsothatwhen t=0,2:=10, andv=20.Substituting these values oft,zt,andvin(a)and(b),weﬁnd cg=10,cl=20.Theparticular solution, therefore, which satisﬁes the given conditions ofthisproblem is (c) 2:=16t3—|—20t—|—10. Deﬁnition 4.71. The nconditions which enable ustodetermine the values ofthearbitrary constants cl,cg,---,c,,inann-parameter family, ifgiven interms ofonevalue oftheindependent variable, arecalled initial conditions. Intheexample above, thegiven conditions were initial ones. Both the value ofthefunction andofitsderivative were given interms oftheone value t=0. Comment 4.72. Normally thenumber ofinitial conditions must equal theorder ofthedifferential equation. There are, asusual, excep- tional cases where thisrequirement canbemodiﬁed. Forourclasses of differential equations, however, thisstatement willbeatrue one. Example 4.8. Find a1-parameter family ofsolutions ofthedifferen- tialequation (a) yy’=(21+1)’. andtheparticular solution forwhich y(2) =0.[This notation, y(2) =0, isashorthand way ofstating theinitial conditions. Here these are2:=2, y=0.Itmeans thatthepoint (2,0) must lieonorsatisfy theparticular solution.] Solution. Ify96—-1,wemay divide (a)by(y—|—1)”andobtain . /m.=/_. U <y+1>2” ‘3"”"’ 1 Lesson 4-—Exercise 37 Performing theindicated integrations gives (e) ,7_[—_-;+les|v+1I=e+¢, yr-6-1. which istherequired 1-parameter family. Toﬁndtheparticular solution forwhich 2=2,y=0,wesubstitute these values in(c)andobtain (d) 1=2+c orc=——1. Substituting (d)in(c),there results therequired particular solution, I(e) y—_,_—1+log|y+1|-2-1, y;-6-1. NoTE. Thefunction deﬁned byy=—1which wehadtodiscard to obtain (c)isalsoasolution of(a). (Verify it.)Hence, (f) 2/+1=0 isalsoaparticular solution of(a). Itisaparticular solution which cannot beobtained from thefamily (c)byassigning anyvalue totheconstant c. EXERCISE 4 Inproblems 1-3,show that each ofthefunctions ontheleftisa2-param- eterfamily ofsolutions ofthedifferential equation onitsright. 3 l.y=c1—|—cge""’+%, y”+y’—22—22=0. =c1e"2‘ +cge"‘ +2e’, —|—3y’+2y—12c" =0. =e12:—|—cg2'1—[- £2log2, 22y" —[—2y’—-y——2:=0. USN00Qﬁ§2 Inproblems 4and5,show that each ofthefunctions ontheleftis a3-parameter family ofsolutions ofthedifferential equation onitsright. 3 4.y=e”(c1—l-cg2:—|—c32:2+ 26-), y”'—- 3y"-l—3y' —-y-e‘=0. 1 9 2—-7'25_,=,,+,,,.+,3,_.+(E+ ,2.’ y"'—y’—e2’sinza: =0. Ineach ofproblems 6-17, ﬁndadifferential equation whose solution is thegiven n-parameter family. 6.y=02+ 03. 12.y=c1e‘1". 7.22-cy+c2=0. l3.y='23+£- 8.y=c1cos32+ cgsin32:. 14.y=c1e2’+ cge'2’. 9.r=Otan (0+ c). 15.(y-e)2=c2:. 10.y=c2:+ 302—4e. 16.r=a(1— cos9). ll.y=V0122 +C2. 17.logy=012:2'+02. 38BAs1c CoNcEr'rs Chapter I Find adifferential equation whose solution is 18.Afamily ofcircles ofﬁxed radii andcenters onthe2axis. 19.Afamily ofcircles ofvariable radii, centers onthe2axisandpassing through theorigin. 20.Afamily ofcircles with centers at(h,lc) andofﬁxed radius. 21.Afamily ofcircles with centers inthe2y-plane andofvariable Hint. Write theequation ofthefamily as2:2—|-y’—2c12 —2cgy+203=0. 22.Afamily ofparabolas with vertices attheorigin andfocionthe2axis. 23.Afamily ofparabolas with fociattheorigin andvertices onthe2axis. 24-.Afamily ofparabolas with fociandvertices onthe2axis. 25.Afamily ofparabolas with axesparallel tothe2axisandwith aﬁxed dis- tance a/2between thevertex andfocus ofeach parabola. 26.Afamily ofequilateral hyperbolas whose asymptotes arethecoordinate axes. 27.Afamily ofstraight lines whose yintercept isafunction ofitsslope. 28.Afamily ofstraight lines thataretangents totheparabola y'*’=22. 29.Afamily ofstraight lines that aretangents tothecircle 22+y“—03, where cisaconstant. 30.Find a1-parameter family ofsolutions ofthedifferential equation dy=yd2: andtheparticular solution forwhich y(3) =1. ANSWERS 4 6.y=2y’+(y')3. 19.22yy' +2:2-—ya=-0. 7-23(2)’ —2222'+42”=0- 20-[1+(1/>213 =4261")’- 8-2”+92=0- 21-y"'[1+ (1/)3]=3r'(11”)’-9.Hr’=02+ r2—|— r. 22.22y’ =y. 10-1/=(w—4)u'+3(z/’)’- 23-I'!(l/)2+212'—1/=0-11-Irv"+r(;1/)2—yr’=0- 24-yr”+(2/)2=0-12-yr”=(v’)3- - 25-ey"+ (1/)3=0-13.y'2:=42:3—-y. 26.2y’—[—y=0. 14.y”=4y. 27.y=2y'+j(y'). 15.42(y’)2 —|—22y’—y=0. 28.22(y')2 -2yy’—[-1=0. 16.(1—cosﬂ) 2%=rsin0. 29.y=2y’=1:c\/(y')2+ 1. 17.2yy” —yy’-—2(y')2 =0. 30.y=ce',y=e‘“3. 18-(I/2')’+2’=<1’- LESSON 5.Direction Field. LESSON 5A. Construction ofaDirection Field. The Isoclines of aDirection Field. Before beginning aformal presentation oftechniques which areavailable forsolving certain types ofdifferential equations, we wish toemphasize thegeometric signiﬁcance ofasolution ofaﬁrst order differential equation. Inmany practical problems, arough geometrical approximation toasolution, such asthose weshall describe below andin later lessons, may beallthat isneeded. Let (5-1) 1/=f(2)erf(r.y) =0 deﬁne afunction of2:,whose derivative y’exists onaninterval I.‘a<2<b. Lesson 5A CONSTRUCTION orADmncrron Fnann. Isocnnms 39 Then y’willgive theslope ofthegraph ofthisfunction ateach point whose atcoordinate isinI,i.e.,y’willgive thedirection ofthetangent tothe curve ateach ofthese points. When, therefore, weareasked toﬁnd a 1-parameter family ofsolutions of (5.11) y’=F(:z:,y), a<:1:<b, weareineffect being asked thefollowing. Find afamily ofcurves, every member ofwhich hasateach ofitspoints aslope given by(5.11). Deﬁnition 5.12. Ify=f(:::) orf(x,y) =0deﬁnes yasafunction of zcwhich satisﬁes (5.11) onaninterval I,then thegraph ofthisfunction iscalled anintegral curve, i.e.,itisthegraph ofafunction which isa solution of(5.11). Therefore even ifwecannot ﬁnd anelementary function which isa solution of(5.11), wecanby(5.11) draw asmall lineelement atanypoint (x,y), forwhich :1:isinI,torepresent theslope ofanintegral curve. And ifthislineisshort enough, thecurve itself over that length willresemble theline. Forexample, letusassume thaty’by(5.11) hasthevalue 2at thepoint (4,3). This means that at(4,3), theslope ofanintegral curve is 2.Hence wecandraw ashort lineatthispoint with slope 2.Inasimilar manner wecandraw, theoretically, such short lines overallthatpartof theplane forwhich (5.11) isvalid. These lines arecalled line elements orsometimes lineal ele- ments. The totality ofsuch lines hasbeen given various descriptive names. Weshall usetheterm direction ﬁeld.* Anycurve which hasat each ofitspoints oneofthese lineelements asatangent willsatisfy (5.11), andwilltherefore bethegraph ofaparticular solution. Example 5.2. Construct adirection ﬁeld forthedifferential equation (a) y’=w+y- Solution. Table 5.21 gives thevalues ofy’fortheinteger coordi- nates from ——5to5.InFig.5.22, wehave drawn thelineelements for these values ofy’andalso oneintegral curve. Itisthegraph ofthepar- ticular solution (b) y=e’——:c—-1 of(a) The construction ofline elements isunquestionably atedious job. Further, ifasuﬂicient number ofthem isnotconstructed inclose prox- imity, itmay bediﬂicult orimpossible tochoose thecorrect lineelement ‘Other names areslope ﬁeld, lineal element diagrarn. 4-0BASIC Concarrs Chapter 1 Table 5.21 N-5-4-3-2-1 0 1 2 3 4 5 U!_10 _<0 ® Q G‘: OI >5_._3 __2__ O Hi €D W Q G9 U! Hi CA7__2__O ;_4 C0 W Q Oi U! uh W Q O P-4 Q Q Q Oi OI >$~ CD Q I-l O p.-4 Q 00 ,_. O3 C7! ah DO Q ;_¢ O |_¢ Q O0 Hi O U! vi CA7 Q ,_¢ O ;-| Q C0 rh U! ;-1 ab 60 Q -1-1 Q D-l Q C0 rﬁ OI O5 Q O0 Q ;_¢ O ‘-1 Q O0 ilk U! Q Q 6&7 Q r—1 O u-1 Q 00 vb OI O5 Q W ii r—1 O o-A Q 09 ii OI Gt Q W QO U! 0 1 2 CO oh U! O3 Q ® <0 10 fortheparticular integral curve wewish toﬁnd. Ifsuch doubt exists ina certain neighborhood, itthen becomes necessary toconstruct additional lineelements inthisarea until thedoubt isresolved. Fortunately there \\‘i \2 1 X 1 2 3 4 5 l0»0) \ \ \ Figure 5.22 exist certain aids which canfacilitate theconstruction oflineelements. One ofthese istomake useoftheisoclines ofadirection ﬁeld. We shall explain itsmeaning below. Lesson 5B ORDINARY ANDSINGULAR Pomrs ory’=F(x,y) 41 In(a),lety’equal anyvalue, say3.Then (a)becomes M w+y=& which ineffect says that theslope y’hasthevalue 3ateach point where theintegral curve crosses thisline. [Look atthetable ofvalues given in 5.21. Atallpoints which satisfy (c)andaretherefore points onthisline, asforexample (5,—2), (0,3), (1,2), etc., y’=3.]Hence wecanquickly draw agreat many lineal elements ontheline(c). Allweneed doisto construct atanypoint onitalineelement with slope 3.This linethere- fore hasbeen called appropriately anisocline ofthedirection ﬁeld. For each different value ofy’,weobtain adifferent isocline. Allthestraight lines drawn inFig. 5.22 areisoclines. Ingeneral, therefore, if (5-23) 2/’=F(w=.y), then each curve forwhich (5.24) F(x,y) =k, where kisanynumber, willbeanisocline _ofthedirection ﬁeld deter- mined by(5.23). Every integral curve willcross theisocline withaslope Is. Remark. Forourillustration, wechose anF(x,y) which, when set equal tolc,could besolved explicitly fory.Wewere therefore able toﬁnd theisoclines ofthedirection ﬁeld without much trouble. Weshould warn you however, that inmany practical cases, (5.24) may bemore difﬁcult tosolve than thegiven differential equation itself. Insuch cases, wemust resort toother means toﬁndasolution. Anintegral curve which hasbeen drawn bymeans ofadirection ﬁeld may belooked upon asifitwere formed byaparticle moving insuch a way that itistangent toeach ofitslineelements. Therefore thepath of thisparticle (which remember isanintegral curve) issometimes referred toasastreamline oftheﬁeldmoving inthe_direction oftheﬁeld. Every student ofphysics haswitnessed theformation ofadirection ﬁeldwhen hehasgently tapped aglass, covered with iron ﬁlings, which had been placed over abarmagnet. Each iron ﬁling assumes thedirection ofa lineelement, andtheimaginary curve which hastheproper lineelements astangents isastreamline. LESSON 5B. The Ordinary and Singular Points ofthe First Order Equation (5.11). Intheexample oftheprevious lesson, each point (x,y) intheplane determined oneandonly onelineal element. Now con- sider thefollowing example. 4-2BASIC Concsrrs Chapter I Example 5.3. Construct adirection ﬁeldforthedifferential equation (a) y’=%!%D» :c¢0. Solution. (SeeFig.5.31.) Weobserve from (a)that wecancon- struct lineelements atevery point oftheplane excepting atthose points whose atcoordinate iszero. Iftherefore wewere attempting toﬁnda particular integral curve of(a)bymeans ofadirection ﬁeldconstruction, wecould dosoaslongaswedidnotcross the2:axis. Forexample, ifwe »B Y A A "-Q (0,1) /*’ (0,0) ‘\ X B Figure 5.31 began atapoint inthesecond quadrant oftheplane andfollowed a streamline, wewould bestopped atthepoint (0,1) since by(a),y’is meaningless there. Even ifwehadconcluded thatanarbitrary assign- ment ofthevalue zerotoy’atthispoint would seem reasonable andgive continuity totheintegral curve made bythestreamline, sothat(a)would now read (b) g/ x¢0 =0» i=0’!/=1» wewould beatalosstoknow which streamline tofollow after crossing (0,1). Ifyouwilldraw suﬁicient lineal elements intheneighborhood of Lesson 5B ORDINARY ANDSINGULAR Pomrs ory’=F(a:,y) 4-3 (0,1), itwillsoon become evident toyou, that with thenew deﬁnition of y’asgiven in(b),aninﬁnite number ofintegral curves liesonthepoint (0,1) with slope zero. Hence after crossing thispoint, onecould follow any ﬁrst, third, orfourth quadrant streamline, oreven another second quadrant streamline. Toovercome thisdiﬂiculty, wecould specify twosetsofinitial condi- tions inplace oftheusual one. Forexample, wecould require oursolu- tion tolieonthepoint (-—3,2), andafter crossing (0,1) togothrough the point (2,—1). These twoinitial conditions would then ﬁxaparticular integral curve. This annoying diﬁiculty arises because ofthenecessity of excluding :1:=0from theinterval ofdeﬁnition. Actually, there aretwo distinct solutions of(a),namely (c) y=c,:c2+l, x<0, y=c2x2—|-1,:c>0. [Besure toverify that each function deﬁned in(c)isasolution of(a).] Since thepoint (0,1) satisﬁes both equations in(c)and since wehave agreed todeﬁne theslope y’asequal tozero atthispoint, wecanwrite (c)as ~ (d) y=c,x2+1, as§0, y==c2x2+l, :z:;0. Inthis form thesolutions (d)include every particular solution ofthe given differential equation with theagreement that y’equals zerowhen x=0,y=1. Intheform ofsolution (d),wecannow make thefurther observation that with theexception of(0,1), nointegral curve liesonanypoint inthe plane whose 2:coordinate iszero. Forexample thepoint (0,3) does not satisfy either equation in(d)nomatter what values you assign tocl andc2. Wecharacterize thedifference between points like(0,1), (0,3), and (2,3) inthefollowing deﬁnitions. Deﬁnition 5.4. Anordinary point oftheﬁrstorder differential equa- tion (5.l1) isapoint intheplane which liesononeandonly oneofits integral curves. Deﬁnition 5.41. Asingular point oftheﬁrstorder differential equa- tion (5.11) isapoint intheplane which meets thefollowing tworequire- ments: 1.Itisnotanordinary point, i.e.,itdoes notlieonanyintegral curve or itliesonmore than oneintegral curve of(5.11). 44BASIC Concsrrs Chapter 1 2.Ifacircle ofarbitrarily small radius isdrawn about thepoint (i.e., the radius may beassmall asonewishes), there isatleast oneordinary point initsinterior. (We describe thiscondition bysaying thesingular point isalimit ofordinary points.) Intheabove Example 5.3,every circle, nomatter how small, drawn about anypoint ontheyaxis, sayabout thepoint (0,3), contains not only oneordinary point butalsoinﬁnitely many such points. Remark. Requirement 2isneeded toexclude extraneous points. For example, ify’=\/1—$2,then only points whose :0coordinates lie between —1and1need beconsidered. If,therefore, wedeﬁned asingular point byrequirement 1alone, then apoint like (3,7) would besingular. This point, however, isextraneous totheproblem. This example hasserved three purposes: 1.Ithasshown youwhat anordinary point andasingular point are. 2.Ithasshown youtheneed forspecifying intervals forwhich adiffer- ential equation anditssolution have meaning. You cannot work auto- matically andblindly andwrite asasolution of(a) (e) y=cx2+1, —-oo<:v<oo. Ifyoudidthat, andeven ifyoudeﬁned y’=0at(0,1), youcould get from (e)only theparabolic curves assolutions. Oneofthese isshown in Fig.5.31. Itismarked A,andwasobtained from (e)bysetting c=1 [equivalent tosetting c1=1andC2=1in(d)]. Ifin(d),wesetcl=2 andC2=-2,wegettheintegral curve marked Binthegraph. And if in(d)wesetcl=1andC2=-2,wegetthecurve which ismarked A inthesecond quadrant andBinthefourth quadrant. 3.Itshows once more that notevery ﬁrst order differential equation has a1-parameter family ofsolutions foritsgeneral solution. Thedifferen- tialequation inthis example requires two 1-parameter families to include allpossible solutions. EXERCISE 5 1.Construct adirection ﬁeldforthedifferential equation y’=21¢. Draw anintegral curve. 2.Construct adirection ﬁeldforthedifferential equation 2:1 y’_$ Where areitssingular points? How many parameters arerequired toinclude allpossible solutions? Draw theintegral curve thatgoesthrough thepoints (—1,2) and(2,—1). Lesson 5—Exercise 45 3.Construct adirection ﬁeldforthedifferential equation 1+y Draw anintegral curve thatgoesthrough thepoint (1,1). 4.Find theisoclines ofthedirection ﬁeld andindicate theslope ateach point where anintegral curve crosses anisocline, (a)forproblem 2,(b)forproblem 3. 5.Describe theisoclines ofthedirection ﬁeld, ify’=2:2+2y”. ANSWERS 5 2.Online2:=0;twoparameters. 4.(a)Isoclines arethefamily ofstraight lines through theorigin: 2y=ex. Slope ofanintegral curve ateachpoint where itcrosses anisocline isequal toc. (b)Isoclines: a:(1-c)=y(1+c),slope c. 5.Isoclines arethefamily ofellipses 2:2+2y2=c.Ateach point where an integral curve crosses oneofthese ellipses, theslope oftheintegral curve isc. Chapter 2 Special Types ofDifferential Equations oftheFirst Order Introductory Remarks. Inthischapter webegin thestudy offormal methods ofsolving special types ofﬁrst order differential equations. A fewpreliminary observations however should beinstructive andhelpful. 1.Itisunfortunately true that only very special types ofﬁrst order differential equations possess solutions (remember asolution isafunction) which canbeexpressed interms oftheelementary functions mentioned inLesson 2E. Most ﬁrstorder differential equations, infact, onecould say almost all,cannot bethus expressed. 2.There isnoconnection between theappearance ofadifferential equation andtheease ordifficulty ofﬁnding itssolution interms ofele- mentary functions. The differential equation d3%=$2+y does notlook lesscomplicated than gig_2 2 d_l/_:2dz-:1: +y ordz-e . Yet theﬁrst hasanelementary function foritssolution; theother two donot. 3.Ifthesolution youhave found canbeexpressed onlyintheimplicit form f($.11) =0,itwillusually beoflittle practical value. Animplicit solu- tion isfrequently such acomplicated expression that itisalmost impos- sible toﬁnd theneeded function g(x) which itimplicitly deﬁnes, (see Deﬁnition 2.81). And without aknowledge ofthefunction g(a:) orat least aknowledge ofwhat arough graph ofg(:c) looks like, thesolution willnotbeofmuch usetoyou. While weshall show you, therefore, in thelessons which follow, formal techniques forﬁnding solutions ofaﬁrst order differential equation, keep inmind, ifthesolution isanimplicit one, 4-6 Lesson 6A DIFFERENTIAL orAFUNCTION 47 that other forms ofsolutions weshall describe later, such asgeometric solutions, series solutions, andnumerical solutions, willbeoffargreater practical importance toyou. 4.Ifyou start with analgebraic equation and follow acertain pro- cedure toﬁndasolution foravariable 1:,itispossible that thevalue thus obtained isextraneous. Forexample, theusual procedure followed to solve theequation \/x2 +4:1:—3=1—2:2;istosquare both sides and then factor theresulting equation. Ifyoudothisyouwillobtain thesolu- tions :0=2anda:=§-.However, both values areextraneous. Neither solution satisﬁes thegiven equation. (Verify it.) Similarly, inshowing youaprocedure that willlead youtoasolution ofadifferential equation, itispossible that thefunction thus obtained willbeextraneous. Hence, tobecertain afunction isasolution ofagiven differential equation, you should always verify that itdoes infactsatisfy thegiven equation. 5.Finally, andwecannot emphasize thispoint toostrongly, examples generally found intextbooks are“textbook” examples. They areinserted asillustrations inorder toclarify thesubject matter under discussion. Hence they arecarefully selected toyield “nice, relatively easy”solutions. Actual practical problems areavoided since they may require, forin- stance, thedetermination oftheimaginary roots ofafourth degree equa- tion,orthesolving ofasystem offourormore equations inacorresponding number ofunknowns—burdensome andtime-consuming problems tosaythe least. LESSON 6.Meaning oftheDifferential ofaFunction. Separable Differential Equations. LESSON 6A. Differential ofaFunction ofOne Independent Variable. Weassume inthis Lesson 6Athat allfunctions arediffer- entiable onaninterval. Lety=f(a:) deﬁne yasafunction ofx.Then itsderivative f’(a:) willgive theslope ofthecurve atany point P(x,y) onit,i.e.,itistheslope ofthetangent linedrawn tothecurve atP. Itisevident from Fig.6.12, that (6.1) f’(x) =tana= Hence, (6.11) dy=f’(x) Ax. Wecalldythedifferential ofy,i.e.,itisthedifferential ofthefunction deﬁned byy=f(a:). From (6.11) wenote that thedifferential ofy, namely dy,isdependent ontheabscissa at(remember asthepoint P changes, f’(a:) changes), and onthesizeofAx. Wesee,therefore, that whereas y=f(x) deﬁnes yasafunction ofoneindependent variable 2:, 48Srscuu. Tyres orFmsr Orumn Eqoarrons Chapter 2 Y y-rc) y-re) P(==.y) Ay L (0,0) x z+Ax X Figure 6.12 thedifferential dyisafunction oftwoindependent variables :0andAx. Weindicate thisdependence ofdyon2:andAxbywriting itas (dy)(rm)-Hence, Deﬁnition 6.13. Lety=f(x)deﬁne yasa.function of2:onaninter- valI.Thedifferential ofy,written asdy(ordf)isdeﬁned by (6-14) (dy)(1.41!) =f'(=v)Ar- Note. Weshall want toapply Deﬁnition 6.13tothefunction deﬁned byy=x.Therefore, inorder todistinguish between thefunction deﬁned byy=2:andthevariable az,weplace thesymbol Aoverthea:sothat (6.15) y=It willdeﬁne thefunction that assigns toeach value oftheindependent variable :1:thesame unique value tothedependent variable y. Theorem 6.2. If (6.21) y=2, then (6.22) (dy)(:c,A1:) E(d:i:)(x,Ax) =Ax. Proof. Since (a) y=i deﬁnes thefunction thatassigns toeachvalue oftheindependent variable 2:thesame unique value tothedependent variable y,itsgraph isthe straight linewhose slope isgiven by (b) y’Ef'(¢) =1- Substituting f’(:c) =1in(6.14), weobtain (6.22). Lesson 6A DIFFERENTIAL orAFrmcrron 49 Comment 6.23. Ifin(6.14) wereplace Axbyitsvalue asgiven in (6.22), itbecomes (6-24) (dy)(#=.A¢) =f’(w)(d~i)($.A1=)- Inwords (6.24) saysthatify=f(a:)deﬁnes yasafunction ofx,thenthe differential ofyistheproduct ofthederivative ofthefunction fandthe differential ofthefunction deﬁned byy=:8.Therelation (6.24) isthe correct one, butintliecourse oftime, itbecame customary towrite (6.24) inthemore familiar form <6-25> o=re)ax.-jg=re). Example 6.26. If (1%) y=1’. deﬁnes yasafunction of1:,ﬁnddy. Solution. Here f(x)=2:2.Therefore f'(x) =2x.Hence by(6.24) (b) (dy)(I.A1=) ==2-'vd¢(1=.-'31). which iscustomarily written as (c) dy=2xdx. Theimportance ofthedeﬁnition ofthedifferential asgiven in6.13lies inthefollowing theorem. Theorem 6.3. Ify=f(a;)deﬁnes yasafunction of:1:and:1:=g(t), y=f[g(t)] =F(t), deﬁne atandyasfunctions oft,then (6-31) (dy)($.41) =f'(w)(d1)(¢.Ai)- Proof. Since 2:=g(t)deﬁnes xasafunction oft,wehave by(6.24) (a) (div)(1.4!) =9'0)<if(i./ii). where ac=fdeﬁnes thefunction which assigns toeach value oftheinde- pendent variable t,thesame value tothedependent variable x.By hypothesis y=f[g(t)] deﬁnes yasa.function oft.Therefore by(6.24) andthechain ruleofdifferentiation, (b) (<11/)(i,/ll) =f’l9(l)][a'(i) df(L401- In(b)replace thelastexpression inbrackets ontheright, byitsequal leftsideof(a)andreplace g(t)byitsequal :0.Theresult is(6.31). 50SPECIAL Twas orFmsr ORDER EQUATIONS Chapter 2 Tosummarize: 1.Ify=f(a:), then (dy)(:z:,Ax) =f’(:c) d£(:z:,Aa:). 2.Ify=_f(a:) and :1:=g(t) sothat y=f[g(t)], then (dy)(t,At) = f'(:z:)(d:c)(t,At), where dyanddzaredifferentials ofyandacrespectively. The ﬁrst isthedifferential off[g(t)]; thesecond isthedifferential ofg(t). Inboth cases 1and2,weshall follow theusual custom andwrite (6.32) dy=f’(x) dxor%=f'(:c). Comment 6.33. Ify=f(:::)andax=g(t),then y=f[g(t)] deﬁnes y asafunction oft.Theindependent variable istherefore t;thedependent variables a:andy.Ingeneral ifyisadependent variable, theincrement Ay;-6dy,seeFig.6.12. Itfollows therefore thatAx95da:since herea:is also adependent variable. Thus there isnojustiﬁcation inreplacing an increment A2:byda:in“dy=]"(:c) Ax.” However, ifboth dyandda:are differentials asdeﬁned in6.13, then asweproved inTheorem 6.3,“dy= f'(x) dz”even when :1:isitself dependent onathird variable t. LESSON 6B. Diﬁerential ofaFunction ofTwo Independent Variables. Letz=f(:v,y) deﬁne zasafunction ofthetwoindependent variables a:andy.Then, following theanalogy oftheoneindependent variable treatment, wedeﬁne thedifferential ofzasfollows. Deﬁnition 6.4. Letz=f(:c,y) deﬁne 2asafunction ofanandy. The differential ofz,written asdzordf,isdeﬁned by (6.41) <dz><w.y.Aw.Ay> = Ax+ Ay-3/ Note that whereas 2isafunction oftwo independent variables, the differential of2isafunction offour independent variables. Theorem 6.42. If (6-43) Z=f(1,y) =1, where ihastheusual meaning, then (6.44) (dz)(x,y,A:::,Ay) Ed:E(x,y,Ax,Ay) =Ax. Proof. Here z=f(x,y) =5:.Hence ‘ﬂail/l =g-1 ‘E__0 Ox _Bx_’ 6y_' Substituting these values in(6.41), weobtain (6.44). Similarly, itcanbeshown, ifz=f(:e,y) =ﬁ,that (6-45) (d12)(w,y,Arv,Ay) =A11, Lesson 6C DIFFERENTIAL EQUATIONS wrrn SEPARABLE VARIABLES 51 where 3]hastheusual meaning. Substituting (6.44) and (6.45) in(6.41), weobtain (6.46) <e><»,y,Aw,Ay> = <de><z,y.Ax.Ay> + (dz2)(r,y,Aw,Ay)- This relation (6.46) isthecorrect one. However, inthecourse oftime, itbecame thecustom towrite (6.46) as (6.47) dz= dx+ dy. Keep inmind thatda:anddymean di:anddg]andaretherefore differentials, notincrements. With thisunderstanding ofthemeaning ofdz:anddywe shall now state, butnotprove, animportant theorem analogous toTheo- rem6.3inthecase ofoneindependent variable. Thetheorem asserts that (6.47) isvalid even when reandyareboth dependent onother variables. Theorem 6.5. Ifz=f(x,y) deﬁnes zasafunction ofxand y,and 1:=whys: ''')ay=I/(T231 ''')1Z =fl-x(Tss: ''')sl/(7:81 '' =F(rvsv '' deﬁne 2:,y,and2asfunctions ofr,s,andaﬁnite number ofother variables (indicated bythedotsafter s),then <6-51>(dam.---.Aw,--->= (dam,---,Ar.As---> + <dy><r,s,---,AT,/18,--->- Here alsoweshall follow theusual custom andwrite (6.51) as a, a,(6.52) dz=% dz+lgjﬂ dy. Example _6.53. Find dzif (a) Z=f(w,y) =$3+31211+ya+5- Solution. Here a, a,(b) -% =3x2+cw,% =3:112+31/2. Hence by(6.52) (c) dz=(3:e2 +6a:y) dx+(3x2 +3y2) dy. LESSON 6C. Differential Equations with Separable Variables. The ﬁrst order differential equations weshall study inthischapter willbe 52SPECIAL Trras orFIRST Onm-:3 EQUATIONS Chapter 2 those which canbewritten intheform <6-6) onegg+Pay)=0- Written inthisform itisassumed thatyisthedependent variable and2: istheindependent variable. Ifwemultiply (6.6) bydz,itbecomes (6-61) P(w.y) dw+Q(r,y) dz)=0- Written inthisform, either atorymay beconsidered asbeing thede- pendent variable. Inboth cases, however, dyanddzaredifferentials and notincrements. Although (6.6) and(6.61) arenotthemost general equations ofthe ﬁrstorder, theyaresuﬂiciently inclusive tocover most oftheapplications which youwillmeet. Examples ofsuch equations are (a) %=2w+at (b) y’==logw+y. (6) (56-21/)d:v+($+2‘!l+1)d!/=0» (d) e’cosydx+:csinydy =0. Ifitispossible torewrite (6.6) or(6.61) intheform (6-62) f(w)d1= +0(1))dz)=6. sothatthecoeﬁicient ofdz:isafunction of:1:alone andthecoeﬁcient of dyisafunction ofyalone, then thevariables arecalled separable. And after theyhave been putintheform (6.62), theyaresaidtobeseparated. A1-parameter family ofsolutions of(6.62) isthen (6-63) /f(w) dw+few) dy=6‘. where Cisanarbitrary constant. Example 6.64. Find a1-parameter family ofsolutions of (a) 2:1:dx—91/2dy=0. Solution. Acomparison of(a)with (6.62) shows that thevariables areseparated. Hence, by(6.63), itssolution is (b) 2:2—3;/3=C. Example 6.65. Find a1-parameter family ofsolutions of (a) \/1-—x2dIv+V5+ydy=0, —-1§x§1,y>—'5. Lesson 6C DIFFERENTIAL EQUATIONS wrrn SEPARABLE VARIABLES 53 Solution. Acomparison of(a)with (6.62) shows that thevariables areseparated. Hence itssolution by(6.63) is (b)§z\/1-''@=+§AmsiM+§(5+y)“”=c, -1;1.s1,y>-5. Comment 6.651. Because ofthepresence oftheinverse sine, (b)im- plicitly/defmes amultiple-valued function. Byourdeﬁnition ofafunc- tion itmust besingle-valued, i.e.,each value ofzshould determine one andonly onevalue ofy.Forthisreason wehave written theinverse sine withacapital Atoindicate thatwemean onlyitsprincipal values, namely those values which liebetween —1r/2 and1r/2. Example 6.66. Find a1-parameter family ofsolutions of (a) zx/1——ydz—\/1-—z2dy=0; alsoaparticular solution notobtainable from thefamily. Solution. We note ﬁrst that (a)makes sense only ify§1and —-1§z§1.Further ify;£1,z;-6=!;1, wecan divide (a)by \/1—— y\/1— 1:2andobtain dz dy(b) L--——=0,-1<x<1,y<1.V1——z” V1—y '2hisequation isnow oftheform (6.62). A1-parameter family ofsolu- tions by(6.63) is (c) \/1—z2—2\/l—y=C, —1<z<1,y<1. Thefunction y=1,which wehadtoexclude toobtain (c)alsosatisﬁes (a)forvalues ofzbetween —1and1.(Besure toverify it.)Itisapar- ticular solution of(a)that cannot beobtained from thefamily (c). Remark. InFig.6.67wehave y indicated thesetintheplane for y=1 11'h(11It‘(c)' l'd.It""“/"’//I/ V//’(L1) Z; 1,and z=--1. Byy(c), when 4’/”"'V"/é X z==|=1andy<1,aunique value % ' oftheconstant Candtherefore a %/ unique particular solution of(a)is determined. However, ify<1, Figufg 6.67 dy/dz —>=|=oo asz —>=|=1. Hence ify<1,thelines z==!=1aretangents tothefamily ofintegral curves. Thecorner points (1,1) and(—1,1) canbemade part oftheset. Forthese twopoints, weobtain from (c)thesolution (<1) x/1-z2—2\/1—y=0. 54-Srncnu. Twas orFmsr Onnnn EQUATIONS Chapter 2 Squaring (d)andtaking thederivative oftheresulting function, weobtain <e> y'=;- By(a),y’ismeaningless atthetwocorner points (1,1) and(—-1,1). How- ever, because of(e)weareledtodeﬁne theslope oftheintegral curve at these twocorner points as=}and ——§. Weseenow that every particular solution of(a)obtained from (c)liesintheregion below theliney=1. Theliney=1however, is,asweobserved previously, alsoaparticular integral curve of(a)andispart ofthesetforwhich solutions of(a)are valid. InFig.6.67 wehave drawn theintegral curve (d)andanintegral curve of(c)through (0,0). Example 6.68. Find a1-parameter family ofsolutions of (a) zcosydz+\/rﬁsinydy-=0; alsoaparticular solution notobtainable from thefamily. Solution. Wenote ﬁrstthat (a)makes sense only ifz>—1. Fur- ther,ifz sf--l,y ¢:1:-g» =|=gig»---,wecandivide(a)by\/zﬁcosy andobtain . ' \3(b)%:+Idz+%dy=0, :c>—-1, y¢¢§.\ :1:-7—;=|=,... Theequation isnow oftheform (6.62). A1-parameter family, by(6?63\), is (c) 2(%m \/at+1—logIcosyl =C, z>—1,y¢=l=g» ¢%=|=,--- Thefunctions 1r 31rl/"55?’ :|:_§_’ =|=:"'i which wehadtoexclude toobtain (c),alsosatisfy (a).They areparticular solutions of(a)which cannot beobtained from thefamily (c). Remark. InFig. 6.69 wehave indicated thesetintheplane for which thesolutions (c)arevalid. Itisbounded ontheleftbytheline z==-1, and excludes thelines y==1;5-These lasttwo lines, how- ever, alsoareparticular solutions of(a)notobtainable from (c),andare therefore also part ofthesetforwhich solutions of(a)arevalid. Each liney=31r/2, —31r/2, etc., isalsoasolution of(a). Lesson 6—Exercise 55 y=%I-1 I (-1,; %L6\ /. .--3," Figure 6.69 Example 6.7. Find aparticular solution of (a) wy”dz +(1—w)dz)=0, forwhich y(2) =1. Solution. Ifysf0andz;-51,wecanobtain from (a) (b) %dz+y_2dy=0, “£1, y#O, which wecanwrite as 1 _(0) (1—_-;—1)dz+y2dy=0, z¢1,y¢0. By(6.63), afamily ofsolutions of(a)is (d) log|1—z|+z+%=C, z¢1,y;-$0. Toﬁndtheparticular solution forwhich z=2,y=1,wesubstitute these values in(d)and obtain 0+2+1=C’,orC=3.Hence (d) becomes (e) log|l-—z|+z-I-5-=3, z¢1,y#0. EXERCISE 6 Find a1-parameter family ofsolutions ofeach ofthedifferential equa- tions 1-16listed below. Becareful tojustify allsteps used inobtaining a solution andtoindicate intervals forwhich thedifferential equation and thesolution arevalid. Also trytodiscover particular solutions which are notmembers ofthefamily ofsolutions. l.y'=y. 2.zdy—-ydz=0. 3.%%=—-sin0. 4.(y2+1)dz—(:c2+1)dy =0. s.%e<>to-1 =2. 56Sracmn Trrns orFinsr ORDER Equxrrons Chapter 2 6.yz2dy—yadz=21:2dy. 7.(ya—1)dz—(2y+zy) dy=0. 8.zlogzdy—|—\/1—|—y2dz=0. 9.e'+1tanydz+c0sydy=0. 10.zcosydz+ z2sinydy =a2sinydy. ll.%=rtan0. 12.(z—1)cosydy=2zsiny dz. 13.y'=ylogycotz. 14.zdy+(1—|— yz)Arctaiiy dz=0. 15.dy+1(y+1)dz =0. 16.e"’(z2 +2::+1)dz-1-(zy—|—y)dy=0. Find aparticular solution satisfying theinitial condition, ofeach ofthe following differential equations 17-21. The initial condition isindicated alongside each equation. dy _ _17.E+y -0,y(1) -1. 18.sinzcos2ydz+coszsin2ydy=0,y(0) =1r/2. 19.(1—-z)dy=z(y+1)dz, y(0) =0. 20.ydy+ zdz =3zy2 dz, y(2) =1. 21.dy=e‘+"dz, y(0) =0. 22.Deﬁne thedifferential ofafunction ofthree independent variables; ofn independent variables. ANSWERS 6 l.y=ce’. 2.y=cz. 3.r=cos0+c. 4.Arctanz =Arctany—|- corz—y=C’(1—|—zy), when-eC’ =tanc. 5.r=csec0-2, r¢——2, o¢§+n1;r=-2. 6.(cz+1)y2%(y—1)z, z#0, 1/#0; y=0. 7.z—|-2= y2—1,z#-2,;/#:|;1;y=:|:l. 8.log|z|(y+\/y2—|-1)=c, za-$0, zaél. 9.e‘+‘—|—log(cscy —coty) +cosy =c,y95mr; y=mr. 10.a2-—z2=ccos2y, zzaéaz, y94%+n1r; y=%+n-Jr. ll.rcos0=c, r#0, 0;¢%+n1r;r=0. 12.siny =(z-1)2e2""‘, z#1,y94n1r; y=n1r. 13.y=e“““, z94n1r, yaé0. 14.y=tan(c/z), z#0. 15.y=ce"'/2 -1, y#—1; y=-1. 16.z2+ 2z=e-"’+ c,z94——1. 17.y=cl". 18.coszzcos 2y=—1. 19.(y+1)(1 -—z)=e"‘. 20.3112=1+2e3""l2. 21.e‘+ e""=2. Lesson 7A Dnrmmon orAHouoom-woos FUNCTION 57 LESSON 7.First Order Differential Equation with Homogeneous Coeﬁicients. LESSON 7A. Deﬁnition ofaHomogeneous Function. Deﬁnition 7.1. Letz=f(z,y) deﬁne zasafunction ofzandyina region R.The function f(z,y) issaid tobehomogeneous oforder nif itcanbewritten as (7-11) f(w.y) =r"y(u). where u=y/zand g(u) isafunction ofu;oralternately ifitcanbe written as (7-12) f(1.y) =1/"h(u). where u=z/yandh(u) isafunction ofu. Example 7.13. Determine whether thefunction (a) f(z,y)=$2+y2log%> R:z>0,y>0, ishomogeneous. Ifitis,give itsorder. Solution. Wecanwrite theright sideof(a)as s/2 2/(b) z2(1+FlogQ)- Ifwenow letu=y/z, itbecomes (c) z2(1 +u2logu)=z2g(u). Hence, byDeﬁnition 7.1, (a)isahomogeneous function. Comparing (c)with (7.11), weseethat itisoforder 2. Orifwewished, wecould have written (a)as (d) 112+ 10.5)=fa’—1<>gu> =yzhc->. where u=z/y. Hence byDeﬁnition 7.1,(a)ishomogeneous oforder 2. Example 7.14. Determine whether thefunction (a) ray)=\/ism(j) ishomogeneous. Ifitis,give itsorder. Solution. With u=z/y, wecanwrite theright sideof(a)as (b) yl/2sinu=y”2h(u). Hence, byDeﬁnition 7.1,(a)ishomogeneous oforder %. 58Sracmr. Tress orFmsr Oannn EQUATIONS Chapter 2 Follow theprocedure used inExamples 7.13 to7.14 tocheck theaccu- racy oftheanswers given intheexamples below. Answers 9""°°!°!"1-:WeeI5"/”+tan(y/z). Homogeneous oforder zero. 2—|—sinzcosy. Nonhomogeneous. Homogeneous oforder .z—|—y. .\/1:2 +3zy+2y’. Homogeneous oforder 1. —3z3y +5y2z2 —2y‘. Homogeneous oforder 4. Comment 7.15. Analternate deﬁnition ofahomogeneous function is thefollowing. Afunction f(z,y) issaidtobehomogeneous oforder nif (7-16) f(tw.i1/) =t"f(w.y). where l>0andnisaconstant. Byusing thisdeﬁnition, (a)ofExample 7.13 becomes t(a) f(tz,ty) =t2z2-1-t2y2log% =t2(zz+1/2log =t2f(x1l/)- Hence thegiven function :02+yzlog(y/z) ishomogeneous oforder two. Byusing thisdeﬁnition, (a)ofExample 7.14 becomes (b) f(tz,ty) =(ty)1/zsin =)1”(ymsin =t”’f(w. y)- Hence thegiven function isoforder 1}.Asanexercise, use(7.16) totest theaﬁuracy oftheanswers given forthefunctions 1to5after Example 7.14. LESSON 7B. Solution ofaDifferential Equation inWhich the Coefficients ofdxand dyAre Each Homogeneous Functions of theSame Order. Deﬁnition 7.2. The differential equation (7-3) P(r.y) dz+Q(w.y) dz)=0. where P(z,y) andQ(z,y) areeach homogeneous functions oforder nis called aﬁrst order differential equation with homogeneous coeffi- cients. Lesson 7B EQUATIONS wrrn Honooamzous COEFFICIENTS 59 Weshall nowprove thatthesubstitution in(7.3) of (7.31) y=uz, dy=udz-1-zdu willalways lead toadifferential equation inzanduinwhich thevariables areseparable andhence solvable forubyLesson 6.The solution ywill then beobtainable by(7.31). The proof isincorporated inthefollowing theorem. Theorem 7.32. Ifthecoeﬁicients in(7.3) areeachhomogeneous func- tions oforder n,then thesubstitution initof(7.31) willleadtoanequation inwhich thevariables areseparable. Proof. Byhypothesis P(z,y) andQ(z,y) areeach homogeneous func- tions oforder n.Hence byDeﬁnition 7.1with u=y/z, each canbe written as (11) P(z,y) =1v"91(u). Q(rv.1/) ==v".<J2(u)- Substituting in(7.3) thevalue ofdyasgiven in(7.31) andthevalues of P(z,y), Q(z,y) asgiven in(a),weobtain (b) z”g1(u) dz+z"g2(u) (udz+zdu)=0, which simpliﬁes to (6) l(11(u) +uyz(u)l dz+$9264) du=0. dz g(u) _?+ du— 0.1'F"0.91(u)+WJ2(") #50» anequation inwhich thevariables zanduhave been separated. Prove asanexercise that thesubstitution in(7.3) of (7.33) z=uy, dz=udy—|—ydu willalsoleadtoaseparable equation inuandy. Remark. Ifthedifferential equation (7.3) iswritten intheform Q1_P(I.y) _<1-4) d,-QM,-P(z,y). then thestatement that P(z,y) andQ(z,y) areeach homogeneous oforder nisequivalent tosaying F(z,y) ishomogeneous oforder 0.Forby(7.4) andDeﬁnition 7.1 (7.41) F(z,y)=%'£ ="5-,‘%;)E%; =z°G(u). 60Srncnu. Tress orFmsr ORDER EQUATIONS Chapter 2 Example 7.5. Find a1-parameter family ofsolutions of (a) (\/w2—2/2+1/)dz—rdy=0; alsoanyparticular solution notobtainable from thefamily. Solution. Weobserve ﬁrst that (a)makes sense only if|y|§|z|,or |y/z| §1,z960.Second wenote byDeﬁnition 7.1that (a)isadifferen- tialequation with homogeneous coefficients oforder one. Wehave a choice therefore ofeither ofthesubstitutions (7.31) or(7.33). Byexperi- menting with both, youquickly willdiscover that theﬁrst ispreferable. Byusing thissubstitution in(a),weobtain (b) (\/z2—u2z2+uz) dz—z(udz+zdu) =0, |u|=|g§1, 2:950. Since z¢0,wecandivide (b)byittoobtain after simpliﬁcation (c) ;l;\/1—u2dz—-zdu=0, z¢0, |u|=|gl§1, where the+sign istobeused ifz>0;the——sign ifz<0.*Further ifu95:l:l,wemaydivide (c)by\/1——u2.Therefore (c)becomes dz du yd ——= i O =— 1. () x =|=\/_1____;5, z# ,Iul < Thevariables arenowseparated. Hence, by(6.63), a1-parameter family ofsolutions of(d)is (e) logz =Arcsinu +c,|u|<1,z>0, —log (—z) =Arcsinu+c,Iul<1,z<0. Replacing uin(e)byitsvalue asgiven in(7.31), wehave (f) logz=Arcsing+c, <1,z>O, —-log(—z) =Arcsing+¢, <1,z<0. Inobtaining thesolution (f),wehadtoexclude thevalues =|y/z| = 1.This means wehadtoexclude thefunctions y=;l:z. You canand should verify that these twofunctions alsosatisfy (a).They areparticular solutions of(a)notobtainable from thefamily (f). ‘For realz,\/F =zifzZ0and\/F =—zifzé0.Forexample, ifz=2, \/F=2andifz =-2,\/(-2)= =-(-2) =2. Lesson 7—Exercise 61 EXERCISE 7 1.Prove thatthesubstitution in(7.3) of z=uy, dz=udy—l—ydu, g/#0, leads toaseparable equation. Find a1-parameter family ofsolutions ofeach ofthefollowing equa- tions. Assume ineach case that thecoefficient ofdy75O. 2.2zydz+(z2+y2)dy=0. 3.(z+\/y2 —zy)dy —ydz =0. 4.(z+y)dz -(z—y)dy =0. 5.zy'-—y-—zsin (y/z) =0. 6.(2z2y —|—y3)dz+(zyz—22:3)dy=0. 7.y2dz+ (z\/F? ——zy)dy =0. —cosydz --(£siny+cosg)dy =0.z y z z HQ9° 9.ydz+zloggdy —2zdy =0. 10.2ye‘/"dz+(y—22:0"/1/)dy =0. ll.(zev/I —ysin dz+zsin5dy=0. Find aparticular solution, satisfying theinitial condition, ofeachofthe following differential equations. 12.(z2+y2)dz=2zydy, y(—1) =0. 13.(ze”/1+ y)dz=zdy, y(1) =0. 14.y’—g—|—cscg =0,y(1) =0. 15.(zy—y2)dz—z2dy=0,y(1) =1. ANSWERS 7 2.3z2y+ ya=c. 3.y=cc-2‘/1"”, y>0,z<y;y=ce2‘1_"'", y<0,z>y. 4.Arctan(y/z) —1}log(zz—|—y2)=c. 5.y=2zArctancz. 2 6.55-1-logzy =c,z9!0,yaé0. 7.y2—ez=y\/y2—2:2,orequivalently, e(y—l— \/yz —zz)=zy,y2>z2. .y_8.ysin I-c. 9.y=c(1—|—logz/y). 10.2e‘/"+ logy =c. ll.logzz—e_"I’ (sinZ+cos =c. 12.y2=z2+z. 14.logz—cosg+1=0. 13.logz+ 5""=1. 15.z=¢"’"’*‘. 62Srnonu. Trrrs orFmsr Oman EQUATIONS Chapter 2 LESSON 8.Differential Equations with Linear Coeﬁicients. LESSON 8A. AReview ofSome Plane Analytic Geometry. The ﬁrstdegree equation az+by+c=0represents astraight line. Forthis reason itiscalled alinear equation. (Nora. Thepresence ofthecon- stant cintheequation prevents thefunction deﬁned byitfrom being homogeneous.) Ifthecoefficients ofzandyinonelinear equation are proportional tothezand ycoefficients inanother, thetwo lines they represent areparallel. Forexample, thetwolines 3z—-2y+7=0, 6:1:-—4y—|-3=0, areparallel since 3:—-2=6:-4. Ifthethree constants inonelinear equation areproportional tothethree constants respectively inasecond linear equation, thetwolines coincide, i.e.,they arethesame line. For example, thetwolines 2z+3y+1=0, 4z+6y—|—2=0, arecoincident. (Doyouseewhy?) Another concept ofanalytic geometry thatweshall need forthislesson isthat of“translation ofaxes.” Let(z,y) bethecoordinates ofapoint P withrespect toanorigin (0,0) (Fig. 8.1), andletustranslate theorigin to P(1»J‘)"(5-5')Y <1»1-) 7(0.0) ‘ y k h 2 (0,0) X x Figure 8.1 anewposition whose zandydistances from (0,0) arehandlcrespectively. Todistinguish theneworigin from theoldone,wecallitscoordinates (6,6). The point Pwillthen have twosetsofcoordinates, onewith re- spect to(6,6), which wedesignate by(z,y), andtheother with respect to Lesson 8B EQUATIONS WITH LINEAR COEFFICIENTS 63 (0,0), which wehave already designated as(z,y). This means that ifa point ismeasured from (0,0), itscoordinates have nobars over them; if itismeasured from (6,6) itscoordinates have bars over them. The ques- tion wenow askandwhose answer weseek isthis. What istherelation- ship between thetwosetsofcoordinates (z,y) and(z,y)? Ifyouwillexamine Fig.8.1carefully, youwillseethat (8.11) z=2Z+h, y==y+k. Hence by(8.11), (8.12) 3t=z—-h, y=y—k. These aretheequations oftranslation. Their purpose, youmay recall, is tochange more complicated second degree equations into simpler ones by eliminating theﬁrst degree terms. Weshall now demonstrate how a translation ofaxes can help solve adifferential equation with linear coefficients. LESSON 8B. Solution ofaDifferential Equation inWhich the Coefficients ofdxand dyAreLinear, Nonhomogeneous, and When Equated toZero Represent Nonparallel Lines. Consider thediffer- ential equation (8-2) (2191 —|—bi?!+¢1)d1’3 +(2293 +623/+62)dll=0. inwhich thecoefficients ofdzanddyarelinear andwhen equated tozero represent nonparallel lines. Weassume also that both cland02arenot zero. (Ifboth c1=0andC2=0,then (8.2) isadifferential equation with homogeneous coefficients which canbesolved bythemethod of Lesson 7.)Since thecoefficients in(8.2) areassumed todeﬁne nonparallel lines, thepair ofequations (8-21) 111$-l"bi?!-l"61=0. (Z223-1-bgy-1-c2=0, formed with them, have aunique point ofintersection andtherefore a unique solution forzandy.Letuscallthispoint (h,Ic). Ifwenow trans- latetheorigi_n_to (h,k), then by(8.11), (8.2) becomes, with respect tothis new origin (0,0), (3-22) [6107+h)+bi(?2+k)+011113 +l<12(!¥+h)+b2(l7+ls)+02]dy=9. which simpliﬁes to (8.23) [alz+b,y+(alh +bllc-1-c1)]dz +[GQE +bgy +(Ugh +bgk +02)] = 64Srncmr. Tyres orFnzsr Onnnn Eooxrrons Chapter 2 But (h,lc) isthepoint ofintersection ofthetwolines in(8.21) andthere- foreliesonboth ofthem. Hence theterm intheparentheses ineach bracket of(8.23) iszero. This equation therefore reduces to (3-24) (11117+b11J)d?B +(<12?+bz17)dl7 =0, which isnow ahomogeneous type solvable forEBandybythemethod of Lesson 7.By(8.11) wecanthen ﬁndsolutions interms of:1:andy. Note. 1.The left-hand members ofthesystem (8.21) bywhich hand kare determined arethecoefficients inthegiven differential equation (8.2). 2.Equation (8.24) which isequivalent to(8.2) with respect toanew origin translated tothepoint (h,k), canbeeasily obtained from (8.2). Omit theconstants clandc2andplace bars over asandy. Example 8.25. Find a1-parameter family ofsolutions of (a) (2x—y+1)dx+(:c+y)dy=0. Solution. The coefficient ofda:islinear butnonhomogeneous, and thetwo lines deﬁned bythecoefﬁcients ofdzand dyarenonparallel. Hence theprocedure outlined above applies. Solving simultaneously the twoequations determined bythecoeﬁicients ofdz:anddy,namely, (b) 2:z:—y+l=0, a:+y=0, weﬁnd that their point ofintersection is(—§-,§-). Hence, in(8.11) h=—§, lo= Translating theorigin tothepoint (—-§,§), theequa- tions oftranslation areby(8.11) and(8.12) (0)r=1'-t. y=!7+§, 1‘=w+t, !7=1/—%- By(8.24), (a)becomes with respect tothisnew origin (seeNote 2above) (d) (22z—’;)d:+(r+y)dy= 0. Tosolve it,weapply themethod ofLesson 7.Let (e) fl]=uiif, dy=udfi +Idu, Bysubstituting (e)in(d)andfollowing theprocedure outlined inLesson 7,weobtain __i. _"__1 2 (f) loglfifl-c1 \/§Arctan\/5 2log|2—|—u|, 2960, Lesson 8B Equxrrons wrrn Lmmn COEFFICIENTS 65 which isa1-parameter family ofsolutions of(d).In(f),wereplace uby itsvalue in(e)andmultiply by2.There results 2 (s) log2z’<2+%) =c—\/2Arctan#:, z¢0. Substituting thelasttwoequations of(c)in(g)gives ﬁnally 3:v+1)’ (3y—l)’_ __ 3y—1(h) logi2(—-——3 +ii; -c\/2Arctan\/_—f3x+ 1)» :c¢—%-- Comment 8.26. Thesolution (h)above isanexcellent example of thepoint made atthebeginning ofthischapter inintroductory remark No.3.Here isasolution written inimplicit form, which haslittle value forpractical purposes. Totrytoﬁndthefunction g(a:)implicitly deﬁned bythisrelation would beanextremely laborious ifnotahopeless task. Ingeneral a1-parameter family ofsolutions ofadifferential equation with linear coefficients orwith homogeneous coefficients willusually bea complicated expression ofthiskind. Inthese cases, more important for practical purposes than animplicit solution isaknowledge oftheapproxi- mate behavior oftheintegral curves. There arefortunately means avail- able bywhich itispossible todetermine thecharacter ofthese integral curves from thediﬂerential equation (8.2) itself, without theneed tosolve it.Since differential equations with homogeneous orlinear coefficients arise inpractical problems when trying toﬁnd anapproximation tothe behavior ofthemotion ofaparticle whose velocities inthe:0andydirec- tions aregiven bythetwodifferential equations <1if=fay), d-6%=a(w,y), wehave deferred toLesson 32afurther discussion ofthisimportant topic. Weshall show there, how itispossible toﬁnd anapproximation ofthe particle’s motion bychanging thetwoequations into oneequation with linear coefficients, andthen showing how more useful information canbe obtained from theresulting diﬂerential equation itself than from itsusual complicated implicit solution. Weshall thus beable tolearn what the solution (h)intheabove example approximately looks like, notfrom this solution, butfrom thegiven differential equation (a);seeExample 32.44. 66Sracnu. Trrns orFmsr Oman EQUATIONS Chapter 2 LESSON 8C. ASecond Method ofSolving theDifferential Equa- tion (8.2) with Nonhomogeneous Coefficients. In(8.2), let (8-3) u=air+biy+cl, v=(1256+bgy+Cg. Therefore (8.31) du=a1dx+b1dy, dv=a2dz+b2dy. Now solve (8.31) forda:anddy.Thesubstitution in(8.2) of(8.3) andthese values ofda:anddywillalsoleadtoadifferential equation with homo- geneous coefficients solvable bythemethod ofLesson 7. Example 8.32. Find a1-parameter family ofsolutions of (=1) (2w~y+1)dw+(rv+y)dy=0- Solution. Asindicated in(8.3), welet (b) u=2:v—y-I-1, v=x+y. Therefore (c) du=2d:c-—dy, dv=dz+dy, Thesolution of(c)forda:anddyis Substituting (b)and(d)in(a),weobtain (6) u<du 3-dv)_v<du -32dv) =0, which simpliﬁes to (f) (u-v)du+(u+2v)dv=0. This equation isnow ofthetype with homogeneous coefficients. Follow- ingthemethod ofLesson 7,welet (g) u=tv, du=tdv+vdt. Lesson 8D Coarrrcunrrs PARALLEL onCOINCIDENT Lmas 67 Substituting these values in(f),weobtain (h) (tv—v)(tdv +vdt)+(tv+2v)dv=O, which reduces to . d t—1(1) '-by‘-l-5,‘;--_{—__-é'dl=0, U750. Itssolution is (j) log|v|+§log(t2+2)—:/L-§Arctan——=c, v#0, §._. log[v2(t2 +2)]=C+\/2Arc tan——» v;-6O. §.. By(s)and(b), _1i__2z—y+1(k) t—v— ! Substituting (k)in(j),weobtain (1)log[(2¢-y+1)’+2(z+3/)2]=0+\/2Arctan z+y9'60. LESSON 8D. Solution ofaDifferential Equation inWhich the Coefficients ofdzand dyDeﬁne Parallel orCoincident Lines. Ifthelines deﬁned bythecoeﬂicients ofdzanddyin(8.2) areparallel, the method ofLesson 8Bwillnotwork. Parallel lines donothave apoint of intersection andtherefore (8.21) hasnosolution forzandy.Inthiscase wemust resort toadifferent substitution. Itisillustrated inthefollowing example. Example 8.4. Find a1-parameter family ofsolutions of (ii) (2rv+3y—1)dz+(4w+6y+2)dy=0, alsoanyparticular solution notobtainable from thefamily. Solution. Weobserve that thelines deﬁned bythecoefﬁcients ofdz anddyareparallel butnotcoincident lines. Inallsuch cases, thesubsti- tution ofanew variable forthecoefficient ofdzorofdywilltransform theequation into onewhich isseparable. Wetherefore let -d(b)u=2z+3y——1, du=2dz+3dy, dx=‘%”- Then by(b) (c) 2u+4=4z—|—6y-[-2. 68SPECIAL Tress orFmsr ORDER Eqnxrrons Chapter 2 Substituting (b)and(c)in(a),weobtain (<1) u +<21»+4)dy=0, which simpliﬁes to (e) udu+(u+8)dy=0, anequation Whose variables areseparable. Ifu#5-8, (e)canbewritten as (f) u-_'§§du+dy=0, 14¢-s. Integration of(f)gives (8) u-—8log|u+8|+y=c, usé-8. Finally, replace in(g)thevalue ofuasgiven in(b),noting atthesame time that theexclusion ofu=——8implies theexclusion oftheline 22:+3y+7=0.Hence (g)becomes (h) 2z+3y—1—8log|2z+3y+7|+y=c, 2z+3y+7#0, which isa1-parameter family ofsolutions of(a). The function deﬁned by (i) 2x+3y+7=0, which hadtobeexcluded inobtaining (h)alsosatisﬁes (a).(Besureto verify it.)Itisaparticular solution notobtainable from thefamily (h). Example 8.41. Find a1-parameter family ofsolutions of (=1) (2w+3y+2)drv+(4w+6y+4)d2/=0; alsoanyparticular solution notobtainable from thefamily. Solution. Weobserve thatthecoefficients in(a)deﬁne thesame line. Ifweexclude values ofzandyforwhich (b) 2w+3y+2=0. wemay divide (a)byitandobtain (c) dz+2dy=0. Itssolution is (d) as+2y=c, Lesson 8—Exe|-cise 69 which isvalid forthose values ofzandywhich donotlieontheline 2:0+3y+2=0.Itistherequired 1-parameter family. However, the function deﬁned by (6) 2:23+3y—|—2=O alsosatisﬁes (a).Itisaparticular solution notobtainable from thefamily. EXERCISE 8 parameter family ofsolutions ofeach ofthefollowing equations. -w - =o=0. -- - =0. -— =0. 509:-zampmpy+++-l-I-+++1,new==~===@to,.. N559,-,vgumI“self+++‘"11-:_|_g,;_—;C;P“Q‘/\/-\,\&8Q8-“es;++~’\/'\ ~:::§@§§‘Q‘§\./+L;ea.+@+=§‘‘=wl°+e5fiws'o<g"O'§C',tQ‘:\;‘Qt:-"§l._;"~§".=>.o =0. — -— 71/—-1)dy=0. Findaparticular solution, satisfying theinitial condition, ofeach of thefollowing differential equations. 11-(I+y)d-'¢+(31+31/-4)d@/= 11(1)=0-12.(3z+ 2;/+3) dz—- - =0,y(—2) =1. m@+na+m+ = ﬂ®=L14.(z+y+2)dz—(z-— —— =0, y(1)=0.‘Q/-\+8 @g»+\-/N) é“:r#\./9..‘€Q:‘ac,§-. ANSWERS 8 1.log[4(y—-1)2+(z—-2)2]~—2Arc tan%Z =c. 2.log|15z:+10y -—1|+§(z —-y)=c. 3.z+2y =c. 4-.z+2y+log|z+y—2| =c. 5.(t_ v2 =6e2Arotn1[v/(z—1)]. 6.z+2y+log|z+y— 1|=c. 7.7log|2z+1|+2log|7y—— 3|=c. 8.z+3y—3log|z+2y+3| =c. 9-[(1-i"2)/(I-l"!/+1)l+1°8|$+1'/+1l= 6- 10.82:2—4zy+8z —7y2—|—2y =c. ll.z+3y+2log(2—z——y) =1. 12.(2z+2y+1)(3z --21/+9)‘ =—1. 13.(1/+7)2(3z+y+ 1)=12s. 14-.log[(22——1)2+(y+3)2]—|—2Arc tanﬁg =2log3. 70SPECIAL Tvrns orFmsr Onnsn Eooxrrons Chapter 2 LESSON 9.Exact Differential Equations. Before beginning astudy ofthistype ofequation, weshall review those concepts from thetheory ofintegration which weshall need. s\\\\\§lf(xk) | I /// I xo=a x1 x2 x,,_, xk x,,_1 x,,=b Figure 9.1 1.Letf(z) beafunction ofzdeﬁned onaninterval I:a§z§b. LetIbedivided into nsubintervals and callAzhthewidth ofthekth subinterval, (Fig. 9.1). Then if (9.11) lim£3f(z).) Ax). "-’°°r==1 exists asthenumber ofsubintervals increases insuch amanner that the largest subinterval approaches zero, wesaythat n b (9.12) limZfa.)Ax),=/'f(z)dz.n—m kzl a This limit iscalled theRiemann integral off(z) over I.Ifthislimit does notexist, wesayf(z) isnotRiemann-integrable over I. 2.Iff(z)isacontinuous function ofzonaninterval I:a§z§b,and I (9.13) Fa)=/fa)du, then bythefundamental theorem ofthecalculus (9.14) F’(z) =f(z), a<z<b, orequivalently I (9.15) %/gof(u)du=f(z). 3.LetP(z,y) and Q(z,y) befunctions oftwo independent variables, z,y,both functions being deﬁned onacommon domain D.InLesson 2C Lesson 9 Exxcr DIFFERENTIAL EQUATIONS 71 (wesuggest your rereading thislesson), weshowed that atwo-dimensional domain may assume various shapes. Hence ifwewish toperform, for example, thefollowing integration, theﬁrstwith respect toz(yconstant), andthesecond with respect toy(zconstant), 97 U wemust besure that (z0,yo) isapoint ofDandthat therectangle deter- mined bythelinesegments joining thepoints (z0,y0), (z,;4/0) and (z0,y0), (z0,y) liesentirely inD.Ifthedomain Diseither oftheregions shown 0 y 0 y ;(rayo) / / (x /1/_//0» ) Figure 9.17 Figure 9.18 intheshaded areas inFig. 9.17 or9.18, then youcannot integrate (9.16) along thestraight linefrom zotozorfrom yotoybecause part ofthese lines arenotinD. There isalso another factor tobeconsidered. Ifthedomain Disthe region shown inFig. 9.17, andtherectangle determined bythelineseg- ments joining thepoints (zo,yo), (z,yo) and (z0,y0), (z0,y) liesentirely in D,then (9.16) canbeintegrated. If,however, thedomain Disaregion with ahole initsuch asinFig. 9.18, then even iftherectangle were en- tirely inD,(9.16) cannot beintegrated because ofthishole initsinterior. Therefore, wemust insome way distinguish between these twotypes of regions. The region inFig. 9.17, i.e.,theonewhich hasnohole init,is called asimply connected region. Itsformal deﬁnition isthefollowing. Deﬁnition 9.19. Aregion iscalled asimply connected region if every simple closed curve lying entirely intheregion encloses only points of theregion. Tosummarize: Wecanperform theintegrations called forin(9.16) only if: (a)The common domain ofdeﬁnition ofthefunctions isasimply con- nected region R. 72SPECIAL Trees orFrasr Ommn Eqmmons Chapter 2 (b)Thepoint (zo,yo) isinR. (c)Therectangle determined bythelinesjoining thepoints (z0,y0), (z,y°) and(zo,yo), (z0,y) liesentirely inR. LESSON 9A. Deﬁnition ofanExact Differential and ofanExact Differential Equation. Weshowed inLesson 6Bthatif,forexample, (9-2) Z=f(=v.y) =3w’y+My+ya+5. then thedifferential ofz[see(6.47)] is (9.21) ¢= dz+ dy= (6zy+5y)dz+(3z2+5z+3y2)dy. Iftherefore, wehadstarted with thedifferential expression (9.22) (6zy +5y)dz+(3212 +5z+3y2) dy, wewould know that itwasthetotal differential ofthefunction f(z,y) deﬁned in(9.2). Differential expressions ofthetype (9.22); i.e., those which arethetotal differentials ofafunction f(z,y), arecalled exact differentials. Hence: Deﬁnition 9.23. Adifferential expression (9-24) P($.11)dw+Q(1=.!/) dy iscalled anexact differential ifitisthetotal differential ofafunction f(=v.y). i-8-.if (9241) P(z,y)=,"’;r<x,y> andcan=§r<x.y>. Setting thedifferential expression (9.22) equal tozero, weobtain the differential equation (9.25) (ﬁzy +5y)dz+(3:02 +5z+33/2) dy=0, whose solution, by(9.2), is (9-26) f(w.y) =3w’y+My+ya=6. valid forallvalues ofzforwhich (9.26) deﬁnes yasanimplicit function ofzandforwhich dy/dz exists. [Ifyouhave anydoubt that (9.26) isan implicit solution of(9.25) orthat 8f(z,y)/8z isthedzcoefficient in(9.25) orthat 6f(z,y)/By isthedycoefficient in(9.25), verify these statements] Adifferential equation ofthetype (9.25), i.e.,onewhose dzcoefficient isthepartial derivative with respect tozofafunction f(z,y) andwhose dy Lesson 9B SOLUTION orANExncr EQUATION 73 coefficient isthepartial derivative with respect toyofthesame function, iscalled anexact differential equation. Hence: Deﬁnition 9.27. The differential equation (9-28) P(z,y) dw+Q(w.y) dy=0 iscalled exact ifthere exists afunction f(z,y) suchthatitspartial deriva- tivewith respect tozisP(z,y) anditspartial derivative with respect to yisQ(z,y). Insymbolic notation, thedeﬁnition says that (9.28) isan exact differential equation ifthere exists afunction f(z,y) such that (9.29) =Pay), =Q(r.y-)- A1-parameter family ofsolutions oftheexact differential equation (9.28) isthen (9.291) f(z,y) =c. Intheexample used above, weknew inadvance that(9.25) wasexact andthatthefunction deﬁned in(9.26) wasitssolution, because westarted with thisfunction f(z,y) andthensetitstotal differential equal tozeroto obtain thedifferential equation (9.25). Ingeneral, however, ifaﬁrstorder differential equation were selected atrandom, twoquestions would present themselves. First, how would weknow itwasexact, andsecond, ifitwere exact, how could weﬁnd thesolution f(z,y) =c?The answer toboth questions isincorporated inthetheorem andproof which follow. LESSON 9B. Necessary and Sufficient Condition forExactness and Method ofSolving anExact Differential Equation. Theorem 9.3. Anecessary andsuﬂicient condition thatthedifferential equation (9-31) P(z,y) dw+Q(mI) dy=0 beexact isthat (9.32) 5%P(z,y)=5’;cat). where thefunctions deﬁned byP(z,y) andQ(z,y), thepartial derivatives in (9.82) and8P(z,y)/oz, 6Q(z,y)/6y ezist andarecontinuous inasimply connected region R. Norm. Although thetheorem isvalid asstated, theproof willbegiven only forarectangular domain contained entirely within thesimply con- nected region R. 74SPECIAL TYPES orFmsr Oanan Eouurons Chapter 2 Proof ofnecessary condition, i.e.,given (9.31) isexact, toprove (9.32). Since (9.31) isexact, itfollows from Deﬁnition 9.27 that there isafunc- tionf(z,y) such that (9.33) 5.-<.,.> ='P<-9). ,;’3r(-».y> =eat). Because oftheassumptions about thefunctions PandQstated after (9.32) andbyatheorem inanalysis, wearepermitted toassert that 1.1% f(z,y)> and 6% f(z,y)) exist. 6 8 3 8 i.e.,theorder inwhich wetake theﬁrst andsecond partial derivatives of f(z,y) isimmaterial. Substituting (9.33) in2,above, weobtain <9-34> 5’;Pay)=5’;eat). which is(9.32). Proof ofsufficient condition, i.e.,given (9.32), toprove (9.31) isexact. ByDeﬁnition 9.27, theproof that (9.31) isexact isequivalent toproving theexistence ofafunction f(z,y) such that 8f/élz =P(z,y) and6f/6y = Q(z,y). [Intheproof ofthis sufficient condition, weshall atthesame time discover themethod ofﬁnding f(z,y).] Hence thefunction f(z,y), if itexists, must have theproperty that (9.35) =P(z,y). Therefore, with yconstant, f(z,y), by(9.14) and(9.13), must beafunction such that I (9-36) f(I.1/) =£0P(@,1/) dw+R(y). where zoisaconstant andR(y) stands forthearbitrary constant ofinte- gration. [Remember that ingoing from (9.36) to(9.35), R(y) andyare constants.] Butthisfunction f(z,y) must alsohave theproperty, byDeﬁnition 9.27, that 6 (9-37) a—yf(@=.2/) =Q(=v.y)- Therefore, differentiating (9.36) with respect toyandsetting theresult Lesson 9B SOLUTION orANExxcr Eqnzrrron 75 equal toQ(z,y), weobtain insymbolic notation (9-38) ,-1;0P(-.9)at+rm)=cat)- Byhypotheses P(z,y) iscontinuous. Hence, byatheorem inanalysis, we can, in(9.38), putthesymbol 0/8y inside theintegral sign. Itwillthen read (9.981) Pay)dz+19(9)=Q(r.y)- By(9.32), wecanwrite (9.381) as (9.99) /5’;cat)<19+19(9)=Q(r.y)- Study theintegral in(9.39) carefully. Inwords, itsays: differentiate Q(z,y) with respect toz,with yﬁxed, andthen integrate thisresult with y stillﬁxed. The netresult istogetback thefunction Q(z,y). [Try itfor thefunction Q(z,y) =zzywith yconstant.] Hence (9.39) becomes (9-4) Q(r.z/)l;°,, +R’(y) =Q(w,y). which Simpliﬁes to[remember Q(w.y)li,, =Q(w,y) —Q(1'3o)1/)l (9-41) R’(y) =Q(wo.y)- Integration of(9.41)gives,by(9.14)and(9.13), U (9-42) R(y)=IQ(w<).u)dy,U0 where yoisaconstant. Substituting (9.42) in(9.36), weobtain ﬁnally 1 u (9-43) f(z,y) =IP(r.y)dr +/iQ(9<).y)dy. where (z0,y0) isapoint inR,and thelinesegments joining thepoints ($0.90). ($.90) and($0.1/0). ($0.11) lieentirely inR- This function f(z,y) weshall now show istheoneweseek. Since the second integral in(9.43) isafunction ofy,wehave, by(9.13) and(9.14), (9.44) =59,;0P(z,y)dz+0=P(z,y). 76SPECIAL Trras orFmsr Oman Eooxrrons Chapter 2 And byfollowing thesteps from (9.38) to(9.4), weobtain (9.441) ,,-Z;0P(z,y)39=Q(9.y)—can-). By(9.13) and(9.14) II‘ (9-443) 5’;Q(wo,y)dy=Q(w<>.u)- Hence, by(9.43), (9.441) and(9.442), itfollows that (9.443) 5’;/(3.9) =9(-,9). Wehave thus notonly proved thetheorem, buthave shown atthesame time how toﬁnd a1-parameter family ofsolutions of(9.31). Itis,by (9.291) and(9.43), I u (9-45) f(z,y) =AP(z,y) dz+/LQ(w<).y) dy=9. where (z0,yo) isapoint inRandtherectangle determined bythelineseg- ments joining thepoints (z0,y0), (z,y0) and(z0,yo), (a:0,y) liesentirely inR. Prove asanexercise that ifinplace of(9.35) wehadstarted with (9.46) %my)=cat). wewould have obtained forthesolution of(9.31) ﬂ I (9-47) f(w.y) =IQ(w.y) dy+IP(3=.@/0) dz=9- Remark. Both (9.45) and (9.47) will give a1-parameter family of solutions of(9.31). Youmayusewhichever youﬁndeasiest inaparticular problem. Example 9.5. Show that thefollowing differential equation isexact andﬁnda1-parameter family ofsolutions. (a) cosydz --(zsiny-—y2)dy=0. Solution. Comparing (a)with (9.31) weseethatP(z,y) =cosyand Q(z,y) =-zsiny+yz.Therefore 8P(z,y)/6y =—sin yand6Q(z,y)/Oz = -sin y.Since 6P/8y =8Q/Bz, theequation, byTheorem 9.3,isexact and since P(z,y) andQ(z,y) aredeﬁned forallz,y,theregion Risthewhole plane. Hence wemay take zo=0and yo=0.With zo=0, Lesson 9B SOLUTION orANExscr Eqnxrron 77 Q(zo,y) =Q(0,y) =yz.Thus (9.45) becomes 47 ll (b) /Icosydz-1-/Iyzdy=c.o o Integration of(b)gives (remember intheﬁrst integration yisaconstant) 3 (c) zcosy+%=c, which istherequired 1-parameter family. Example 9.51. Show that thefollowing differential equation isexact andﬁndaparticular solution y(z)forwhich y(1)=0. (a) (z—2zy-1-e”)dz+(y—zz—|—zel’)dy=0. Solution. Comparing (a)with (9.31) weseethat P(z,y) =z—2zy +e"and Q(z,y) =y—zz+ze”. Therefore 6P/é)y =—2z +e"and 6Q/8z =—2z —|—e".Since 8P/6y =8Q/82:, theequation, byTheorem 9.3,isexact. And since theregion Risthewhole plane, wemay take zo=0andyo=0.With zo=0,Q(zo,y) =y.Hence (9.45) becomes 3 ll (b) /(z-—2zy+e”)dz-}-/0ydy=c.0 Integration of(b)gives 2 2 (c) %—z2y+ze"+%—=c. Toobtain aparticular solution forwhich z=1,y=0,wesubstitute these values in(c)andﬁndc=§.Hence therequired particular solution is (d) zz—2z2y +2:00” —|—y2=3. Example 9.52. Show that thefollowing equation isexact, andﬁnd a 1-parameter family ofsolutions. (a) (zs—|—zyzsin2z+y2sin2z)dz+(2zy sin2z)dy=0. Solution. Here 8P/by =2zysin2z—|—2ysin’zand QQ ___ . .2 __ . .2 ax-2y(2z sinzcosz—|—sinz)-2zysin2z+2ysinz. Since 6P/8y =8Q/6z, theequation byTheorem 9.3isexact. Inthiscase, itwillbefound easier touse(9.47). Since theregion Risthewhole plane, 78Sracnu. Trrns orFmsr Oanan EQUATIONS Chapter 2 wemay take zo=0andyo=0sothat P(z,0) =za.Hence (9.47) becomes u I (b) f2xysin2 zdy+I$3dz=6.o 0 Integration of(b)gives (remember thatthistime zisaconstant inthe ﬁrstintegration) 4 (c) zy2sinzz+2-=c. which istherequired 1-parameter family. Remark. Formulas have theadvantage ofenabling onetoobtain a result relatively easily, buthave thedisadvantage ofbeing easily forgot- ten. Wetherefore outline amethod bywhich youcansolve anyofthe above differential equations directly from Deﬁnition 9.27. Inouropinion, thismethod isthepreferable one. Example 9.6. Solve Example 9.5,using Deﬁnition 9.27. Solution. The differential equation is (a) cosydz —(zsiny--y2)dy=0. Wehave already proved (a)isexact. ByDeﬁnition 9.27, therefore, there exists afunction f(z,y) such that (b) '-2% =P(z,y) =909y- Hence integrating (b)withrespect toz,weobtain (9) f(9.y)=I9991/dw +R(y)=99999 +R(y)- Again byDeﬁnition 9.27, thisfunction f(z,y) must alsohave theproperty that (3) =9(3))=-9sin1)+9’. Hence differentiating thelastexpression in(c)partially with respect toy andsubstituting thisvalue in(d)weobtain (9) —zsiny+R'(?/) =“"31sin1/-1'1/2) which simpliﬁes to (f) R(y)=[1/zdy =933- Lesson 9—Exercise 79 Substituting (f)in(c)gives (s) f(w.y) =39991/+ By(9.291), a1-parameter family ofsolutions of(a)istherefore a (h) zcosy+%=c, justaswefound previously. Asanexercise, solve theother twoExamples 9.51and9.52directly from thedeﬁnition aswedidabove. EXERCISE 9 1.Prove formula (9.47). 2.Solve Example 9.51bythemethod ofExample 9.6. 3.Solve Example 9.52bythemethod ofExample 9.6. Show that each ofthefollowing differential equations 4-13 isexact and ﬁnda1-parameter family ofsolutions using formula (9.45) or(9.47), and alsothemethod outlined inExample 9.6. 4.(337211 +8zy2) dz—[—(za+8z2y +12y2) dy=0. 5-(2.2)-+ (1.1-“)9-~-6.2zydz+(z2—[—yz)dy=0. 7.(e'siny +e"")dz-(ze'" —e‘cosy dy)=0. 8.cosydz -—(zsiny -yz)dy=0. 9.(:2:—2zy+e")dz+ (y——z2+ze")dy =0. 10.(:02—z+1/2)dz-(e"—2zy)dy=0. ll.(2z—|—ycosz)dz—|—(2y+sinz -siny)dy=0. 2 12.:r\/3+y=d¢-——3—”-—-.1 =0. x u——V9”+1/2y13.(4213—-sinz+ ya)dz-—(yz+1 —39:;/2) dy=0. 14.Iff(z,y) isthefunction deﬁned in(9.43), prove (9.44). Find aparticular solution, satisfying theinitial condition, ofeach ofthe following differential equations. 15.e‘(y3 +zy3+1)dz+3y2(ze‘ —6)dy=0,y(0) =1. 16.sinzcosydz+coszsinydy=0,y(1r/4) =1r/4. 17.(y%=~’ +44*)119+(ziyw/= -aw)dy=0,y(1)=0. ANSWERS 9 4.zay+4z2y2 —|—4y3=c. 7.e‘siny+ze" =c. 5-I2-I-E-I-10g|!/I =6- 8.3zcosy+y3=c. 6-3121/ '11/3=6- 9.zz—2z2y-|— ya—[-2ze" =c. 80SPECIAL Trras orFmsr ORDER Eousrrons Chapter 2 10.22:3—3z2—l- 6zy2 -6e"=c. ll.z2+ ysinz—[- y2+ cosy =c. 12_(I2+1,2):/2 +ya=c_ 13.3z4+ 3cosz+ 3y3z -—ya-—3y= 17.e="’+ z‘—y=2.c. 15.ze‘y3 +e”—6y3=-5. 16.2cosz cosy =1. 3 LESSON 10. Recognizable Exact Differential Equations. Integrating Factors. LESSON 10A. Recognizable Exact Differential Equations. Itis sometimes possible torecognize thesolution f(z,y) =cofanexact differ- ential equation without thenecessity ofresorting tothemethods of Lesson 9b.F01‘example, iftheexact differential equation is (10.1) 2zy2 dz-l-2z2y dy=0, youmight beable torecognize that itssolution is (10.11) z2y2 =6. Ifyoucannot, youmust ofcourse usethemethod ofsolution outlined in theprevious lesson. Welistbelow anumber ofexact differential equations andtheir solu- tions. Itwillbeproﬁtable foryoutoverify some ofthem bytaking the total differential ofthefunction ontheright and seeing ifityields the differential equation ontheleft. Exact Dijferential Equation (10.2) ydz—[-zdy=0 (10.21) (10.22) (10.23) (10.24) (10.25) (10.26) (10.27) (10.28) (10.29) (10.3) (10.31) (10.32)2zydz—|—zzdy yzdz-1-2zydy 2zy2 dz—l-2z2ydy0 0 =0 3z2y3 dz+3z3y2 dy= 3z2y dz+z3dy=0 ycoszdz—l- sinzdy =0 sinydz+zcosydy = ye"dz—|—ze"dy=0 dz d_+_l/=0 3 ll ydz -—zdy _ ya _ydz —zdy _._.___;2i_ 2zydy —y2dzw 0 0Solution zy=c zzy=c zyz c z2y2=c 223113 =c z3y=c ysinz =c zsiny =c e"=c log(zy) =c QS§¢§l~lN)_.=c _=c =0 -=5 H Lesson 10A Rncoomzxnnn Exxcr Drrraasmun EQUATIONS 81 Exact Di_ﬂ'erential Equation Solution 2 2 (1933) _§! =0 1;;=c (10.34) ii%@ =0 Arctan5=6 .1-.11(10.35) 2”—;,{-_-% =0 10;gig =1 2 3 (1036, _PM/ ,—H9 =0 1=.$3 2 (1037) lfizrqfl/ill =() 5%=¢ __ydz+zdy_ 1_(10.38) -———-$21,, -0 -c303 2 3 rd3—11/<11/_ 1/_- —0 '3?" —C dz+1dy0.39 ‘i = \/33= (1 1) W 0 z—[-y c (l0.392) ea’dy+3e3’y dz=0 e3'y =c Theexact differential equations ontheleftaresometimes referred to asintegrable combinations. Weencourage youtoaddtothislist whenever youdiscover thesolution ofanew integrable combination. Ifadifferential equation isexact, butnointegrable combination is readily apparent toyou, youcanalways fallback onthemethod ofsolu- tion outlined inLesson 9B. However, itissometimes possible, ifanequa- tion isexact, tosolve itmore readily andeasily byajudicious rearrange- ment ofterms soastotake advantage ofanyintegrable combinations it may possess. Note. Hereafter when weusetheword solve, inconnection with ﬁrst order equations, weshall mean “ﬁnd a1-parameter family ofsolutions of thegiven differential equation." Example 10.4. Solve (4) E,-,i1de+%dy=0, 1/#0. Solution. By(9.31), P(z,y),Q(z,y) aretherespective coefficients of dz,dy.Therefore, 8P____l 8Q___1_(b) ay__ 2/2and ax— yz Hence theequation isexact. Asolution therefore canbeobtained by Theorem 9.3. However, ifwerewrite theequation as 1 2 z—dz -——-d=0 0 (9) zd.'v+y +yd9 y,9.1/9*. 82Smzcnu. 'I‘1m:s orFmsr Omma Eomvrrons Chapter 2 which canbeputintheform 2 dx—- d(d) a:da:+§dy-§-Vi?/2:0, y#0, weobserve thatthelastterm oftheleftsideis,by(10.30), d(x/y). Each oftheother terms canbeintegrated individually. Hence integration of (d)yields 2 (e) §5+2l<>s|1/|+§=¢,z/#0, which istherequired solution. Example 10.41. Solve (a) (3e3‘y -—22:)dx+ea‘dy=0. Solution. You caneasily verify that (a)isexact. Ittherefore can besolved bythemethod ofLesson 9B.However, ifwerewrite theequa- tionas (b) 3e3"y dz+e3‘dy—-22:da:=0, weobserve that thesum oftheﬁrsttwoterms isby(10.392), d(e3‘y), andthatthethird term canbeintegrated individually. Hence integration of(b)yields thesolution (c) e3‘y —1:2=c. Example 10.42. Solve (a) (x—-2xy+e")dx+ (y—-a:"'+a:e")dy=0. Solution. Wehave already shown (seeExample 9.51) that this equation isexact, andsolved itbyuseofLesson 9B. However, ifwe rewrite itas (b) radar: —(2:z:yd:c +1:2dy)—|—(e"da: +:ve”dy) +ydy =0, thesecond expression, by(10.21), isd(a:2y) andthethird expression is recognizable asd(:z:e"). Thesolution of(b)istherefore 2 2 (c) %~—a:2y+a:e"+%=c. just aswefound previously. LESSON 10B. Integrating Factors. Deﬁnition 10.5. Amultiplying factor which willconvert aninexact differential equation intoanexact oneiscalled anintegrating factor. Lesson 10B INTEGRATING Facrons 83 Forexample, theequation (ya+y)dx-—:1:dy=0isnotexact. If, however, wemultiply itbyy_2, theresulting equation 1@+Qa-%@=cy¢a isexact. (Verify it.) Hence byDeﬁnition 10.5, 3/-2 isanintegrating factor. Remark. Theoretically anintegrating factor exists forevery differ- ential equation oftheform P(z,y) div+Q($,y) dy=0,butnogeneral rule isknown todiscover it.Methods have been devised forﬁnding integrating factors forcertain special types ofdifferential equations, but thetypes aresospecial that themethods areoflittle practical value. It isevident that ifastandard method ofﬁnding integrating factors were available then every ﬁrst order equation ofthisform would besolvable bythismeans. Unfortunately thisisnotthecase. However, inthenext lesson weshall discuss aspecial, important, ﬁrstorder differential equation forwhich anintegrating factor isknown andforwhich astandard method forﬁnding itexists. Inthemeantime, weshall show byafewexamples how anintegrating factor, ifyouareshrewd enough todiscover one,canhelp yousolve a differential equation. Example 10.51. Solve (a) (y’+y)dw—My=0- Solution. Weproved toyouabove that1/'2isanintegrating factor of(a). Hence multiplication of(a)by3/'2willconvert itinto theexact differential equation 1@ Q+Qm-%@=ay¢c Wecannowsolve (b)bymeans ofLesson 9Bor10A. Ifwerearrange the terms toread ydz:—:1:dy (<5) div+*1/Ti =0» thesecond term by(10.3) isd(:c/y). Hence byintegration of(c)we obtain thesolution (<1) w+§=¢; y=;'_%—,,' y#0- NOTE. Thecurve y=0alsosatisﬁes (a).Itisaparticular solution notobtainable from thefamily (d). 84Srscuu. Txrss orFmsr Onnsn EQUATIONS Chapter 2 Example 10.52. Solve (a) ysecxd:c+sin:z:dy=0, x#g»-'3§1-r»-- Solution. First verify that (a)isnotexact. However, secx isan integrating factor. Multiplication of(a)byitwill therefore yield the exact equation (b) ysec2xdx+tan:cdy=0, :c¢%-»%,---. [Now verify that (b)isexact.] Thesolution of(b)bymeans ofLesson 9B orbyrecognizing that itsleftsideisanintegrable combination is (c) ytan:c=c ory=ccota:. LESSON 10C. Finding anIntegrating Factor. Asnoted inthe remark following Deﬁnition 10.5, astandard method ofﬁnding aninte- grating factor isknown only forcertain very special types ofdifferential equations. Weshall discuss some ofthese special types below. Weassume that (10-6) P($.1/) drv+Q(@=,y) dy=0 isnotanexact differential equation andthat hisanintegrating factor of (10.6), where hisanunknown function which wewish todetermine. Hence, byDeﬁnition 10.5, (10.61) hP(:z:,y) dx+hQ(a:,y) dy=0 isexact. Ittherefore follows, byTheorem 9.3,that <10-62> 5[hP<x,y>1 =%[how]. Weconsider ﬁvepossibilities. 1.hisafunction only ofx,i.e., h=h(x). Inthiscase weobtain from (10.62), 6 6 dh(10.63) ho)5Pct)=hm5can+can which wecanwrite as 8 5 —P(w,y) ——Q(w,y)(10.64) ‘$21)=L” Q(x’y‘;” ]d:c. Inthespecial case that thecoeﬂicient ofdxin(10.64) alsosimpliﬁes toa Lesson 10C FINDING ANINTEGRATING Facron 85 function onlyofx,letuscallitF(:c), sothat 0 8 .__.__.._______.__ , Q(w,y) then,by(10.64) and(10.66), log[h(:c)]=fro)dx.Hence,(10.65) F(a:) = (10.66) ho.)=elm"=, where wehave omitted theconstant ofintegration, isanintegrating factor of(10.6). Example 10.661. First show that thedifferential equation (a) (e"—-siny)dz+cosydy=O isnotexact andthen ﬁndanintegrating factor. Solution. Comparing (a)with (10.6), weseethat (b) P(z,y) =e’—-siny, Q(:v,y) =cosy. Therefore (C) ~9P§r;,y) =_c0Sy’ 8Q§;,y) =0_ Hence, byTheorem 9.3,(a)isnotexact. By(b),(c),and(10.65), (6) Fa)=%"‘y” =-1. Therefore, by(d)and(10.66), (e) h(x)=eI‘“"" =e“" isanintegrating factor of(a).(Verify it.) 2.hisafunction only ofy,i.e.,h=h(y). Inthiscaseweobtain from (10.62), (10.67) ho)5’;P(z,y)+P(z,y)%=ho)51-cot). which wecanwrite as 8 6 (10.6s) M5= PW) dy. Ifthecoefficient ofdyin(10.68) alsosimpliﬁes toafimction only ofy— 86SPECIAL Trrns orFmsr Osman EQUATIONS Chapter 2 letuscallitG(y)—so that §can~52,-P(z,y) P(m() ’(10.69) G(y) = then, by(10.68) and(10.69), log[h(y)] =fG(y) dy.Hence, (107) My)=eloondu, isanintegrating factor of(10.6). Example 10.701. First show thatthedifferential equation (a) xydx+ (1+a:2) dy=0 isnotexact andthen ﬁndanintegrating factor. Solution. Comparing (a)with (10.6), weseethat (b) PM/) =11/. Q(¢,y) =1+x’- Therefore (c) ' =as, ( =2:0. Hence, byTheorem 9.3,(a)isnotexact. By(b),(c),and(10.69), 2 1(<1) cc)="—,,,,—“=5- Therefore, by(d)and(10.7), (B) h(y)=6%“ =¢'°"’=y isanintegrating factor of(a).(Verify it.) 3.hisafunction ofxy,i.e., h=h(u), where u=xy. Inthis case weobtain from (10.62), (10.71) ho)5";Pct)+P(z,y) hm]=ho),,3,;cot)+cot)E,hon]- . 6 8 6 dSmce u=xy,a—Z=x. Therefore, tﬁh(u) =h'(u) ai;=:1:Eh(u). Similarly, g=yand58;h(u)=h'(u)%=yi1.6.).Substituting in Lesson 10C F1NmNG ANINTEGRATING FAc-roa 87 (10.71) thesevalues of%h(u)and%h(u)andsimplifying theresult, weobtain 1P(z,y)-1oat) 0°”) dig]=it/Q(t,t) -:i><t,y> "“" Ifthecoeﬂicient ofduin(10.72) alsosimpliﬁes toafunction ofu==zy- letuscallitF(u) EF(:ty)-——so that §’-Putt)—5%Q(r,y)(10.13) F(u)= then, by(10.72) and(10.73), log[h(u)] =fF(u) du.Hence, (10.74) h(u)=t”"“>‘“, isanintegrating factor of(10.6), where u=xy. Example 10.741. First show thatthedifferential equation (a) (z/’+wy’+y)dw+(w3+w’y+w)dy=0 isnotexact andthen ﬁndanintegrating factor. Solution. Comparing (a)with (10.6), weseethat (b) P0021) =2/3+wy”+y.Q(w.y) =w“+$221+x. Therefore <c)@=3y’+2¢z/+1, ‘E5’-’l=at’+2@~t+1. y 6x Hence, byTheorem 9.3,(a)isnotexact. By(b),(c),and(10.73), _3y2-I-2xy+1—3:c’—2:cy—1 _-sot”-y’) (d)H“)—w“u+w’u”+wy—wy“—¢”y’—wz/_ 1"y(w’—y”) __§._ u Therefore, by(d)and(10.74), (e) h(u) ___=e.f."% J“=__e—8 logu =u-3 =(xy)—3 isanintegrating factor of(a).(Verify it.) 88S1=Ec1A1. Tvrss orFmsr Onnsa EQUATIONS Chapter 2 4.hisafunction ofx/y, i.e., h=h(u), where u=x/y. Here u=z/ysothatélu/6y =-—x/ya and6u/8:c=1/y. Therefore 8 8u :1:d(10.75) 56h(u) --h'(u) E-—Fd—uh(u), 8 Bu 1dgh(u) -h'(u) 5;--1;Eh(u). Substituting (10.75) in(10.71) andsimplifying theresult, weobtain y.[@P<t,t> _0000)] (10.76) d[h(“)] = 6” 8”‘du.h(u) wP(w,y) +0Q(w,z/) Ifthecoefficient ofduin(10.76) alsosimpliﬁes toafunction ofu=at/y— letuscallitG(u) EG(a:/y)—so that y.[6P(w.y) _0000)] (‘°'") GM= then, by(10.76) and(10.77), log[h(u)] =_[G(u) du.Hence, (10.78) h(u)=J"<">"'", isanintegrating factor of(10.6), where u=z/y. Example 10.781. First show that thedifferential equation (a) 3yd:t——:tdy=0 isnotexact andthen ﬁndanintegrating factor. Solution. Comparing (a)with (10.6), weseethat (b) P(1=,y) =31/, Q(w.y) =~1- Therefore Hence, byTheorem 9.3,(a)isnotexact. By(b),(c),and(10.77), 2 Therefore, by(d)and (10.78), 2,, 2 (e) h(u) =ef"d =e1°"“I=uz=% isanintegrating factor of(a). (Verify it.) Lesson 10C FINDING ANINTEGRATING FAcroR 89 5.hisafunction ofy/x, i.e., h=h(u), where u=y/x. Inthis case, weleave ittoyouasanexercise—follow themethod used in4-—to show that anintegrating factor of(10.6) is (10.79) h(u)=t"<<">'1'", where u=y/2:and ,.[a0<x.t) _aP(t,y)]8 0_mg . (ms) K“)‘xP(xi/)+700.0) Example 10.81. First show thatthedifferential equation (a) ydx—-3:cdy=0 isnotexact andthen ﬁndanintegrating factor. Solution. Comparing (a)with (10.6), weseethat (b) P($,y) =y. Q(Iv,y) =~31- Therefore 6P , 0 ,(6) --xi’) =1,Qggl =-3. Hence, byTheorem 9.3,(a)isnotexact. By(b),(c),and(10.8), _t*[-3-1]_gu__g_(<1) K<u)-———,y_3,,y -y-u Therefore, by(d)and(10.79), Z,, 2 (e) hfu) =ef"d =el°‘"2=u2=Z; isanintegrating factor of(a). (Verify it.) Ifadifferential equation canbeputinthespecial form (10.82) y(Aa:"yq +B:t'y') dx+x(Cx"yq +Da:'y") dy=0, where A,B,C,Dareconstants, then itcanbeshown that anintegrating factor of(10.82) hastheform x“y" where aand baresuitably chosen constants. Weillustrate, byanexample, themethod ofﬁnding aninte- grating factor of(10.82). Example 10.83. First show that thedifferential equation (a) y(2x2y3 +3)da:+x(1:2y3 ——1)dy=0 isnotexact andthen ﬁndanintegrating factor. 90SPEc1A1. TYPES orFmsr Onnsa EqUAT10Ns Chapter 2 Solution. Comparing (a)with (10.6), weseethat (b) P(w,y) =210221‘ +30. Q(m1) =way“-—1- Therefore P 8(C) Q__5(. Z 8x22/3 + 3’ L Z 3x2?/3 1 1- Hence, byTheorem 9.3,(a)isnotexact. Since (a)hastheform of(10.82), anintegrating factor willhave theform :c"y". Multiplying (a)byx°y", we have (d) (2xu+2yb+4 +3xu,/0+1) dz+(xa+3yb+3 __xa+1yb) dy=0_ ByTheorem 9.3,(d)willbeexact if (e)2(b+4)w“+2y"+" +(b+1)3w“y°=(<1+3)w"+’y"+“ —(a+1)w“y"- Multiplying (e)by1/(a:"y"), weobtain (f) (20+SW0“+30+3=(<1+3)r/0'2/3—(0+1)- Equation (f)willbeanequality ifwechoose aandbsothat (g) 2b+8=a+3, 3b+3=—a-—1. Solving (g)foraandb,weﬁnd (h) <1=1}) b=-2" Hence, 2:7/5y'9/5 isanintegrating factor of(a). (Verify it.) EXERCISE 10 Test each ofthefollowing equations 1-19 forexactness. Ifitisnotexact, trytoﬁndanintegrating factor. (Integrating factors fornonexact equa- tions aregiven intheanswers.) After theequation ismade exact, solve by looking forintegrable combinations. Ifyoucannot ﬁndany, usemethod of Lesson 9. 1.(2zy —|—x2)dz—|—($2+1/2)dy=0. 2.(:02—|—ycosz)dz—|—(ya+sin2:)dy=0. 3.(x2+y2+:c) d:t—|—a:ydy =0. 4.(:2:—2a:y+e") dx+ (y—2:2+xe'/)dy =0. 5.(e’siny-1-e"")dx—(xe"" —e‘cosy)dy=0. 6.(12-——y2—-y)dz—(x2-—y2—:|:)dy =0. 7.(x41/2 —y)dz+(2:21/4 —-:0)dy=0. 8.y(2a:+y3)dz—-:z:(2:c —-y3)dy=0. 1/-w — 9At +§i/.2: ,1_|_ d -0 .rcanxy l+z2y2 2: 1_|_$2y2 y-. Lesson 11A DEFINITION orAFmsr Onnan LINEAn EoUATIoN 91 10.e‘(x+ 1)da:+ (1/e"-:ce')dy =0. 1 2-11.ﬁg-—dx+—z%§dy =0. 12.(ya-—-3:011—2x"')dx+ (my—2:2)dy=0. 13.y(y+2:v+1)dx—a:(2y+a:-- 1)dy =0. 14.y(2:r—-y—1)da:+:r(2y——a:-- 1)dy =0. 15.(y’—l— 12121/)dx+ (2:01/+4:v3)dy =0. 16.3(y+x)’d:c+a:(3y+ 22>)dy=0. 17.1/dx— (1/2+:c2+x)dy =0. 18.2:ryda:+(x"’+y"’+a)dy =0. 19.(2:cy—I—x2+b)d:r+(y2+a:2+a)dy =0. ANSWERS 10 1.3:021; +2:3+ya=c. 2.41:3-1-31/4+121/sin2:=c. 3.3:04+41:3+62:2;/2 =c;integrating factor :02. 4-.2:2—2:021] +yz+2:re" =c. 5.e'siny+:te_" =c. 6.:1:—-31+logx/at +y—-log\/:1: —-y=c;integrating factor 1/(22 —-1/2). 7.:t4y+avg/4-czy=-3;integrating factor 1/2:211/2. 8.2:2+mg/3=c112;integrating factor 1/ya. 9.:0Arctanxy—log(1+2:21/2) =c. 10.2:ce*_" +112=c;integrating factor e"". 211.:c2—|-—5+4log|y| =c. 12.ray’ —2x3y —x4=c;integrating factor 22:. 13.(y—-at+1)3=cry;integrating factor (xy)‘4/3. 14.(2+y+1)3=cry;integrating factor :t‘1y'1(:t +1/+1)"1. 15.42:31] +:01/2=c. 16.6:z:2y2 +8:031] +32:4=c;integrating factor zt. 17.y+Arctan5=c;integrating factor 1/($2+yz). 18.yi‘+31:21;+3ay=c. 19.ya+2:3+3(:t2y +ay+bx)=c. LESSON ll. The Linear Differential Equation ofthe First Order. Bernoulli Equation. LESSON 11A. Deﬁnition ofaLinear Differential Equation ofthe First Order. The important differential equation which weareabout todiscuss hasmany theoretical andpractical applications. Itisaspecial type ofﬁrst order differential equation inwhich both thedependent vari- ableanditsderivative areoftheﬁrstdegree. Anequation ofthistype is called alinear differential equation oftheﬁrst order. Hence: 92SPECIAL TYPES orFIRST ORDER EqUATIoNs Chapter 2 Deﬁnition 11.1. Alinear differential equation oftheﬁrst order isonewhich canbewritten as (11.11) %+Pew=00), where P(z) andQ(z) arecontinuous functions ofzover theintervals for which solutions aresought. (Note that yand itsderivative both have exponent one.) Forthisdifferential equation weshall prove inLesson 11Bthat the 1-parameter family ofsolutions weshall obtain isactually atrue general solution aswedeﬁned theterm in4.7. Every particular solution of(11.11) willbeobtainable from this1-parameter family ofsolutions. LESSON 11B. Method ofSolution ofaLinear Differential Equa- tion oftheFirst Order. Asmentioned inLesson 10B, anintegrating factor isknown forthistype ofequation (11.11). Itis (11.12) eI”""”, where theconstant ofintegration istaken tobezero. Themotivation andmeans bywhich thisrather terrifying looking inte- grating factor wasobtained have been deferred toLesson 11C. Inthe meantime letusverify that (11.12) isindeed anintegrating factor for (11.11). Multiplying (11.11) by(11.12) andchanging theorder inwhich theterms appear, weobtain (11.13) [P(t)JP<""y -Q(u)t"’<'>“’] at+t-"’<'>"” at=0. Theterms P(x)efP"”"‘ andQ(z)eI")”" arefunctions of2:.Hence the partial derivative with respect toyofthecoefficient ofdzin(11.13) is (11.14) P(u)u"’<’>“. By(9.15) thepartial derivative with respect tozofthecoefficient ofdy din(11.13) is[remember ie“ =e“—u ;hereu=/P(z) dz]dz dz (11.16) P(t)JP<’>‘=. Since thefunctions in(11.14) and (11.15) arethesame, byTheorem 9.3, (11.13) isexact. Hence byDeﬁnition 10.5, ef-P(‘)‘i" isanintegrating factor. Letusnow rewrite (11.13) intheform (11.16) t"’<=>"’= at+P(u)JP<’*'"y at=Q(x)JP<=>‘= dz. Since weknow (11.16) isexact, wecansolve iteither bythemethod of Lesson 9B,orbytrying todiscover anintegrable combination. Ashrewd Lesson 11B SoLUTIoN orAFIRST ORDER LINEAR EQUATION 93 observer now discerns that theleftsideof(11.16) isindeed (11.17) d(JP<'>'"y). [Besure toverify thisstatement. Remember d(uy) =udy-1-ydu.Here u=efP(")‘l‘.] Hence (11.16) becomes (11.18) d(e"’<=>‘=y) =elP<=>'"Q(u) at whose solution is (11.10) JP<*>"=y =ftlP<'>"‘Q(t) at+t. Proof that (11.19) isatrue general solution of(11.11). The argu- ment proceeds asfollows. Since eIP(")""’ 9'60,[forproof seeLesson 18, (18.86)], (11.11) holds ifandonly if(11.13) holds; (11.13) holds ifandonly if(11.16) holds; (11.16) holds ifandonly if(11.18) holds. Finally (11.18) holds ifandonly if(11.19) holds. Wehave thus demonstrated that (11.11) istrue ifandonly if(11.19) istrue. And since eIP(")""’ aé0,wecanin (11.19) divide bythisfactor toobtain a1-parameter family ofsolutions of(11.11) intheexplicit form (11191) y=e“IP")d‘/eIP("d’Q(z) at+tt-lP<=>‘“. Hence (11.11) holds ifand only if(ll.l91) holds. The “ifand only if” clause isequivalent tosaying that if(11.11) holds, then (11.191) holds, andif(11.191) holds, then (11.11) holds. This means that if(11.11) is true, then itssolutions are(11.191), andif(l1.191) istrue (i.e., true for each c),then (11.11) issatisﬁed. Example 11.2. Find thegeneral solution of (a) y’—-2zy=e1’. Solution. Bycomparing (a)with (11.11), weseethat theequation islinear, P(z) =—2z andQ(x) =e".By(111.12) anintegrating factor of(a)istherefore ef—2:r dx=6-H. Multiplication of(a)bye“”’gives (c) e_" dy—2e_"zy dz=dz, anequation which now corresponds to(11.16). Hence by(11.17), theleft sideof(c)should be(and is) (<1) d(¢"2v)- 94SrEcIAI. Trras orFIRST ORDER EoUATIoNs Chapter 2 Replacing theleftsideof(c)by(d),andintegrating theresulting expres- sion, weobtain forthegeneral solution of(a), (e) e_"2y =/6dz=x+c, which wecanwrite as (1) 1/=¢"(1+6)- Comment 11.21. After theintegrating factor e"" hasbeen deter- mined, wecould by(11.19) goimmediately from (a)to(e).For(11.19) says, “ytimes theintegrating factor =I(integrating factor) Q(z)dz+c.” Example 11.3. Find theparticular solution of dy sinz(a) z%+3y=?T) z¢0, forwhich y(1r/2) =1. Solution. Since :1:#50,wemaydivide (a)byitandobtain .13 '(6) %+;y=§‘;‘,‘,§. “-0. Comparing (b)with (11.11) weseethat (b)islinear, P(.'c) =3/:1:and Q(z) =sin:1:/0:3. Hence by(11.12) anintegrating factor is 1.2dz 3103 x logma 3 (c) e =e =e =z. (Bytaking thelogarithm ofboth sides, youcanshow that 01°" =u.) Multiplication of(b)bytheintegrating factor 2:3willgive, by(11.19) [seeComment (11.21)], (d) z3y=/sinzdz+c=—cosz+c. Toﬁnd theparticular solution forwhich :2:=1r/2, y=1,wesubstitute these values in(d)andobtain a (e) 61 4% - Replacing (e)in(d),therequired solution is a (f) yzs+cosz = LESSON 11C. Determination ofthe Integrating Factor efpmd‘. Weoutline below amethod bywhich theintegrating factor for(11.11) canbedetermined. First werewrite (11.11) as (11-4) [P(=v)1/ —Q(1)ldw +dy=0. Lesson 11D BERNGULLI EQUATION 95 andthen multiply itbyu(z). There results (11.41) [u(z)P(z)y —u(:t)Q(z)] dz+u(z) dy=0. Wenow askourselves thisquestion. What must u(z) look likeifitisto beanintegrating factor for(11.11)? Weknow byDeﬁnition 10.5that it willbeanintegrating factor ifitmakes (11.41) exact. And byTheorem 9.3,weknow that (11.41) willbeexact if (11-42) 51u(»)P(1)t -1000(1)] =511(1). Taking these derivatives [observe that u(z), P(x), andQ(z) arefunctions onlyofz],weobtain (11.43) u(z)P(z) =50(1), which wecanwrite as (11.44) P(z)at=$- Integration of(11.44), with theconstant ofintegration taken tobezero, _ dugives remember 7;=logu (11.45) logu(z) =/P(z) dz, which isequivalent to (11.46) tot)=e[*’<*>". fleincf) ifu(z) hasthevalue e'[P(°’)""‘, itwillbeanintegrating factor for 1.1. LESSON 11D. Bernoulli Equation. Aspecial type ofﬁrst order dif- ferential equation, named fortheSwiss mathematician James Bernoulli (1654—1705), andsolvable bythemethods ofthislesson isthefollowing. (11.5) 5+P001=0(1))». Ifn=1,(11.5) canbewritten asdy/dz =[Q(z) -—P(z)]y, anequation inwhich thevariables areseparable andtherefore solvable bythemethod ofLesson 6C.Hence weassume n¢1.Note alsothatthepresence of y“prevents theequation from being linear. Ifwemultiply (11.5) by (11.51) (1——n)y"', weobtain (11-52)(1—or"5+(1—n)P(:»)(t‘-") =(1—100(1). 96SPEcIAL TYPES orFIRST ORDER EqUATIoNs Chapter 2 . .d,_,, .Theﬁrst term in(11.52) isZ:(y ).Hence (11.52) canbewritten as d —n -11 one 5@)+0—mwM)=u—mw Ifwenow think of1/1-" asthedependent variable instead oftheusual y [orifyouprefer youcanreplace y1_" byanewvariable usothat (11.53) d ..becomes d—;+(1—n)P(z)u =(1—n)Q(z)], then, byDeﬁnition 11.1, (11.53) islinear in3/1-" (oru).Itcantherefore besolved bythemethod ofLesson 11B. Example 11.54. Solve (11) v'+wv=$»1/¢0- Solution. Comparing (a)with (11.5), weseethat (a)isaBernoulli equation with n=-3. Hence, by(11.51), wemust multiply (a)by41/3. There results (b) 4;/3y’ +4zy‘ =4z. Because ofthesentence after (11.52), weknow thattheﬁrstterm of(b) should be(and is) d(<1) ,7;1/4- Hence wecanwrite (b)as (0 5wHam=a anequation which isnow linear inthevariable y‘.Anintegrating factor for(d)istherefore, by(11.12), (6) e_[4z dz=e21,‘ After multiplying (d)byeh’, wecantake advantage of(11.19) towrite immediately (f) eh’;/4 =4/zen’ dz=e2‘2+c. Therefore, (8) y"==1+cf” istherequired solution. Lesson ll—Exercise 97 EXERCISE ll Find thegeneral solution ofeach ofthefollowing. 1.zy'—[—y=1:3. 2.y’—[-ay=b. 3.zy'+ y=yzlogz. 4.g—[-2yz=e_”a. Hint. Consider zasthedependent variable. 5.g=(r—|—e_')tan0. dy 2zy _ 6'dz :02+1_1' 7-v’+:11=1:11“- a.(1-13)Q-2(1+4);,=11"’.dz 9.tan0g% —r=tanz9. 10.L%+ Ri=Esinkt.(This istheequation ofasimple electric circuit containing aninductor, aresistor, andanapplied electromotive force. For themeaning ofthese terms andforamore complete discussion ofelectric circuits, seeLessons 30and33C.) ll.y’-1-2y=3e"2‘. 15.y’—|—ycosz =ﬁsin 2z. 12.y’+2y=§e'2‘. 16..ry'—[—y=zsin z. 13.y’-1-2y=sinz. 17.zy’——y=2:2sinz. 14.y’—[-ycosz=e2’. 18.zy’-1-zyz—y=0. 19.zy'—y(2ylogz —1)=0. 20.z2(z —1)y'—yz—z(z-—2)y=0. Find aparticular solution ofeach ofthedifferential equations 21-24. 21.y’-—y=e‘,y(0) =1. 2 22.y'—|—5y =y?y(-——1) =1. 23.2coszdy =(ysinz —3/3)dz, y(0) =1. 24.(z—siny)dy-1-tany dz=0,y(1) =1r/6. 25.Thedifferential equation (11-6) 11'=f0(1>) +fi(1)y +f2(1)y2, f2(I) 7*0, iscalled aRiccati equation. Ify1(z) isaparticular solution ofthisequa- tion, show thatthesubstitution 1 1 (11-61) v=1/1+5» 1/’=vi’—514’. willtransform theequation intotheﬁrstorder linear equation (11-62) 14'+lf1(w) +2f2(1)y1lu =—f2(I)- Hint. Since y1isaparticular solution ofthegiven equation, y1'= f0($) -1-fi(=v)1/1 +f2(1)yi2- 98SPECIAL TYPES orFIRST ORDER EoUA'rioNs Chapter 2 With theaidofproblem 25above, ﬁndthegeneral solution ofeach of thefollowing Riccati equations. 2 1 26-0’=r3+;v -51/2.1/1(1)=-12- 27.y'=2tanzsecz ——yzsinz,y1(z) =secz. 1 y 1 28-v’=;-5-11/2, 1/1(w)=;- 2 _ Z_Z/_ _29.y’-1+9: $21 y1(z) —z. ANSWERS ll 1.4zy 7,z4—|— c. 4.z=e""(y+ c). 2.y=It-—|—ce"". 5.2r=csec0 —e_'(tan0+ 1). 3.ylogz+y+czy=1. 6.y=(Arctanz+c)(z2-I-1). 7.-12;=ce2'—[—z—[-l-y 2 8y-3/2 =_ 3 +¢(1—=v)2 _ ' 4(1+z—[—z2) 1+:t+z2 9.r=sin0[log (sec0—|—tan9)]+csin0. , _ E(Rsin kt—-kLcoskt) 1°-*=6°R'IL+ '11.y=3ze"2‘+ ce‘2". 12.y=§ze-2‘+ ce"2=. 13.y=§(2sinz—cosz)-[-ce'2‘. 14- y=e—sinz(c+ -/‘e21-[-lihldx) _ 2 l5.y =sinz—1—|—ce"'i“. 20.1; =-Ti—i-_ ("-1)¢-['1 l6.y=§E1;£—cosz+§- 2l.y=e'(z+1). 17.y=z(c-—cosz). 22.y=1. 2 l8.y=a:T_%) y=0. 23.secz=y2(tanz—[-1). 19.1—2y(1—[—logz) =czy. 24-.Szsiny =4sin21/+3. 26'+3+2. 0-1 —-1’+--2“2‘z2-'u z _z’ y_ e=’+c . 1 3cos2z27.u'—2utanz=sinz, y=—-—-—[———i—3—-cosz c-—cos z HI-‘RC0 Ho-11-1 M[OH28.’!/—'—1l=l, y=;+;£-;-- 29-’|l."'--’ll=-—2'i 1/=1-[-Q-—_Tl" Lesson 12A EouA'rIoNs PEium'rING ACHOICE orMETHOD 99 LESSON 12. Miscellaneous Methods of‘Solving aFirst Order Differential Equation. LESSON 12A. Equations Permitting aChoice ofMethod. Ifa differential equation isselected atrandom, itmay besolvable bymore than onemethod. Theoneselected willdepend ultimately onyour in- genuity indetermining which willmost readily leadtoasolution. The following examples willillustrate thispoint. Example 12.1. Solve (a) zdy—ydz=y2dz. Solution. Aswasshown inExample 10.51, multiplication by1/"3, yas0,willmake (a)exact. Itwillthen besolvable bythemethod of Lesson 9B,orbymeans of(10.30). Finally ifwedivide byz,(z950), theequation becomes2 (11) 1/'—g=y;»==¢0. which onerecognizes asaBernoulli equation. Hence itissolvable by themethod ofLesson 11D.Ofthethree choices available tosolve (a), youwillﬁndthatuseof(10.30) istheeasiest. Example 12.11. Solve (11) (w’+2/’)dy+2111/dz=0- Solution. This equation isofthetype with homogeneous coefficients andistherefore solvable bythemethod ofLesson 7B.Itisalsoexact andtherefore solvable bythemethod ofLesson 9B.However, theeasiest method istorewrite (a)as (b) zzdy+2zydz+yady=0, andthenmake useof(10.21). Example 12.12. Solve (a) (3e3’y —2z)dz+ea”dy=0. Solution. The equation isexact andistherefore solvable bythe method ofLesson 9B.Ifwedivide (a)bye3’dz,weobtain d _(b) i-1-3y=2ze3’, anequation which islinear andhence solvable bythemethod ofLesson 100 SPEc1AI. TYPES orFIRsT ORDER EoUATIoNs Chapter 2 11B. If,however, (a)isrewritten as (c) 3e3‘y dz+es”dy—2zdz=0, itcanbesolved most easily bymaking useof(10.392). Example 12.13. Solve (a) zzdy—(zy+yx/z” +y”)dz=0. Solution. Theequation isofthetype with homogeneous coefficients and therefore issolvable bythemethod ofLesson 7B. However, the presence ofthecombination z(zdy-—ydz)leads onetotrytomake use of(10.3) or(10.31). Wetherefore divide (a)byz,(z#50)andrewrite ittoread (b) zdy—ydz=-fix/z2+y2dz, z¢0. Wethen divide (b)byyatoobtain d— d 1 1 2(0)2-li=—\/z2+y”dw=—‘[£,+1dz, 22¢0,y950.ll 133/ 17 1/ By(10.3) theleftsideof(c)is-—d(z/y). Hence (c)canbewritten as __ﬂ%L_=@, @) xﬁiﬁﬁﬁ xz¢Qy¢0 Integration of(d)now gives (e)-161[5+~/i1+(1/t)2[=loslrl+1ot|c|. :1-10.1-A0. which canbewritten as 0 -—iL——= ,¢0y¢0( z+\/z’+v” axx Thesolution of(a)istherefore (g) y==cz(z+\/z2+y2), z¢0,1/#0. Comment 12.14. The function y=0which wehad toexclude to obtain (g)also satisﬁes (a). Moreover every member ofthefamily of solutions (g)goes through thepoint (0,0). Observe from (a),however, that dy/dz isundeﬁned when z=0.The point (0,0) isoneofthose singular points wediscussed inLesson 5B. Every member ofthefamily (g)liesonit,butnomember goes through any other point ontheline z=0.This solution, therefore, actually should have been written with Lesson 12B SoI.U'rioN BYSUBSTITUTION ANDOTHER MEANs 101 twoparameters instead ofone,namely y=c1z(z +\/z2 +ya), z§0 y=c2z(z -1-\/z2 +yz), zg,0. LESSON 12B. Solution bySubstitution and Other Means. A ﬁrst order differential equation need notcome under anyoftheheadings mentioned heretofore. This factshould notbetoosurprising. Youyour- selfcould easily write aﬁrstorder differential equation which would not ﬁtanyofthetypes thusfardiscussed. Itispossible insome cases tosolve adifferential equation bymeans ofashrewd substitution, orbydiscovering anintegrating factor, orbysome other ingenious method. Wegivebelow anumber ofexamples which donot, asthey stand, lend themselves to standardization. You should keep inmind that these examples have been specially designed toyield asolution interms ofelementary functions. Itiseasily conceivable, ifadifferential equation were selected atrandom, that onecould spend hours anddays using every known method anddevice atone’s disposal and stillfailtoﬁnd anexplicit orimplicit solution in terms oftheelementary functions. More than likely nosuch solution exists. Example 12.2. Solve (11) (y—v’—w’)dw—wdy=0- The equation asitstands cannot besolved byanyofthemethods out- lined thus far. However, thepresence ofthecombination ydz—zdy leads onetotrytomake useof(10.3) or(10.31). Rearranging terms and dividing byzzgives _ 2 (6) =[1+(g)]dz, z;='5O. By(10.31), theleftsideof(b)is—d(y/z). Hence (b)canbewritten as Q (c) —--dl2=dz.1/ 1+(1) (d) Arctan5=-—(z+c)orZ-= —tan (z-1-c),z¢0.Integration of(c)gives Hence thesolution of(a)is (e) y= —ztan (z+c), z750. NOTE. Inregard tothelinez=0andthepoint (0,0), werefer youto ourComment 12.14. 102 Sracmn TYPES orFmsr Onnsn Eqvxrrons Chapter 2 Example 12.21. Solve (a) (2cosy)y’+siny=2:2cscy,y960. Solution. Theequation asitstands cannot besolved byanyofthe methods outlined thusfar.Ifwemultiply itbysiny,weobtain (b) (2sinycosy)y’+sin2y=1:2,y¢0. dTheﬁrstterm isequal to8;(sin2y).Wetherefore canwrite (b)as <c> 5'5sin’2l)+($iI12 1/)=Z’,y¢0, anequation which isnowlinear inthevariable sin’y.Theintegrating factor by(11.12) isfound tobee”.Hence by(11.19) (d) e‘sin’y=/xze” dx=e’(a:2 —2:1:+2)+c,yaé0. Thesolution of(a)istherefore (e) sin2y=(222——2:0+2)+cc”, y#0. Example 12.22. Solve (a) y’+2x=2(:c2+y——1)2'3. Solution. Theequation asitstands cannot besolved byanyofthe methods outlined thusfar.Apowerful anduseful method frequently used bymathematicians tointegrate functions isthat ofsubstitution. An attempt ismade tosimplify theintegrand byusing anewvariable to represent afunction ofthegiven variable. Insome cases thismethod of substitution canalsobeused proﬁtably toﬁndsolutions ofdifferential equations. Forthisexample, wetrythesubstitution (b) u=x’+y—1, andhope itwillyield adifferential equation inuwhich wecansolve in terms ofelementary functions. Differentiating (b)weobtain d d d d(c) ‘%=2a:+ag-; El;-=d-%—2x. Wenowsubstitute (b)and(c)in(a).There results (<1) %=213'“, “-2/wt =24¢,u950. Lesson 12—Exercise 103 Byintegration of(d)weobtain (e) 3u‘/3 =22:+c,u950. Replacing ubyitsvalue in(b)andcubing, wehave (f) =v’+z/—1=%(2w+v)3. w’+u-1¢0- Thesolution of(a)is.therefore 3 (s) 1/=1-—w’+(l”-”—;%’)—, 1’+y—~1#0- NOTE. Thefunction y=1-—2:2which hadtobeexcluded inobtain- ing(g)alsosatisﬁes (a).Itisaparticular solution of(a)notobtainable from thefamily (g). EXERCISE 12 Solve each ofthefollowing differential equations. 1.2:cy:—: +(1—|— 2:);/2 =0'. 2.cosy 112+ siny =1:2.Hint. -1-(sing) =cosydl-dz dz: dz 8.(:c+1)dy—(y+1)dx =(a:+1)\/g+ld:c. Letu =1/+1. 4-.e'(y'—I- 1)=e’.Hint. $1?e"=e";/. 5.1/siny—|-sinxcosy =sinx. Hint. dimcosy =—-(sing);/. 6.(1-1,)’:-Z =4.Letu=z—y. 7.:c%—y =\/:02-l-yz. 8.(3:c—|—21/+1)dy+(4:c+ 3y+ 2)da: =0. 9.(:02—1/2)dy=2:01/da:. 10.1/dz+(1+11%") dy=0.Lety=e“u. ll.(1211—I-yz)dz+2:3dy=0. 12.(y2e’"’ +41:3)dz+(2:cye‘"’ —3112)dy=0. 13.y’=(x2+2y —1)2/3 —2:.Letu =:c2+2y ~—1. 14.2:gg+ y=:c2(1+e”)y2. MISCELLANEOUS PROBLEMS Inthefollowing setofproblems, classify each differential equation by type before attempting toﬁnda1-parameter family ofsolutions. Some may fallintomore than onetype; some may notfallintoany. Inthe 104 SPECIAL TYPES orFIRST Onnaa EQUATIONS Chapter 2 answers, youwillﬁndhints astomethods ofsolving Always ﬁrst tryto solve theequation before looking atthese hints Also keep inmind that there maybeother methods inaddition totheoneswehave given 15.(2y—mylog1:)dx—2::loga: dy=0. 16.y’+ay=Ice“. 17.1/=(a:+ 1/)2. 18.y’+8z3y3 —|—2zy= 19.(mg/\/2:2 ——y2+x)y’ =1/——a:2\/$2 —yz. 20.y’—|—ay=bsin kx. 21.xy’—y2+ 1=0. 22.(g2+asin 2:)%:=cos2:. 23.my’=we"/" —|—2:+y. 24.y+ycosx =e"“‘ ". 25.my’—y(log xy—1)=0. 26.2:31;’—yz—-2:21;= 27.zy’+ay+b:c" =0,2:>0. 28.:cy'—:csiny-—y =0.2: 29.(my—:c2)y’ +yz—-32y—22:2=0. 30.(6a:y+x2+3)y'+3y2+2zy—|—2::=0. 31.xzy’+y2+xy+2:2=0. 32.(:2——1)y’+22:11—cosz =0. 33.(xzy——l)y'+ :cy2—1=0 34- 2.(2:—1)y’-I-my -3:cy2 =0. 35.(:02——1)y'—2zylogy = (12+;u’+1)y’+21y+w2+8 =0- 36.0. 37.y'cosa:—|—y+ (1+sin:|:) cosa: =0. 38. 39. 40-($2-—y);/’—411/41.zyy'+2:2+y2= 42.2a:yy' +32:2—yz 43.0. 0.(2zy+4:c3)y’ +yz+12221; =0. ($2-:1/)1/’+x= (211/3-r“)y'+Zway—1/‘=0-44-(wy—1)2wy’+(121/2+ 1)u=0-45-(12+1/2)?/’+2=v(2==+y)=0-46.3zy2y' +y3—2:2:= 47.23/3y’ +2:1/2—2:3= 48. 50.y’-e‘_”+e‘= 1.2y2a: = 2.siny =e‘+ce"‘. 4.y=log(§e'+cc“ x2——2z—|—2+ce". 5.cosy =1+ce‘°°"(21y3+I11+$2):/’—$21+1/”=0-49.(2y3-l"U)?!’--22:3—2:= 0.0. ANSWERS 12 3.\/y+1=z—|—1+c\/:c+1. 26-y—log :5:-‘T2 +C. 7.y—|—\/2:2 +y?=cxz, orequivalently, y=C+\/:02 +2x—y— 8.3xy+y2+y+2:z:2+2:c =c. Miscellaneous Pr-oblems—Answers 105 9.2:2+y2+cy=0. 12.e""2+2:4—y3=c. e'2= =2y2(log |y|—c). 133(:2:2+2y—11/3=22+ c. 10. . ) ll.32:2+y=c2:3y. 14.2:y(2:+e‘+c)+1=0. 15. 16. 17 18. 19. 20. 21 22 23 24 25 26 27 28 29 30. 31. 32. 33 36 37 38 39 40 41.34. 35ANSWERS FOR MISCELLANEOUS PROBLEMS Separable. x+2log |y|—210g (log|x|)=c. Linear. k -asy=;ﬁeb'—|-ce ,a+b#0, y=lm'“‘+ ti", a+1»=0. Letu=2:+y.Resulting equation isseparable. 2:+y=tan(2:+c). Bernoulli. y-2=ce2” —42:2—2. Lety=um.Resulting equation canbemade exact. 2:sin[c—'}(2:2+1/2)], 1>0 rSinlv+H12+I/2)], Z<0-y= y% . b . _..,Linear. y=m (asmkw-—lccos kw)—|—ce . Separable. y=(1—c2:2)/(1+ c2:2). Letybetheindependent variable. Bernoulli insin2:;seeproblem 2above. 2 sin2:=ce°'— L-|—Zy-+Z-a a2 a3 Homogeneous. (1-c2:)e"/2 =c2:. Linear. y=(2:—|—c)e"“‘ “’. Letu=xy.Resulting equation isseparable. my=e". Bernoulli. Also 1/2:2y2 isanintegrating factor. 2:2—y=c2:y. Linear y=c:z:"“—L2:”, a#—na+n ' y=c:c"° ~—b2:_“ log2:,a=-—n. H U llomogeneous. cscE—cotE=c2:. Integrating factor 2:.2:2y2 -—22:3y —-:4=c. Exact. 32:y2 +2:2y+3y+2:2=c. Homogeneous. 2:=(y+2:)(log|2:|+c). Linear. Also exact. (2:2—1)y=sin2:+c. Exact. 2:2y2 -2(2:—|—U)=c. Bernoulli. y_1=3+c\/|2:2 ——1|. Separable. y=e°("-1). Exact. 2:3+y3+3(2:2y +y+32:)=c. _ Linear. (1+sinz)y=cos2:(sin2:-—2log1c%2l:-Z +c)- Integrable combinations. 4:c3y —|—2:y2=c. Bernoulli, yindependent variable. 22:2=2y-—1+ce'2". Lety=ua:2. Resulting equation isseparable. (2:2+ y)2=cy. Homogeneous. 2:2(2:2 +2y2) =c. 106 SPECIAL Tvras orFIRST Onnsn EQUATIONS Chapter 2 42.Bernoulli. y2=ca:—32:2. 43.Integrating factor 1/2:2y2. 2:3+ya=cxy. 4-4.Letu=2:y.Resulting equation isseparable. y2=cel’"“‘1/"ll. 45.Homogeneous. Also exact. y3+42:3+32:2y =c. 46.Bernoulli. 2:y3=2:2+c. 47.Homogeneous. Also 2:2+y2isanintegrating factor. ($2+1/2)2(2z/2 —12)=c- 43.Integrating factor 1/2:y2. ya+ylog|2:y|—2:=cy. 49.Exact. Also separable. y4—|—y2=2:4+2:2—|—c. 50.Letu=e".Resulting equation islinear. e"=1+ce"‘. Chapter 3 Problems Leading toDifferential Equations oftheFirst Order LESSON 13. Geometric Problems. Weareatlastready tostudy awide variety ofproblems which leadto diﬂerential equations oftheﬁrst order. Weconsider ﬁrst certain geo- metric problems inwhich weseek theequation ofacurve whose deriva- tivey’hascertain preassigned properties. Example 13.1. Find thefamily ofcurves which hastheproperty that thesegment ofatangent linedrawn between apoint oftangency andthe yaxisisbisected bythe2:axis. Y , -P 1) y (Iy (0,0) A\ ,0) X Q (0!“y) Figure 13.11 Solution (Fig. 13.11). LetP(z,y) beapoint onacurve oftherequired family andletPQbeasegment ofthetangent linedrawn atthispoint. Byhypothesis, PQisbisected bythe2:axis. Hence thecoordinates ofthe point Qare(O,—y). [The mid-point formula fortwopoints (x1,y1) and (x2,y2) is(iii, %—-yl) The equation ofthelinePQis,there- 107 l08 Pnonnams LEADING T0Fmsr ORDER EQUATIONS Chapter 3 fore, given by Z2__Q.(a) 2:_yl—dx ByLesson 6C,thesolution of(a)is (b) y=c:c2, a:¢0,y;-60. Wehave thusshown thatifthere exists afamily ofcurves which meets therequirements ofourproblem, itmust satisfy (a)andhence must satisfy (b). Conversely, wemust show' that ifafamily ofcurves satisﬁes (b),it willmeet therequirements ofourproblem. LetP(:c0,yo) beapoint ona curve of(b).Then by(b) (C) l/0=C5502; l/[(10-llo) =26170- Theequation ofthelineatP(:co,y0) with slope 2c2:0 is (<1) 1/-yo=2¢$o($ —$0)- When 2:=0,weobtain from (d) (e) y=1/o—201:0’. which istheycoordinate oftheintersection oftheline(d)with they axis, (QinFig.13.11). Thecoordinates ofthemid-point ofPQaretherefore (fl (£2>yo_61°02) '2 Replacing yoin(f)byitsvalue in(c),weobtain forthecoordinates of themid-point ofPQ <g> (%»0)- Wehave thus proved that thetangent Y atanypoint P(2:o,y0) ofamember of P(I.y) thefamily (b),drawn totheyaxis is bisected bythe:0axis. y’ y Example 13.2 (Fig. 13.21). Find A thefamily ofcurves with theproperty that thearea oftheregion bounded by the2:axis, thetangent linedrawn ata Xpoint P(z,y) ofacurve ofthefamily O(o’°) Q(¢l.0) R(I,°) andtheprojection ofthetangent line onthe2:axishasaconstant value A.Figure 13.21 Solution. AtP(z,y) draw atan- gent toacurve oftherequired family. CallQ(a,0) thepoint ofinter- Lesson 13 GEOMETRIC Pnoatnus 109 section ofthislinewith the2:axis, R(2:,0) theintersection oftheprojec- tionofthetangent linewith the2:axis. Theequation ofthetangent lineis (a) Ti/_7=y’, 2:#a. Solving thisequation fora,weobtain (b) a=2:—-~‘€ F which represents thedistance OQ. Hence thedistance QRis s »~t-e-tTheregion whose areaisA,istherefore given bytheformula _1_v__ii.<d> A—21/(t)—Hence, 2 (8) y’=Q2;i whose solution, byLesson 6C,is (f) %=—§%+c ory= , 1:;-£2Ac. Wehave thus shown that ifthere exists afamily ofcurves which meets therequirements ofourproblems itmust satisfy (a)andhence must satisfy (f). Conversely, wemust show thatifafamily ofcurves satisﬁes (f),itwill meet therequirements ofourproblem. LetP(:c0,y0) beapoint ona curve of(f).Then by(f) () y=i I=i1.__.2 °2Ac-to ”<"'"°* (2.46-¢t.,)= The equation ofthelineatP(2:°,y0) with slope given in(g)is __2_*‘1__.y_f/0 = (:5_$0) __$0); When y=O,weobtain from (h), _ _ 2(i) x=2Ax0 yg(ﬁAc 2:0) ’ which isthe2:coordinate oftheintersection oftheline(h)with the2:axis 110 PROBLEMS Lemme 'roFmsr ORDER EQUATIONS Chapter 3 (QinFig.13.21). Thedistance QRistherefore G) I_2A2:0 —yo(2Ac —2:o)2=y0(2Ac —2:0)2_ ° 2A 2.4 Hence theareaoftheregion Ais <k> a»Substituting in(k)thevalue of(2Ac ——2:0)2asdetermined bytheﬁrst equation in(g),weobtain 1/0411 <1? 2”“ which simpliﬁes totheconstant A.Hence ourfamily (f)meets there- quirements oftheproblem. Example 13.3. Find thefamily ofcurves such thattheangle from a tangent toanormal atanypoint ofacurve ofthefamily isbisected bythera- diusvector atthatpoint. (Inproblems involving radius vectors, itisusually preferable tousepolar coordinates.) Y , Solution (Fig. 13.31). LetP(r,0) y P("2) bethepolar coordinates ofapoint on -5 A acm've oftherequired family. CallB T“l‘;g:“t theangle measured from theradius r vector rcounterclockwise tothetan- No,-ma11ine gentlineatP.Then byatheorem in 9 thecalculus (mo) X(a) tanI3=r%- Figure 13.31 Byhypothesis rbisects theangle be- tween thenormal andtangent lines. Hence B=45°or—45°. InFig. 13.31 itis—45°. Therefore (a)becomes (using B=45°) d0 dr_1——rE) 7-d0, whose solution is (c) logr =0+c’, r=e'+" =e'e°'=cc’. Weleave ittoyouasanexercise tosolve (a)when B=—45° andto prove theconverse, i.e.,that ifafamily ofcurves satisﬁes (c),then it meets therequirements ofourproblem. Lesson 13 GEOMETRIC Pnonnnms 111 Example 13.4. Find thefamily ofcurves with theproperty thatthe areaoftheregion bounded byacurve ofthefamily, the2:axis, thelines 2:=a,2:=2:isproportional tothelength ofthearcincluded between these twovertical lines. Solution (Fig. 13.41). Theformula forthearclength ofacurve be- tween thepoints A,Bwhose abscissas are2:=aandx=xis I __ 1 Q2 (a)s-[1 1+(dx) dx. Y B =f() Thearea oftheregion bounded by yx thelines, 2:=a,2:=2:,the2:axis, andthearcofthecurve y=f(z) A between these lines, is y I A=re)11.». O Byhypothesis Aisproportional tos; Fadxx=x X therefore, by(a)and(b), Figure 13,41 (o /frown =k/:,/1+(j,’—Z)’d2. where lc>0isaproportionality constant. Differentiation of(c)with respect tox,andreplacing f(z)byitsequivalent y,give[see(9.15)] <12 d2(<1)y=k,(1+(%). y2=k2[1+(a%)]» y>0,k>0, which simpliﬁes to 2—-I02 d d d (2) *\(y k2=dZ’ W22‘ Byintegration of(e),weobtain (f) i=log(21=1:V11’—k’)—log6. which canbewritten as (s) ¢¢""=yiV11’—Iv’. 1/=2lc(c2e'”‘ +k2e“"‘)- Weleave ittoyouasanexercise toprove theconverse, i.e.,ifafamily ofcurves satisﬁes (g),then itmeets therequirements ofourproblem. 112 Pnonnnus LEADING 'roFIRST Onnna EQUATIONS Chapter 3 EXERCISE 13 1.LetP(z,y) beapoint onthecurve y=f(z). AtPdraw atangent anda normal tothecurve. Theslope ofthecurve atPistherefore y’;theslope ofthenormal is-1/y’. Prove (seeFig.13.5) each ofthefollowing. (a)2:—y/y’isthe2:intercept ofthetangent line. (b)y—2:y’istheyintercept ofthetangent line. (c)x—|—yy’isthe2:intercept ofthenormal line. (d)y—|—2:/y’istheyintercept ofthenormal line. (e)|y/y'| isthelength ACoftheprojection ontheanaxis, ofthesegment of thetangent AP. Thelength AC’iscalled thesubtangent. (f)|yy'|isthelength CBoftheprojection onthe2:axisofthesegment of thenormal BP. Thelength CBiscalled thesubnormal. (g)Thelength ofthetangent segment AP=y‘ +ll-ll (h)Thelength ofthetangent segment DP=|2:\/1 +(y’)2|. (i)Thelength ofthenormal segment PB=|y\/1 +(y')2|. (j)Thelength ofthenormal segment PE=2:‘(1—|—# Y I E(0,y + 2-:*<><*<%=1" P(1.y) 0%'/5~57 -sX.Y) °~°@0,.e,,% 09 Q1/ M 3* 2%Subtangent (00)“ |y/y'| I Subnormal |yy'l ’ Ay C B X 2'7'0) (==+:v:v'.0) D(0.y—x.v') Figure 13.5 Ineach ofthefollowing problems 2-20, userectangular coordinates to ﬁndtheequation ofthefamily ofcurves thatsatisﬁes theproperty described. 2.Theslope ofthetangent ateach point ofacurve isequal tothesumofthe coordinates ofthepoint. Find theparticular curve through theorigin. 3.Thesubtangent isapositive constant lsforeach point ofthecurve. Hint. See1(e). Lesson l3—Exercise 113 4.Thesubnormal isapositive constant kforeach point ofthecurve. Hint. See1(f). 5.Thesubnormal isproportional tothesquare oftheabscissa. 6.Thesegment ofanormal linebetween thecurve andtheyaxisisbisected bythe2:axis. Find theparticular curve through thepoint (4,2), with this property. 7.The2:intercept ofatangent lineisequal totheordinate. Hint. See1(a). 8.Thelength ofatangent segment from point ofcontact tothe2:intercept is aconstant. Hint. See1(g). 9.Change theword tangent in8above tonormal. Hint. See1(i). l0.-Theareaoftheright triangle formed byatangent line, the2:axisandthe ordinate ofthepoint ofcontact oftangent andcurve, hasconstant area8. Hint. See1(e). 11.Theareaoftheregion bounded byacin've y=f(2:), the2:axis, andthelines 2:=2,2:=2:,isone-half thelength ofthearcincluded between these vertical lines. 12.Theslope ofthecurve isequal tothesquare oftheabscissa ofthepoint of contact oftangent lineandcurve. Find theparticular curve through the point (—1,1). 13.Thesubtangent isequal tothesumofthecoordinates ofthepoint ofcontact oftangent andcurve. Hint. Seel(e). 14.Thelength ofatangent segment from point ofcontact tothe2:intercept is equal tothe2:intercept ofthetangent line. Hint. See(lg)and1(a). 15.Thenormal andthelinedrawn totheorigin from point ofcontact ofnormal andcurve form anisosceles triangle with the2:axis. 16.Change theword normal in15above totangent. 17.Thepoint ofcontact oftangent andcurve bisects thesegment ofthetan- gent linebetween thecoordinate axis. 18.Change theword tangent in17above tonormal. 19.Thelength ofarcbetween 2:=aand2:=2:isequal to2:2/2. 20.Thearea oftheregion bounded bythecurve y=f(z), thelines 2:=a, 2:=1,andthe2:axis, isproportional tothedifference oftheordinates. 21.Lety=f(z)deﬁne acurve thatpasses through theorigin. Find afamily of curves withtheproperty thatthevolume oftheregion bounded byy=f(z), 2:=0,2:=2:rotated about the2:axisisequal tothevolume oftheregion bounded byy=f(z), y=0,y=yrotated about theyaxis. 22.LetP(r,0) beapoint onthecurve r=1(0). AtPdraw atangent tothe curve. Prove (seeFig.13.6) each ofthefollowing. (a)(Z4;=§r2,where Aistheareaoftheregion bounded byanarcofthe curve andtworadii vectors. _ T2 2 (b)E-, +r,where sisthelength ofarcofthecurve between tworadii vectors. (c)tanﬁ =rg.See(a)ofExample 13.3. Ineach ofthefollowing problems 23-25, usepolar coordinates toﬁndthe equation ofthefamily ofcurves that satisﬁes theproperty described. 23.Theradius vector randthetangent atP(r,0) intersect inaconstant angle B. 114 Pnonnnms Lnmme T0Fmsr Ommn EQUATIONS Chapter 3 24-. 25. S"?§"!° 6. 8. 9. 10. ll. 13. 14-. 15. 16. 19. 20. 21. 23.Theradius vector randthetangent atP(r,0) intersect inanangle which is ktimes thepolar angle 0. dY'/ da= dr2+ (rd9)2 Q5‘~10P‘P(r,0) dA 5’ Tangent line r d0 / A r=r(6) 0 O Figure 13.6 Thearea,oftheregion bounded byanarcofacurve andtheradii vectors to theendpoints ofthearcisequal toone-half thelength ofthearc.Hint.See 22(a) and(b). ANSWERS 13 =e‘——:2:—-1. =ve‘.(y/21’) >0;y"=cf‘,(1//1/') <0- =2kx+c,yy' >0;;/2 =——2k:v+c,yy' <0. =2k:c3 +c,yy’>0;3y’=—-2k:z:3 +c,yy’<0. a:2+2y2=24. 7.:c—|-yl0gy+cy=0. i,,+c We _k1og( >. :0:=ix/k2 —y2+c. zcy=cy-16,y' >0;a:y =cy+ l6,y' <0. 8y=4ce*2“+ c‘1en’. 12.3y=2:3—|—4. I=ylvslvyl, (1//1/') >0;Zn/+ y’=v,(1//1/') <0- x2+ y2=cy. 17.my=c. :c2—y2=c. l8.z2—y2=c. my=0. y=¢[-;\/$2 -1-§l0g(x—|—\/$2 -1)]+¢. y"=ce‘. 24.r"=csink0. a:—y=ca:y. 25.r=1;r=sec(0+c), T751. r=ce'°°‘ﬁ.eeee Lesson 14A Isooozur. TRAJEOTORIEB 115 LESSON 14. Trajectories. LESSON 14A. Isogonal Trajectories. When twocurves intersect in aplane, theangle between them isdeﬁned tobetheangle made bytheir respective tangents drawn attheir point ofintersection. Since these lines determine twoangles, itiscustomary tospecify theparticular onedesired l2 Y ca ma '-'1 d l1 ml I3 _.§ O X Figure 14.1 bystating from which tangent linewearetoproceed inacounterclockwise direction toreach theother. InFig. 14.1, ozisthepositive angle from the curve clwith tangent linel,tothecurve C2with tangent lineI2;Bisthe positive angle from thecurve C2tothecurve cl.Ifwecallmltheslope ofl1andmgtheslope oflg,then byaformula inanalytic geometry _"I2"ml ml—"12(14.11) $8.116! — ; tan B= Deﬁnition 14.12. Acurve which cuts every member ofagiven 1-parameter family ofcurves inthesame angle iscalled anisogonal trajectory ofthefamily. Iftwo 1-parameter families have theproperty that every member of onefamily cuts every member oftheother family inthesame angle, then each family may besaid tobea1-parameter family ofisogonal trajectories oftheother, i.e., thecurves ofeither family areisogonal trajectories oftheother. Aninteresting problem istoﬁnd afamily ofisogonal trajectories that makes apredetermined angle with agiven 1-parameter family ofcurves. Ifwecallyl’theslope ofacurve ofagiven 1-parameter family, y’the slope ofanisogonal trajectory ofthefamily, andatheir angle ofinter- 116 Pnonnsus LEADING T0Fmsr Ormsn EQUATIONS Chapter 3 section measured from thetangent linewith slope y’tothetangent line with slope yl’,then by(14.11) I__ I (14.13) tan<1= Example 14.14. Given the1-parameter family ofparabolas (a) y=M2, ﬁnd a1-parameter family ofisogonal trajectories ofthefamily ifthe angle ofintersection 0:,measured from therequired trajectory tothegiven family, is1r/4. Solution. Differentiation of(a)gives (b) y’=2a:c. From (a)again, weobtain (c) a=y/:02, :0950. Substitution ofthisvalue in(b)gives <d> 1/=3,,”» w#0, which istheslope ofthegiven family atanypoint (z,y), :1:960.Itthere- fore corresponds totheyl’in(14.13). Since byhypothesis a=1r/4, tana=1.Hence (14.13) becomes 2_?/_yI (8) I1+%y! 1v+2l/l/I Simpliﬁcation of(e)gives (1') (1—2y)dw+(Iv+21/)dy=0, whose solution, byLesson 7,is 4 __ (8) 1og\/2y’—a:y+a:2+%Arctan%6=c, 2:750.I6 Comment 14.15. Note that before weused (14.13), weeliminated theparameter in(b)toobtain (d). This elimination isessential. Comment 14.16. Because ofthepresence oftheinverse tangent, (g)implicitly deﬁnes amultiple-valued function. Byourdeﬁnition ofa Lesson 14-B ORTHOGONAL Tnamcronms 117 function itmust besingle-valued, i.e.,each value of:1:should determine oneandonly oney.Forthisreason wehave written theinverse tangent with acapital Atoindicate that wemean only itsprincipal values, namely those values which liebetween —1r/2 and 1r/2. Byplacing thisrestric- tion ontheinverse tangent, wehave thus excluded inﬁnitely many solu- tions, forexample solutions forwhich thearctanliesbetween 1r/2and 31r/2, between 31r/2 and51r/2, etc. LESSON 14B. Orthogonal Trajectories. Deﬁnition 14.2. Acurve which cutsevery member ofagiven 1-param- eter family ofcurves ina90°angle iscalled anorthogonal trajectory ofthefamily. Iftwo 1-parameter families have theproperty that every member of onefamily cuts every member oftheother family inaright angle, then each family may besaid tobeanorthogonal trajectory oftheother. Orthogonal trajectory problems areofspecial interest since they occur inmany physical ﬁelds. Lety1'betheslope ofagiven family andlety’betheslope ofanor- thogonal trajectory family. Then byatheorem inanalytic geometry (14-21> yr;/'=-1.y’=—i.-1/1 Example 14.22. Find theorthogonal trajectories ofthe1-parameter family ofcurves (a) y=cars. Solution (Fig. 14.23). Differentiation of(a)gives (b) y’=5cm‘. From (a)again weobtain (c) c=y/xi’, xaé0 Substituting thisvalue in(b),wehave (d) y’=-5%: 2:;-60, which istheslope ofthegiven family atanypoint (z,y), 2:¢0.It corresponds therefore totheyl’in(14.21). Hence, by(14.21), theslope y’ofanorthogonal family is (e) y’=—%»a:¢0,y#O. 118 Pnoausms LEADING TOFmsr Onnrzn EQUATIONS Chapter 3 Itssolution, byLesson 6C,is (f) x2+5y2=k, :c¢0,y;-60. which isa1-parameter family ofellipses. Y c=_1 ¢=-2 ¢=2 '¢=1 c=0 X c=1 c=2 c=-2 c=—1 Figure 14.23 Comment 14.24. Note that before weused (14.21), weeliminated theparameter in(b)toobtain This elimination isessential. LESSON 14C. Orthogonal Trajectory Formula inPolar Coordi- nates. CallP(r,0) (Fig. 14.3) thepoint ofintersection inpolar coordi- ¢2 Y ¢161 P(r,0) I‘ 0 O X Figure 14-.3 nates oftwocurves c,,c2 which areorthogonal trajectories ofeach other. Call ¢1and¢2therespective angle thetangent toeach curve c1andc2 Lesson 14C ORTHOGONAL Tauscroar FORMULA INPoms COORDINATES 119 makes with theradius vector r(measured from theradius vector counter- clockwise tothetangent). Since thetwo tangents areorthogonal, itis evident from theﬁgure that m=a+§ Therefore (14.31) tan4:1=tan(4>2+ =—$- Asremarked previously inExample 13.3, inpolar coordinates d0(14.32) tan¢2=r$- Therefore (14.31) becomes drtan ¢1-—""IE6‘ Comparing (14.32) with (14.33), weseethat iftwocurves areorthogonal, d0 , _ , d0then r5ofoneisthenegative reciprocal ofrEoftheother. Conversely, ifoneoftwocurves satisﬁes (14.32) andtheother satisﬁes (14.33), then thecurves areorthogonal. Hence toﬁndanorthogonal family ofagiven family, weproceed asfollows. Calculate rﬁgofthegiven family. Replace d0 . . . d . . .rEbyitsnegative reciprocal —-é.The family ofsolutions ofthisr new resulting differential equation isorthogonal tothegiven family. Example 14.34. Find, inpolar form, theorthogonal trajectories of thefamily ofcurves given by (a) r=lcsec0. Solution. Differentiation of(a)gives (b) %=lcsec9tan0. From (a),again, weobtain r <°> hm" Substituting thisvalue in(b),wehave dr d0 1(cl) (T0-rtan0, rE-wT§- 120 Pnostaus LEADING TOFIRST Ormsn EQUATIONS Chapter 3 By(14.33), therefore, thedifferential equation oftheorthogonal family is d 1 d(e) —ri(;6=mT, or—%=cot0d0, whose solution is (f) rsin0=c. Comment 14.35. Weremark once more that before wecould use (14.33), wehadtoeliminate theparameter in(b)toobtain (d). EXERCISE 14 Foreach ofthefollowing family ofcurves, 1-4, ﬁnd a1-parameter family ofisogonal trajectories ofthefamily, where theangle ofintersec- tion, measured from therequired trajectory tothegiven family, isthe angle shown alongside each problem. l.y2=4ka:, a=45°. 2.a:2+y2 =k2,a=45°. 3.y=krc, tana =Q. 4.y2+2:ty—:t2 =k,oz=45°. 5.Given that thedifferential equation ofafamily ofcurves isM(z,y) dz+ N(z,y) dy=0.Find thedifferential equation ofafamily ofisogonal tra- jectories which makes anangle a9*1r/2with thegiven family, where a ismeasured from trajectory family togiven family. 6.Given that thedifferential equation ofafamily ofcurves isofthehomo- geneous type. Prove thatthedifferential equation ofafamily ofisogonal trajectories which makes anangle a94-ir/2with thegiven family isalso homogeneous. Hint. Youwillﬁndaproof inCase 3-2ofLesson 32. Find inrectangular form theorthogonal trajectories ofeach ofthefol- lowing family ofcurves. 7.12+ 2zy—yz=k. 13.1:21; =k. 8.$2+(y——lc)2=k2. 14.$2——yz=I02. 9.(2:—k)2+ 1/2=k2. 15.1/2=k:c3. 10.a:—y=Ice‘. 16.e‘cosy =lc. ll.yz=4px. 17.siny =Ice". 12.xy=ka:—1. Find inrectangular form theorthogonal trajectories ofeach ofthefol- lowing family ofcurves. 18.Afamily ofstraight lines through theorigin. 19.Afamily ofcircles with variable radii, centers onthe:2:axis, andpassing through theorigin. 20.Afamily ofellipses with centers attheorigin andvertices at(i1,0). 21.Afamily ofequilateral hyperbolas whose asymptotes arethecoordinate axes. Lesson 14—Exercise 121 22.Afamily ofparabolas with vertices attheorigin andfocionthexaxis. Ifthedifferential equation ofagiven family ofcurves remains unchanged when y’isreplaced by——1/y’, thefamily iscalled selforthogonal, i.e.,a self-orthogonal family hastheproperty that acurve orthogonal toamem- berofthefamily also belongs tothefamily. Show that each ofthefol- lowing families isselforthogonal. 23.Thefamily ofparabolas that have acommon focus andaxis. Hint. See answer. 24.Thefamily ofcentral conics that have common fociandaxis. Hint. See answer. Find inpolar form theorthogonal trajectories ofeach ofthefollowing family ofcurves. 25.r=kcos0. 30.r2=k(rsin0 —-1). 26.r0=k. 31.r2=lccos 20. 27.r-lc(l+sin0). 32.r"1=sin20+ lc. 28.r=sin0+ lc. 33.1'"sinn0=lc. 29.r=ksin 20. 34.r=e“. ANSWERS 14 S .°s=2|+“IQQ 6l.log|2a:2+my+y2|+—Arctan =c. \/7 2.log|c(:c2 +y2)|—2Arctan(y/z) = 3.log|c(a:2 +y2)|+4Arc tan(y/z) = 4.my=c. 5.y’=(Ntana+ M)/(M tana —N). 7.2:2—2zy—-g2=c. 15.3y2+22:2=c. 8.:c2+ yz=crc. 16.e‘sing =c. 9.2:2+yz=cy. 17.:2:=ccosz y. l0.ce"=y—-a:—l-2. l8.2:2+y2=c. ll.2:2+1/2=c. 19.Seeproblem 9. 12.2:3+3y=c. 20.$2=ce”+"’. 13.2y2—$2=c. 21.2:2—y2=c. 14.xy=c. 22.Seeproblem 11. 23.Equation offamily isg2=4p(x2+ p). 2 24.Equation offamily is-E5+-1% =1,where (=!=c,0) arethecoordinates ofthefoci. 25.r=csi2n0. 30.r=2sin0—|—ccos0. 26.cr=e’/2. 31.r2=csin20. 27.r=c(1—-sin0). 32.tanﬂ =ce2'. 28.e1/'=c(sec 0+tan0). 33.r"cosn0=c. 29.r4=ccos 20. 34.r=e*V°-'”- 122 Pnostams LEADING TOFIRST ORDER EQUATIONS Chapter 3 LESSON 15. Dilution and Accretion Problems. Interest Problems. Temperature Problems. Decomposition and Growth Problems. Second Order Processes. LESSON 15A. Dilution and Accretion Problems. Inthistype ofproblem, weseek aformula which willexpress theamount ofasubstance insolution asafunction ofthetime t,where thisamount ischanging instantaneously with time. Example 15.1. Atank contains 100gallons ofwater. Inerror 300 pounds ofsaltarepoured into thetank instead of200pounds. Tocorrect thiscondition, astopper isremoved from thebottom ofthetank allowing 3gallons ofthebrine toflow outeach minute. Atthesame time 3gallons offresh water perminute arepumped into thetank. Ifthemixture is kept uniform byconstant stirring, how long willittake forthebrine to contain thedesired amount ofsalt? Solution. The usual procedure insolving problems ofthistype isto letavariable, sayx,represent thenumber ofpounds ofsaltinsolution at anytime t.Then anequation issetupwhich willreﬂect theapproximate change inxinanarbitrarily small time interval At.Inthisproblem, 3 gallons ofbrine flow outeach minute. Hence 3Atgallons ofbrine will ﬂow outinAtminutes. Since the100gallons ofsolution arethoroughly mixed, andrerepresents thenumber ofpounds ofsaltpresent inthesolu- tion atanyinstant oftime t,wemay assume that ofthe3Atgallons of 3Al . . brine ﬂowing out,Eacwillbetheapproximate lossofsaltfrom thesolu- tion. (For example, if250pounds ofsaltareinthe100-gallon solution of 3At ,brine attime t,and ifAtissufficiently small, then E250Wlllbethe approximate lossofsaltintime At.) If,therefore, Axrepresents theapproximate lossofsaltintime At,then 100 ’ At 100 (the symbol zmeans approximately equal to). The negative sign is necessary toindicate that xisdecreasing. Since nosaltenters thesolu- tion, weareledtothedifferential equation dz 3:1: dx(3.) E —- '--136! I — The solution of(a)is (b) logz=—0.03t +c’, x=ce"°'°3‘. Lesson 15A DILUTION ANDACCRETION PROBLEMS 123 Inserting in(b)theinitial conditions x=300, t=0,weobtain c=300. Hence (b)becomes (c) :0=300e_°'°3', anequation which gives theamount ofsaltinsolution asafunction of thetime t.And when as=200, weobtain from (c) (d) §=e_°'°3', log%=0.031, from which weﬁnd (e) t=13.5min, i.e.,itwilltake 13.5 minutes fortheamount ofsaltinthesolution tobe reduced to200pounds. Comment 15.11. Amuch shorter and more desirable method of solving theabove problem istoinsert inthesecond equation of(a)the initial andﬁnal conditions aslimits ofintegration. Wewould thus obtain 200 d a (f) / 3=-0.03; dt.===soo 1? l=0 Integration of(f)gives immediately (g) log§=-0.03:, log=§=0.03:, which isthesame as(d)above. Example 15.12. Atank contains 100gallons ofbrine whose saltcon- centration is3pounds pergallon. Three gallons ofbrine whose saltconcen- tration is2pounds pergallon flow into thetank each minute, andatthe same time 3gallons ofthemixture ﬂow outeach minute. Ifthemixture is kept uniform byconstant stirring, ﬁnd thesaltcontent ofthebrine asa function ofthetime t. Solution. Letatrepresent thenumber ofpounds ofsaltinsolution at anytime t.Byhypothesis 6pounds ofsaltenter and3gallons ofbrine leave thetank each minute. Intime At,therefore, 6Atpounds ofsalt flow inandapproximately ﬁ(3At) pounds ﬂow out. Hence intime At, theapproximate change ofthesaltcontent inthesolution is :0 Ax(3.) A23 ~6A3 —'ﬁ63At, It ~6— Wearethus ledtothedifferential equation dx/dt ==6—-0.032;, which wewrite as dx dx (bl m -"‘-i -rd‘- 124 PROBLEMS Lmnmo 'roFIRST ORDER EQUATIONS Chapter 3 Assuggested inComment 15.11, weintegrate (b)and insert theinitial condition asalimit ofintegration. Wethus obtain 3 I dx _f _ (°) LE... 0.03.5-6_.._,, d" whose solution is (<1) :1:=100(2+e-°-°="). NOTE. Wealsocould have solved (b)bythemethod ofLesson 11B. Example 15.13. Same problem asinExample 15.12, excepting that 3gallons offresh water ﬂow into thetank instead ofbrine and5gallons ofthemixture flow outinplace of3. Solution. Since liquid isﬂowing into thetank attherate of3gallons perminute and themixture isﬂowing outattherate of5gallons per minute, there willbe(100 —-2t)gallons ofthebrine inthetank atthe endoftminutes. If:0represents thenumber ofpounds ofsaltinsolution attime t,then theapproximate change inthesaltcontent ofthebrine in asuﬁiciently small time interval Atis Hz_m_5“ . $ A”'“"100-2:(5"‘)' At 100-2: Wearethus ledtothedifferential equation dx 5x dx dt Following thesuggestion incomment 15.11, weintegrate thesecond equa- tion in(a)andinsert theinitial condition asalimit ofintegration. We thus obtain3 C 1fa_f_L.(b) 5==soo I i=0 100—25 Itssolution is (c) :1:=1-0*-1,-,,(100 -21)‘/2. EXERCISE 15A Itisassumed intheproblems below that allmixtures arekept uniform byconstant stirring. 1.Atank initially holds 100galofbrine containing 30lbofdissolved salt. Fresh water flows intothetank attherateof3gal/min andbrine ﬂows out atthesame rate. (a)Find thesaltcontent ofthebrine attheendof10min. (b)When willthesaltcontent be15lb? Lesson 15A—Exercise 125 2.Solve problem 1if2gal/minoffresh water enter thetankinstead of3gal/min. 3.Solve problem 1if4gal/minoffresh water enter thetankinstead of3gal/min. 4.Atank initially contains 200galofbrine whose saltconcentration is3lb/gal. Brine whose saltconcentration is2lb/gal flows intothetank attherateof 4gal/min. Themixture flows outatthesame rate. (a)Find thesaltcontent ofthebrine attheendof20min. (b)When willthesaltconcentration be reduced to2.5lb/gal? 5.Atank initially contains 100galofbrine whose saltconcentration isQlb/gal. Brine whose saltconcentration is2lb/gal flows intothetank attherateof 3gal/min. Themixture flows outattherateof2gal/min. Find thesalt content ofthebrine anditsconcentration attheendof30min. Hint. After 30min, thetank contains 130galofbrine. 6.Atank initially contains 100galofbrine whose saltconcentration is0.6 lb/gal. Brine whose saltconcentration is1lb/gal flows intothetank atthe rateof2gal/min. Themixture flows outattherateof3gal,/min. Find the saltcontent ofthebrine anditsconcentration attheendof60min. Him. After 60min, thetank contains 40galofbrine. 7.Atank initially contains 200galoffresh water. Brine whose saltconcentra- tionis2lb/gal ﬂows intothetank attherateof2gal/min. Themixture ﬂows outatthesame rate. (a)Find thesaltcontent ofthebrine attheendof100min. (b)Atwhat time willthesaltconcentration reach 1lb/gal? (c)Could thesaltcontent ofthebrine everreach 400lb? 8.Atank initially contains 100galoffresh water. Brine whose saltconcentra- tionis1lb/gal ﬂows intothetank attherateof2gal/min. Themixture ﬂows outattherateof1gal/min. (a)Find thesaltcontent ofthebrine and itsconcentration attheendof60min. 9.Atank initially contains 200galoffresh water. Itreceives brine ofan unknown saltconcentration attherateof2gal/min. Themixture ﬂows out atthesame rate. Attheendof120min, 280lbofsaltareinthetank. Find thesaltconcentration oftheentering brine. 10.Two tanks, AandB,each contain 5000 galofwater. Toeach tank 150gal ofachemical should beadded, butinerror theentire 300galarepoured into theAtank. Pumps aresettowork tocirculate theliquid through thetwo /1.’.,1 . ., ,1.~~ / Figure 15.14‘\six \\\‘\ tanks attherateof100gal/min(Fig. 15.14). (a)How longwillittakefor tank Atocontain 200galofthechemical andtank Btocontain 100gal- lons? (b)Isittheoretically possible foreach tank tocontain 150gal? ll.TheCO2 content oftheairina5000-cu-ft room is0.3percent. Fresh air containing 0.1percent CO2 ispumped intotheroom attherateof1000 ft3/min. (a)Find thepercentage ofCO2intheroom after 30min. When willtheCO2content be0.2percent? 126 PROBLEMS LEADING 'roFIRST Onnsn EQUATIONS Chapter 3 12.TheCO2content oftheairina7200-cu-ft room is0.2percent. What volume offresh aircontaining 0.05 percent CO2 must bepumped intotheroom each minute inorder toreduce theCO2content to0.1percent in15min? ANSWERS 15A 1.(a):2:=30e-°-3 =22.2lb. (b)e'°-°3‘ =§,t=23.1min. 2.(a)1=30(1 —0.0103 =30(0.9)3 =21.87 lb. (b)t=20.6 min. 3.(a)2:=30e"°'286 =22.5lb. (b)l=26.0min. 4.(a)1:=200(2+ e'°'4) =534.1 lb. (b)t=50log2 =34.7min. 5.1711b, 1.32lb/gal.6.37.4lb,0.94lb/gal. 7.(a)252.8lb. (b)69.3min. (c)Salt content approaches 400lbast increases without limit. 8.981b, 0.61lb/gal. 9.21b/gal. 10.(a)27.5min. (b)No. Tank Bcontains 150galofthechemical onlyift isinﬁnite. ll.(a)0.10percent. (b)3.47min. 12.527fta/min. LESSON 15B. Interest Problems. LetSAbeinvested at6percent perannum. Then theprincipal Pattheendofoneyear willbe (a) P=A(1+0.06) ifinterest iscompounded annually, 2 P=A(1+ ifinterest iscompounded semiannually, 4 P=A(1+ ifinterest iscompounded quarterly, 0.061”.. .P=A1+W ifinterest iscompounded monthly. And, ingeneral, theprincipal Pattheendofoneyear willbe T "I iftheinterest rate isrpercent perannum compounded mtimes peryear. Attheendofnyears, itwillbe <<=> Al(1+;;)'"l"~Ifthenumber mofcompoundings inoneyear, increases without limit, then <~*>P=A[(1+ill=Al(1+%)'""l"'- Lesson l5B—Exercise 127 m/r Butlim(1+ =e.Hence (d)becomes (e) P=Ac”. Finally, replacing nbyt,weobtain (15.2) P=Ac", which gives theprincipal attheend oftime tif$Aarecompoimded instantaneously orcontinuously atrpercent perannum. Thedifferential equation ofwhich (15.2) isthe1-parameter family, is (15.21) ‘zit’=1P. (Verify it.) Example 15.22. How long willittake for$1.00 todouble itself ifitis compounded continuously at4percent annum. Solution. By(15.21) with r=0.04, weobtain dP(E) $ =0.04dl. Integrating (a)and inserting theinitial andﬁnal conditions aslimits of integration, wehave 2 1 (b) / E=0.04/ dt, log2=0.04t, t=17%years,P=-1P 1-o approximately. Remark 1.Wecould have solved thisproblem byusing (15.2) directly with P=2,A=1, r=0.04. Remark 2.At4percent perannum, compounded semiannually, $1.00 willdouble itself in17%years. Compounded continuously, aswesaw above, itdoubles in171};years. The continuous compounding ofinterest property therefore isnotaspowerful asonemight have believed. EXERCISE 15B Itisassumed intheproblems below thatinterest iscompoimded con- tinuously unless otherwise stated. l.Inhowmany years will$1.00 double itself at5percent perannum? 2.Atwhat interest ratewill$1.00 double itself in12years? 3.How much will$1000.00 beworth at4}percent interest after 10years? 128 PROBLEMS L1-zanmo T0Fmsr Orman EQUATIONS Chapter 3 4-. 5. 6. l. 2. 3. 4. 5. 6.Inawill,amanleftafewmillion dollars, tobedivided among several trusts. Thewillprovided thatthemoney wastobedeposited insavings institutions andheld for500years before being distributed tothedesignated legatees. Thewillwascontested bythegovernment onthegrounds thatthemonetary wealth ofthenation would beconcentrated inthese trusts. Ifthemoney earned anaverage of4percent interest, approximately how much would only $1,000,000 amount toattheendofthe500years? How much money would youneed todeposit inabank at5percent interest inorder tobeabletowithdraw $3600.00 peryearfor20years ifyouwish the entire principal tobeconsumed attheendofthistime: (a)ifthemoney is alsobeing withdrawn continuously from thedate ofdeposit as,forexample, withdrawing $3600/365 each day; (b)ifthemoney isbeing withdrawn at therateof$300.00 permonth beginning with theﬁrstmonth after thede- posit. Hint. After 1month, money leftinthebank equals Ae°-°5/*2 —-300, where Aistheamount atthebeginning ofthemonth. (c)Trytosolve thisproblem assuming themore realistic situation of interest being credited quarterly and$900.00 withdrawn quarterly, begin- ningwith theﬁrstquarter after thedeposit. Nora. This problem nolonger involves adiﬁerential equation. Hint. Attheendoftheﬁrstquarter, money leftinthebank equals A(1+ —900. You plan toretire in30years. Attheendofthat time, youwish tohave $45,500.00, theapproximate amount needed—see problem 5(b)—in order towithdraw $300.00 monthly for20years after retirement. (a)What amount must youdeposit monthly at5percent? Hint. Attheendofonemonth P=A;attheendoftwomonths P=Ae°'°5“2+ A,where Aisthe monthly deposit. (b)What amount must youdeposit semiannually ifinterest iscredited semiannually at5percent instead ofcontinuously. N0'r1:. This problem nolonger involves adiﬁerential equation. ANSWERS 15B 13.86 years. (Compounded semiannually at5percent interest, $1.00 doubles itself in14years.) 5.78percent approximately. $1568.31. $485,165,195,400,000. (a)dP=(rP)dt-—(3600) dt,A=$45,513. 1_ (b)A=300 =$45,418. _ 1-(1.o125>-8°] _ (°)A'mi 0.0125 ‘“5’348' l__ 1.5 (8.)$45,500 =A A=$54.57monthly. 60 (b)$45,500 =Al ]. A=$234.58 semiannually. Lesson 15C TEMPERATURE PROBLEMS 129 LESSON 15C. Temperature Problems. Ithasbeen proved experi- mentally that, under certain conditions, therate ofchange ofthetem- perature ofabody, immersed inamedium whose temperature (kept constant) differs from it,isproportional tothedifference intemperature between itandthemedium. Inmathematical symbols, thisstatement is written as <15-3) %=—k<T5—rt). where k>0isaproportionality constant, TBisthetemperature ofthe body atanytime t,and TMistheconstant temperature ofthemedium. Comment 15.31. Insolving problems inwhich aproportionality constant kispresent, itisnecessary toknow another condition inaddition totheinitial condition. Inthetemperature problem, forexample, we shall need toknow, inaddition totheinitial condition, thetemperature ofthebody atsome future time t.With these twosetsofconditions, it willthen bepossible todetermine thevalues oftheproportionality con- stant lcand thearbitrary constant ofintegration c.And ifwewish to take advantage ofComment 15.11 wemust use(15.3) twice, once to ﬁndk,thesecond time toﬁndthedesired answer. Example 15.32. Abody whose temperature is180° isimmersed ina liquid which iskept ataconstant temperature of60°. Inoneminute, thetemperature oftheimmersed body decreases to120°. How long will ittake forthebody’s temperature todecrease to90°? Solution. LetTrepresent thetemperature ofthebody atanytime t. Then by(15.3) with TM=60, dT dT(a) E;=-Mr-50), -Tic =-rat, where thenegative sign isused toindicate adecreasing T.Writing (a) twice assuggested inComment 15.31, integrating both equations, and inserting allgiven conditions, weobtain 120 1 90 ¢ (b) /r-iso Taﬁ =nk/r-0 dt; /r'=1ao =_k/1-o dt' From theﬁrstintegral equation, weobtain (c) log0.5=—-k, lc=log2. With theproportionality constant kknown, weﬁndfrom thesecond inte- 130 PROBLEMS LEADING 'roFmsr Onnna Equxrrons Chapter 3 gralequation ____ __log4_2log2__(d) log0.25 - (log2)t, t-Tg2-Z-—log2-2. Hence itwilltake 2minutes forthebody's temperature todecrease to90° EXERCISE 15C. Intheproblems below, assume thattherateofchange ofthetempera- ture ofabody obeys thelawgiven in(15.3). 1.Abody whose temperature is100°isplaced inamedium which iskept ata constant temperature of20°. In10minthetemperature ofthebody falls to60°. (a)Find thetemperature Tofthebody asafunction ofthetime t. (b)Find thetemperature ofthebody after 40min. (c)When willthebody's temperature be50°? 2.Thetemperature ofabody differs from thatofamedium, whose temperature iskeptconstant, by40°. In5min,thisdifference is20°. (a)What isthevalue ofkin(15.3)? (b)Inhowmany minutes willthedifference intemperature be10°? 3.Abody whose temperature is20°isplaced inamedium which iskept ata constant temperature of60°. In5minthebody's temperature hasrisen to30°. (a)Find thebody's temperature after 20min. (b)When willthe body's temperature be40°? 4-.Thetemperature inaroom is70°F. Athermometer which hasbeen kept in itisplaced outside. In5minthethermometer reading is60°F. Five minutes later, itis55°F. Find theoutdoor temperature. The speciﬁc heat ofasubstance isdeﬁned astheratio ofthequantity ofheat required toraise aunit weight ofthesubstance 1°tothequantity ofheat required toraise thesame unitweight ofwater 1°.Forexample, it takes 1calorie tochange thetemperature of1gram ofwater 1°C(or1 British thermal unit tochange thetemperature of1lbofwater 1°F). If, therefore, ittakes only 3%ofacalorie tochange thetemperature of1gram ofasubstance 1°C(or-{L6ofaBritish thermal unit tochange 1lbofthe substance 1°F), then thespeciﬁc heat ofthesubstance is Inproblems 5-7below, assume thattheonly exchanges ofheat occur between thebody andwater. 5.A50-lb ironballisheated to200°F andisthen immediately plunged intoa vessel containing 100lbofwater whose temperature is40°F. Thespeciﬁc heat ofironis0.11. (a)Find thetemperature ofthebody asafunction of time. Hint. Thequantity ofheat lostbytheironbody intime tis50(0.11) (200 -—TB),where TBisitstemperature attheendoftime t.Thequantity ofheat gained bythewater—remember thespeciﬁc heat ofwater isone— is100(1)(Tw -—40),where Twisthetemperature ofthewater attheend oftime t.Since theheatgained bythewater isequal totheheatlostbythe ball, 50(0.11)(200 ——TB)=100(Tw —40). Solve forTwandsubstitute thisvalue forTMin(15.3). Solve forTB.(b)Find thecommon temperature approached bybody andwater ast—>w. Lesson 15D DEcom>osrr1oN moGsowm Pnoannus 131 6.Thespeciﬁc heat oftinis0.05. A10-lb body oftin,whose temperature is 100°F, ispltmged intoavessel containing 50lbofwater at10°F. (a)Find thetemperature Tofthetinasafunction oftime. (b)Find thecommon temperature approached bytinandwater ast—>w. 7.Thetemperature ofa100-lb body‘whose speciﬁc heat is115is200°F. Itis plunged intoa40-lb liquid whose speciﬁc heat isQandwhose temperature is50°. (a)Find thetemperature Tofthebody asaflmction oftime. (b)To what temperature willthebody eventually cool? ANSWERS 15C 1.<5)T=2o(1+ 45-°-°°°"‘). (1.)25'. <5)14.2mm.2.(5)1.=-§log(0.5)=0.1385. (1.)1=10min.3.(a)47.3°. (b)12min. 4.50°F. 5.(5)T3=ﬁ(51+l60e"1‘055“). (5)48.3°F. 5.<5)T=1%(11+cor‘-°““). (b)10.9"F. 1.<5)T=10o(1+ e-3"“). (1.)100°F. LESSON 15D. Decomposition and Growth Problems. These prob- lems willalsoinvolve aproportionality constant andwilltherefore re- quire anadditional reading after aninterval oftime t.Themethod of solution isessentially thesame asthatusedtosolve theproblem discussed inLesson 1 Example 15.4. Thenumber ofbacteria inayeast culture grows ata ratewhich isproportional tothenumber present. Ifthepopulation ofa colony ofyeast bacteria doubles inonehour, ﬁndthenumber ofbacteria which willbepresent attheendof3Qhours. Solution. Letxequal thenumber ofbacteria present atanytime t. Then inmathematical symbols, theﬁrstsentence oftheproblem states .1 a(5) if=kw, -f=lcdt, where lcisaproportionality constant. Writing (a)twice, integrating both equations, andinserting allthegiven conditions, weobtain (omitting percent signs) 200 d 1 5 d 1/2 (b) f -”=hfdt; f -”=15/.11.z-100 17 c-0 z-100 1? c-o From theﬁrstintegral equation, weobtain (c) k=log2, 132 Pnonuzms LEADING 'roFmsr Onnsn Eouxrrons Chapter 3 andfrom thesecond integral equation (d) 15;(5/100) =grog2,T35=2”’, 5=1131. Hence 1131 percent or11.31 times theinitial number ofbacteria willbe present attheendof3%hours. Example 15.41. Thedeath rateofanantcolony isproportional to thenumber present. Ifnobirths were totake place, thepopulation at theendofoneweek would bereduced byone-half. However because of births, therateofwhich isalsoproportional tothepopulation present, theantpopulation doubles in2weeks. Determine thebirth rateofthe colony perweek. Solution. Inthis problem, wemust determine two proportionality constants, oneforbirths which wecallkl,theother fordeaths which we callI02.Using ﬁrstthefactthat thedeath rateisproportional tothe number present andthatdeaths without births would reduce thecolony inoneweek byone-half, wehave 0.5 1dz: dx (=1) E—-7621, Lad j-—*-k2/‘-0 dt) where :1:represents thepopulation ofthecolony atanytime t,and:1:=1 stands for100percent. Thesolution of(a)is kg=10g 2. Hence, thedifferential equation which takes into consideration both births anddeaths ofthecolony is d .1(5) i=15,5-(log2)a:, f=(r,-15;2)dt. Integrating (c)andinserting thegiven conditions which reﬂect thenet change inthepopulation, weobtain 2 2 (.1) fhliiif =(k,-log2)ft_0 .11. Thesolution of(d)is (e) log2 =2101—log4, kl=§log8 =1.0397. Hence thebirth rate is103.97 percent perweek. Lesson l5D—Exe1'cise 133 EXERCISE 15D Inproblems 1-10below, assume thatthedecomposition ofasubstance is proportional totheamount ofthesubstance remaining andthat thegrowth ofpopulation isproportional tothenumber present. (Asuggestion: review Lesson 1.) 1.Thepopulation ofacolony doubles in50days. Inhowmany days willthe population triple? 2.Assume thatthehalflifeoftheradium inapiece ofleadis1500 years. How much radium willremain intheleadafter 2500 years? 3.If1.7percent ofasubstance decomposes in50years, what percentage ofthe substance willremain after 100years? How many years willberequired for10percent todecompose? 4-.The bacteria count inaculture is100,000. In2§hours, thenumber hasin- creased by10percent. (a)Inhowmany hours willthecount reach 200,000? (b)What willthebacteria count bein10hours? 5.Thepopulation ofacountry doubles in50years. Itspresent population is 20,000,000. (a)When willitspopulation reach 30,000,000? (b)What will itspopulation bein10years? 6.Tenpercent ofasubstance disintegrates in100years. What isitshalflife? 7.Thebacteria count inaculture doubles in3hours. Attheendof15hours, thecount is1,000,000. How many bacteria were inthecount initially? 8.Bynatural increase, acity, whose population is40,000, willdouble in50 years. There isanetaddition of400persons peryear because ofpeople leaving andmoving intothecity. Estimate itspopulation in10years. Hint. First ﬁndthenatural growth proportionality factor. 9.Solve problem 8,ifthere isanetdecrease inthepopulation of400persons peryear. 10.Aculture ofbacteria whose population isN0will,bynatural increase, double in4log2days. Ifbacteria areextracted from thecolony attheuniform rateofRperday, ﬁndthenumber ofbacteria present asafunction oftime. Show thatthepopulation willincrease ifR<N0/4,willremain stationary ifR=N0/4, willdecrease ifR>No/4. ll.Therateoflossofthevolume ofaspherical substance, forexample amoth ball, duetoevaporation, isproportional toitssurface area. Express the radius oftheballasafunction oftime. 12.Thevolume ofaspherical raindrop increases asitfallsbecause ofthead- hesion toitssurface ofmist particles. Assume itretains itsspherical shape during itsfallandthattherateofchange ofitsvolume with respect tothe distance yithasfallen, isproportional tothesurface areaatthatdistance. Express theradius oftheraindrop asafunction ofy. ANSWERS 15D 1.79days. 5.(a)29years. (b)22,970,000 approx. 2.31percent. 6.658years. 3.96.6percent; 307years. 7.31,250. 4-.(a)18.2hours. (b)146,400. 8.50,240. 9.41,660. No 1/410.:1:=4R+ T—Re - 134 PROBLEMS LEADING roFmsr ORDER Equurxons Chapter 3 11.r=ro—kt,where roistheinitial radius andkisapositive proportionality factor. 12.r=ro+Icywhere roistheinitial radius andkisapositive proportionality factor. LESSON 15E. Second Order Processes. Anewsubstance Cissome- times formed from twogiven substances AandBbytaking something away from each; thegrowth ofthenewsubstance being jointly propor- tional totheamount remaining ofeach oftheoriginal substances. I.et s1ands2betherespective amounts ofAandBpresent initially andlet xrepresent thenumber ofunits ofthenewsubstance Cformed intime t. If,forexample, oneunitofC’isformed bycombining 2units ofsubstance Awith three units ofsubstance B,thenwhen :1:units ofthenewsubstance arepresent attime t (51—2x)istheamount ofAremaining attime t, (52—-3:21)istheamount ofBremaining attime t. Bytheﬁrst sentence above, therefore, thedifferential equation which represents therateofchange ofCatanytime tisgiven by d3%=k(s1 -2x)(s2 —31:), where Icisaproportionality constant. Ingeneral, ifoneunitofC’isformed bycombining munits ofAandnunits ofB,then thedifferential equation becomes <15-5) §=us.—m><s.—M). where s1ands2aretherespective number ofunits ofAandBpresent initially andxisthenumber ofunits ofCpresent intime t. Asubstance may alsobedissolved inasolution, itsrateofdissolution being jointly proportional to: 1.Theamount ofthesubstance which isstillundissolved. 2.Thedifference between theconcentration ofthesubstance inasatu- rated solution andtheactual concentration ofthesubstance inthe solution. Forexample, if10gallons ofwater canhold amaximum of 30pounds ofsalt,itissaidtobesaturated when itholds thisamount ofsalt. Theconcentration ofsaltinasaturated solution isthen 3 pounds pergallon. When therefore thesolution contains only 15 pounds ofsalt, theactual concentration ofthesaltinsolution is1.5 pounds pergallon or50percent ofsaturation. Letxrepresent theamount ofthesubstance undissolved atanytime t, 1:0theinitial amount ofthesubstance, andvthevolume ofthesolution. Lesson 15E SEcoNn ORDER Pnocnssns 135 Then atanytime t, (mo—az)istheamount ofthesubstance dissolved inthesolution, -x—0-;—-2 istheconcentration ofthesubstance inthesolution. Ifcrepresents theconcentration ofthesubstance inasaturated solution, then thedifferential equation which expresses mathematically conditions 1and2above is dx _ (15.51) E=Icx(5- Problems which involve joint proportionality factors areknown as second order processes. Example 15.52. Anewsubstance Cistobeformed byremoving two units from eachoftwosubstances whose initial quantities are10and8units respectively. Assume thattherateatwhich thenewsubstance isformed isjointly proportional totheamount remaining ofeach oftheoriginal substances. Ifatisthenumber ofunits ofCformed atanytime tand .1:=1unitwhen t=5minutes, ﬁnda:when t=10minutes. Solution. In(15.5), s1=10,s2=8,m=n=2.Hence (15.5) becomes (a) %=k(10 —2:z:)(8 —2x)=4k(5 —a:)(4 —1:). Therefore dx 1 14,0dl=z =(see LBSSOII 26) - Writing (b)twice, integrating both equations andinserting allthegiven conditions, weobtain 5 1 1 1 _(C) 4,0‘/t-odi-1!’-0(4—;-—:v'—-5-—:?)dZ, 10 2 5/=/<-L-~—1—>d.t-odt 1-0 4--15 5-9? x From theﬁrstintegral equation, weﬁnd 1 5—51_116 <“> '°-a(‘°@ -§a‘°g15' andfrom thesecond integral equation, 116 _ 5-5 5__ 45—-x)_(e) 4(-§6logT5)(10)-1og;1——_1—i—logZ-log5(——4_x 136 PROBLEMS LEADING T0Frnsr ORDER Eouxrrons Chapter 3 Simpliﬁcation of(e)gives 16”_4(5-5) 5-5_g1_ _ _ Hence 1.63units ofthesubstance a:areformed in10minutes. Example 15.53. Sixgrams ofsulfur areplaced inasolution of100cc ofbenzol which when saturated willhold 10grams ofsulfur. If3grams ofsulfur areinthesolution in50minutes, howmany grams willbein thesolution in250minutes? Solution. Letatrepresent thenumber ofgrams ofsulfur notyetdis- solved atany time t.Then, attime t,(6—x)istheamount ofsulfur dissolved and (6—1:)/100 istheconcentration ofsulfur inbenzol. Here c,theconcentration ofsulfur inasaturated solution ofbenzol, isgiven as 10/100 ==0.1,theinitial amount xoofthesubstance isgiven as6and v=100. Hence (15.51) becomes Therefore k dz 1dx dz (bl 10o°‘—5(5+4)”Z(e_5+4)' Writing (b)twice, integrating both equations, andinserting allthegiven conditions, weobtain k 50 3 1 1 <°’ a/.-.."”=/i...<5“ do k/‘250 1 '/>1 1 ) 25‘=0dt—‘=6 x x+4dx. From theﬁrstintegral equation, weﬁnd 1 313 5 15(<1) r=§15g;;f-46=2-(log;-10gﬁ)=51<>g7. andfrom thesecond integral equation, 115 5’(B) l0g 250 -——log—m =logi— log§= log§(L)-:v+4 5 31+4 Lesson l5E—Exe1-cise 137 Simpliﬁcation of(e)gives 5° 5 :0 .(f) 7=§$ 1ac=0.5gram approximately, which istheamount ofsulfur notyetdissolved attheendof250minutes. Therefore since 6grams ofsulfur were undissolved inthesolution origi- nally, 5.5grams areinthesolution attheendof250minutes. EXERCISE 15E Inproblems 1-8,assume allreactions aregoverned byformulas (15.5) or (15.51), with theexception ofproblem 4which isamodiﬁed version of (15.51). 1.In(15.5) take s1=10,82=10,m=1,n.=1.If5units ofC’areformed in10min,determine thenumber ofCunits formed in50min. 2.In(15.5) take s1=10,82=8,m=1,n=1.If1unitofC’isformed in5 min, determine thenumber ofCunits formed in10min. 3.In(15.5) take m=1,n=1.(a)Solve for:1:asafunction oftime when s1as82andwhen s1=82.(b)Show that ast—>w,x—+s1ifs2Qs1 andx->s2if82§s1. 4.Inacertain chemical reaction, substance A,initially weighing 12lb,iscon- verted intosubstance B.Therateatwhich Bisformed isproportional tothe amount ofAremaining. Attheendof2.5min, 4lbofBhave been formed. (a)How much oftheBsubstance willbepresent after 6min? (b)How much time willberequired toconvert 60percent ofA? Work thisproblem intwoways: 1.Letting xrepresent amount ofAremaining attime t. 2.Letting :2:represent amount ofBformed attime t. Chemical reactions ofthistype arecalled ﬁrst order processes. 5.Anewsubstance Cistobeformed from twogiven substances AandBbycom- bining oneunitofAwithtwounits ofB.Initially Aweighs 20lbandBweighs 40lb.(a)If12lbofC’areformed in,1;hr,express :1:asafunction oftime in hours, where :0:isthenumber ofunits ofCformed intime t.(b)What isthe maximum possible value of2:? 6.Asaturated solution ofsaltwater willhold approximately 3lbofsaltper gallon. Ablock ofsaltweighing 60lbisplaced intoavessel containing 100 galofwater. In5min, 20lbofsaltaredissolved. (a)How much saltwillbedissolved in1hr? (b)When will45lbofsaltbedissolved? 7.Five grams ofachemical Aareplaced inasolution of100ccofaliquid B which, when saturated, willhold 10gofA.If2gofAareinthesolution in1hr,howmany grams ofAwillbeinthesolution in2hr? 8.Fifteen grams ofachemical Aareplaced into50ccofwater, which when saturated willhold 25gofA.If5gofAaredissolved in2hr,howmany grams ofAwillbedissolved in5hr? 9.Asubstance containing 10lbofmoisture isplaced inasealed room, whose volume is2000 cuftandwhich when saturated canhold 0.015 lbofmoisture percubic foot. Initially therelative humidity oftheairis30percent. Ifthe 138 Pnonu-zms Lnxnmo TOF1ns'r Oannn EQUATIONS Chapter 3 substance loses 4lbofmoisture in1hr,howmuch timeisrequired forthesub- stance tolose80percent ofitsmoisture content? Assume thesubstance loses moisture ataratethat isproportional toitsmoisture content andtothe difference between themoisture content ofsaturated airandthemoisture content oftheair. ANSWERS 15E 1.811;units. 2.1.80units. s1s[e"("_")' 1 zkt3_,,= , 1=‘Ii.81e"('1—'2)' _S2 l+81lCl 4.(5)7.4715. (b)5.6min. 5.(5)5=1801/(2+ 91). (b)2011». 6.(a)59.2lb. (b)18.2min. 7.3.04. 8.8.9g. 9.dx/clt =kx[30 —(19—:c)];3.8hr. LESSON 16. Motion ofaParticle Along aStraight Line— Vertical, Horizontal, Inclined. Inthis lesson wediscuss awide variety ofproblems involving the motion ofaparticle along astraight line. InLesson 34,weshall discuss themotion ofaparticle moving inaplane. ByNewton's ﬁrstlawofmotion, abody atrestwillremain atrest,and abody inmotion willmaintain itsvelocity, (i.e., itsspeed anddirection), unless acted upon byanoutside force. Byhissecond law, therate of change ofthemomentum ofabody (momentum =mass Xvelocity) is proportional totheresultant external force Facting upon it.Inmathe- matical symbols, thesecond lawsays (a) F=km% where misthemass ofthebody, vitsvelocity, andk>0isapropor- tionality constant whose value depends ontheunits used. Ifthese are foot fordistance, pound forforce, slug formass (=1/32 pound), second fortime, thenk=1and(a)becomes 11 <1’(15.1) F=mg':=nta=mJi1 where aistherateofchange invelocity, commonly called theacceleration oftheparticle, andsisthedistance theparticle hasmoved from aﬁxed point. Aforce of1lbtherefore willgive amass of1slug anacceleration of1ft/secz. Remember that F,a,andvarevector quantities, i.e.,they notonly have magnitude butalsodirection. (For adiscussion ofavector quantity, seeLesson 16C.) Hence itisalways essential inaproblem to indicate thepositive direction. Lesson 16A VERTICAL Morron 139 Ifwewrite do dods (b) atdsat andrecognize that v=ds/dt, then (b)becomes .1,.1(15.11) 3%=vi- Hence wecanalsowrite (16.1) as do(16.111) F-moa- Newton alsogave usthelawofattraction between bodies. IfmlandW62 arethemasses oftwobodies whose centers ofgravity arerdistance apart, theforce ofattraction between them isgiven by _ mlmg (15.12) F_15-T, . where k>0isaproportionality constant. LESSON 16A. Vertical Motion. Let, seeFig. 16.13, =mass oftheearth, assumed tobeasphere, =mass ofabody intheearth’s gravitational ﬁeld, =theradius oftheearth, =thedistance ofthebody above theearth’s surface. J/=0 T+@2053 my R M Figure 16.13 By(16.12) theforce ofattraction between earth andbody is(weassume their masses areconcentrated attheir respective centers) _114m_.(R+1/)2 The proportionality constant Gwhich wehave used inplace oflciscalled thegravitational constant. The negative signisnecessary because the(16.14) F=-0 140 Pnonmams LEADING T0Fmsr Onnmn EQUATIONS Chapter 3 resulting force acts downward toward theearth’s center, andourpositive direction isupward. Ifthedistance yofthebody above theearth’s sur- face issmall compared totheradius Roftheearth, then theerror in writing (16.14) as GMm(16.15) F-——RT isalsosmall. [R=4000 miles approximately sothat even ifyisashigh as1mile above theearth, thedifference between using (4000 X5280)2 feetand(4001 X5280)2 feetinthedenominator isrelatively negligible.] By(16.1) with yreplacing s,wecanwrite (16.15) as dzy_ GMm Since G,M,andRareconstants, wemay replace GM/R2byanew con- stant which wecallg.Wethus ﬁnally obtain forthedifferential equation ofmotion ofafalling body inthegravitational ﬁeld oftheearth, dzy dv(16.17) mag; =—gm, ma =—-gm, where v=dy/dt. Theminus signisnecessary because wehave taken the upward direction aspositive (seeFig.16.13) andtheforce oftheearth’s attraction isdownward. From (16.17), wehave d2 (16.18) ﬁg=—-g. Theconstant gisthustheacceleration ofabody duetotheearth’s attrac- tive force, commonly known astheforce ofgravity. Itsvalue varies slightly fordifferent locations ontheearth andfordifferent heights. For convenience weshall usethevalue 32ft/secz. Integration of(16.18) gives thevelocity equation d(16.19) v(=7?)=——gt+cl. And byintegration of(16.19), weobtain thedistance equation 93(16.2) y=-—-2-—|—c1t+ C2. Example 16.21. Aball isthrown upward from abuilding which is 64feetabove theground, with avelocity of48ft/sec. Find: 1.How high theballwillrise. 2.How long itwilltake theballtoreach theground. 3.The velocity oftheballwhen ittouches theground. Lesson 16A VERTICAL MOTION 14-1 Solution (Fig. 16.22). By(16.2), with g=32, (a) y=-16:2 +C,»+C2. . .. . y=64Differentiation of(a)gives (b) U=~32:+¢,. 1+ Iftheorigin istaken atground level, the Ground initial conditions aret=0,y=64, y=0 v=48.Inserting these values in(a)and (b),weobtain Figure 16.22 (c) c2=64, c,=48. Hence (a)and(b)become respectively (d) y=-16¢’ +48¢+64, v=-32¢+4s. The ballwillcontinue toriseuntil itsvelocity iszero. By(d),when v=0,t=1.5seconds, andwhen t=1.5seconds, y=100feet. Hence theballwillrise100feetabove theground. When theball isatground level, y=0,and by(d)when y=0, t=4seconds. Hence theball willreach theground in4seconds. Its velocity atthat moment willthen be,bythesecond equation in(d), v=(—32)(4) +48=-80 ft/sec. Thenegative signindicates that theballismoving inadownward direction. Comment 16.23. Intheabove example, weignored thevery im- portant factor ofairresistance. Inarealsituation, thisfactor cannot be thus ignored. Airresistance varies, among other things, with airdensity andwith thespeed oftheobject. Furthermore, airdensity itself changes with height andwith time. Itisdifferent fordifferent heights andmay bedifferent from daytoday. Thefactor ofairresistance inarealproblem isthus acomplicated one. When, therefore, weassume intheexamples which follow, aconstant atmosphere andanairresistance which isdependent only onthespeed of theobject, wehave simpliﬁed thepractical problem enormously. And when inaddition wesuppose that thissimpliﬁed airresistance ispropor- tional toanintegral power ofthespeed, wehave simpliﬁed theproblem considerably further. There isnovalid reason why airresistance may not beproportional tothelogarithm ofthespeed ortothesquare root ofthe speed, etc. Inallcases, however, airresistance always actsinadirection tooppose themotion. 142 PROBLEMS LEADING TOFIRST ORDER EQUATIONS Chapter 3 Example 16.24. Abody ofmass mslugs isdropped from aheight of 5000 feet. Find thevelocity andthedistance itwillfallintime t.Assume that theforce oftheairresistance isproportional totheﬁrstpower of thevelocity, theproportionality constant being m/40. Solution (Fig. 16.241). Theforce oftheairresistance isgiven as (m/40)v. Thedownward force duetotheweight ofthemass mismg pounds. Hence thedifferential equation of y=° motion (16.17) must bemodiﬁed toread, with thepositive direction downward (remem- bermass Xacceleration ofabody =thenet 1+forces acting upon it), (a) mg -m—-ﬂuy=5,000 Ground dg"'79 4()‘ Figure 16.241 Note thattheforce ofgravity gmisnowposi- tivesince itactsinthechosen positive direc- tion. This equation canbesolved bythemethod ofLesson 6Cor11B. Using thelatter method, wewrite (a)as <1») §+g1»=9- Theintegrating factor by(11.12) ise"‘°. Thesolution of(b)istherefore (c) v=40g+c1e“‘/4°. Integration of(c)gives (d) y=40gt——40c;e_'/4° +cg. Iftheorigin istaken atthepoint where thebody isdropped, then the initial conditions aret=0,v=0,y=0.Substituting these values in (c)and(d),weﬁnd (e) c1=——40g, cg=-—1600g. Hence thetworequired equations are (f) v=40g(1 —e“"°), y=40g(t +40c-"‘° —40). Comment 16.25. We seefrom (f),that ast—+oo,thevelocity v—>40g. This means thatwhen aresisting force ispresent, thevelocity does notincrease indeﬁnitely with time butapproaches alimiting value beyond which itwillnotincrease. This limiting velocity iscalled the terminal velocity ofthefalling body. Inthisexample, itis40gft/sec. Lesson 16A VERTICAL MOTION 14-3 Example 16.26. Theproblem andinitial conditions arethesame asin Example 16.24 excepting thattheforce oftheairresistance isassumed to beproportional tothesecond power ofthevelocity. Find thevelocity of thebody asafunction oftime andalsotheterminal velocity ofthebody. Solution. Here theforce oftheairresistance is(m/40)v2. Hence (a)ofExample 16.24 must bemodiﬁed toread <1 m at40g——v2 at 1 (“)"‘i=’""“E"2' E: 40’40g-»2=40‘”' Integrating thelastequation in(a)andinserting theinitial conditions as limits ofintegration, weobtain ' dv __1 t “’> -/i:=0 _ 40/»-ed"Itssolution is 4\/10g 2\/10g -v4° 2\/10g -1»T Solving thelastequation forv,weobtain <~/—/10> _((1) ,,=2./109 ii ,eh/10-0/10): +1 which gives thevelocity ofthebody asafunction oft. Ast—-> oo,weseefrom (d)that v—>2\/10g. This istheterminal velocity ofthebody. Example 16.27. Araindrop falls from amotionless cloud. Find its velocity asafunction ofthedistance itfalls. Assume itissubject toa resisting force which isproportional tothesecond power ofthevelocity. Also ﬁnditsterminal velocity. Solution. Taking thedownward direction aspositive, thedifferential equation ofmotion (16.17) must bemodiﬁed toread (a) 111% =mg—kvz. where k>0isaproportionality constant. Since wewish toﬁnd vasa function ofthedistance y,wereplace dv/dt byitsequal asgiven in(16.11). Hence (a)becomes dv vd dy(b) mv@=1w—Iw’. W_:')T,,2=;' Ifthecloud istaken astheorigin, then theinitial conditions arey=0, 144 PROBLEMS LEADING 'roFmsr ORDER EQUATIONS Chapter 3 v=0.Integration of(b)andinsertion oftheinitial conditions give 9 ll vdv 1fii =__ d (0) /i-=0 "'9—M2 mu-0 y’ whose solution is L"W—16"’)_1 "-2klog( mg -m2 mg ___kvﬂ = ,”Lge—2ky/m’ v2=-770-—q(1—e"'2"”/"‘). Ast—>oo,thedistance ytheraindrop falls approaches inﬁnity, andas y—>oo,weseefrom (d)that v2—>mg/lc. Hence theterminal velocity is (e) T.v.=\/mg/lc. Note. Since thebody isfalling andthedownward direction ispositive, thepositive square rootmust betaken forthevelocity inthelastequation of(d). Comment 16.28. The terminal orlimiting velocity hasnoyinit, andistherefore independent oftheheight from which theraindrop falls. Itisalsoindependent oftheinitial velocity. From actual experience we know that araindrop reaches itslimiting velocity inaﬁnite andnotin aninﬁnite time. This isbecause other factors also operate toslow the raindrop’s velocity. Comment 16.29. Abody falling inwater encounters aresistance justasdoesthebody falling inair.Ifthemagnitude ofthevelocity issmall, theresistance ofthewater isapproximately proportional totheﬁrst power ofthevelocity. The differential equation ofmotion (16.17) there- forebecomes, with thedownward direction positive, d(a) mF'Z= mg—kv, which issimilar to(a)ofExample 16.24. Example 16.3. Aman with aparachute jumps atagreat height from anairplane moving horizontally. After 10seconds, heopens hispara- chute. Find hisvelocity attheendof15seconds andhisterminal velocity (i.e., theapproximate velocity with which hewillﬂoat totheground). Assume that thecombined weight ofman andparachute is160pounds, andtheforce oftheairresistance isproportional totheﬁrst power ofthe velocity, equaling Q12when theparachute isclosed and 10vwhen itis opened. Lesson 16A VERTICAL MOTION 145 Solution. Fortheﬁrst10seconds offall,thedifferential equation of motion (16.17) ofthemanis,with positive direction downward, d(a) mi=ma—iv- Here thedownward force mgisequal to160pounds andthemass m= 160/32. Hence (a)becomes 160dv dv 1(b) -52-'Ft'-—160'—‘}1), gt‘-i-E1)-32. Itssolution, bythemethod ofLesson 11B, is (c) v=320+ce‘°'“. Ifwetake theorigin atthepoint ofjump, then t=0,v=0.Hence by (c),weﬁndc=-320 sothat (d) v=320(1 —-e'°'“). Whent =10 (e) v=320(1 —e_1) =320(0.6321) =202.3 ft/sec. Starting with thetenth second, thedifferential equation (16.17) be- comes (remember theresistance isnow10v) 160.1 .1(r) §§£=160-10.), J';+2»=32, whose solution is (g) v=16+ce_2‘. Inserting in(g)theinitial condition which, by(e),ist=0,v=202.3, weﬁndc=186.3. Hence (g)becomes (h) v=16+186.3e_2‘. When t=5,i.e.,5seconds after theparachute opens and15seconds after hisjump, (i) v=16+1s6.3¢-1° =16+186.3(0.000045) =16+0.008 =16.008 ft/sec. The terminal velocity is[in(h)lett—>co]16ft/sec. Weseefrom (i), therefore, that only 5seconds after theparachute isopened, theman is already ﬂoating toearth with apractically steady velocity of16ft/sec. 146 PROBLEMS LEADING TOFmsr ORDER EQUATIONS Chapter 3 Comment 16.31. Inderiving formula (16.17) foravertically falling body, weignored thedistance yoftheobject above theearth’s surface, since weassumed ittoberelatively small incomparison with theradius Roftheearth. If,however, thedis- "‘ tance oftheobject isvery farabove T+ theearth’s surface, thenthisdistance ' cannot bethusignored. Inthiscase (16.14) becomes Mm(16.32) F=-G7, M r=0 where Misthemass oftheearth con- Figure 16.33sidered asbeing ‘concentrated atits center andristhedistance ofthe body ofmass mfrom thiscenter (Fig. 16.33). Replacing in(16.32) thevalue ofFasgiven in(16.1), weobtain dv Mm dv GM Since Gand Mareconstants, wecanreplace GM byanew constant k. There results2(16.35) dv_ Ic dr_ lc ?1i__r_2’ W“—F’ where v=dr/dt. From (16.35) wededuce that the acceleration of abody inthegravitational ﬁeldoftheearth varies inversely asthesquare ofthedistance ofthebody from thecenter oftheearth. Example 16.36. Abody isshot straight upfrom thesurface ofthe earth with aninitial velocity vo.Assuming noairresistance, ﬁnd: 1.The velocity vofthebody asafunction ofthedistance rfrom the center oftheearth. 2.Itsvelocity when itis4000 miles above theearth’s surface. 3.How highthebody willrise. 4.Themagnitude oftheinitial velocity voinorder that thebody may escape theearth, i.e.,inorder that itmay never return totheearth. 5.The time tasafunction ofthedistance rofthebody from theearth’s center. Solution (Fig. 16.361). Wetake theorigin atthecenter oftheearth, andcallRtheradius oftheearth. Then, by(16.35), dv k(8,) a:Ft-;——'f2- Lesson 16A VERTICAL MOTION 147 Substituting in(a),theinitial conditions r=R,a=—g,weﬁndIc=gR2. Hence (a)becomes dv_ gR2 ‘bl s"-.—2' Since wewish toﬁndvasafunction ofthedistance r,wereplace dv/dt by itsequivalent value asgiven in(16.11). Hence (b)becomes at QR2 gR2(O) t)$=-7: 1)£l1)=——7_§~llT. Integration of(c)and insertion oftheinitial conditions v=vo,r=R, givell I‘ d(d) /lgvovdv =-—gR2‘£=RTr,;, whose solution is 2R2 R(e) v2=v02+—g-r———2gR=vo2+2gR(T—1)- Hence theanswer toquestion 1is (f) v==|=,iv02+2gR(€i—1); thepositive sign istobeused when thebody isrising, thenegative sign when itisfalling. When thebody is4000 miles above theearth’s surface, 1+ Surface oftheearth r=R 7' R r=0, center ofearth Figure 16.361 r=8000 (R=4000 miles approximately). Inserting thisvalue in(f),we obtain (g) v==|=1,1202 +2gR ——1)==:!=\/v02 —4000g, which istheanswer toquestion 2. 148 Pnonmams LEADING roFIRST ORDER EQUATIONS Chapter 3 The body willcontinue toriseuntil v=0.Hence by(f) R 2R’0=f)o2-j-2QR('T—'1)r T= ; which isthedistance thebody willriseabove thecenter oftheearth if ﬁred with aninitial velocity vo.Subtracting Rfrom thisvalue willgive thedistance thebody will riseabove theearth’s surface. This isthe answer toquestion 3. Thebody willescape theearth, i.e.,itwillnever return totheearth, ifrincreases with time. This means wewant rtobecome inﬁnite asthe velocity vofthebody approaches zero. By(e)weseethatif1202=2gR, then r—>coasv—>0.Hence theanswer toquestion 4is (1) to=\/2gR =\/(2)(s2)(4oo0)(s2s0) =36,765 ft/sec =7mi/sec,* approx., =25,100 mi/hr, approx., which istheescape velocity ofabody ifairresistance isignored. The answer toquestion 5issomewhat more difficult toobtain. In(e) replace vbydr/dt andlet (j) a==2gR2, b=v02—2gR. Hence (e)becomes <1.)..=§=i,/-§+b=i,/93-;l=s.}./.;;Tz72. When thebody isrising, thevelocity ispositive andwecan,therefore, write (k)as (D dt= rdr =1(a+2br)dr_a dr _ var +br2 2b\/ar —|—br2 2bVar +brz Ifweassume b<0,i.e.,ifweassume [see(j)]v02<2gR, sothat the body cannot escape theearth, thenintegration of(l)gives (m)t=c+%\/ar+br2 - rcsin b<0, where a,bhave thevalues given in(j). Substituting in(m)theinitial conditions t=0,r=R,weobtain 1 a .—2bR—a(n) c=—FVaR+bR2+m)i\/:EArcsin(——7-i)» b<0. With thisvalue ofc,(m)deﬁnes tasafunction ofrforarising body. ‘With g=32ft/sec’, vo=6.96mi/sec instead of7mi/sec. However, giscloser to 32.17 ft/sec’. With thisvalue ofg,vo=6.98mi/sec. Lesson 16A VERTICAL MOTION 149 Remark. When thebody isfalling, visnegative. Hence forafalling body, wemust, in(k),take <0) »=—\/Q. inorder toarrive atanequation comparable to(1)above. Or,ifyouwish, youmayuseformula (d)ofExample 16.38 following. Itgives thetimeof afalling body asafunction ofrwith initial conditions t=0,v=0, T=1'0. Ifweassume thatb=0,i.e.,ifweassume v02=2gR[see(j)]sothat thebody willescape theearth, then (e)becomes (p) v=%€=——%gR, rl/2dr=\/2gRdt, §r3/2=\/%Rt+C'. When t=0,r=R.Hence c=§R3'2. Therefore thelastequation in (p)becomes 2 () t: i_ 3/2 _ R3/2). q 3R\/260 Comment 16.37. 1.Note from (16.35) that, asr-—>oo,theaccelera- tiondzr/dt2 duetothegravitational force oftheearth approaches zero. This means that theinﬂuence oftheearth’s gravitational ﬁeld, although never zero, becomes insigniﬁcant. 2.From (e)weobserve that when 1202=2gR, theescape velocity of thebody, thevelocity equation reduces tov2=2gR2/r. Hence asrgets larger, thevelocity ofthebody willcontinue togetsmaller until such time asitenters thegravitational ﬁeld ofanother heavenly body. And if v02>2gR sothat v02——2gR equals apositive constant I02,then the . , 2R2velocity equation (e)reduces tov2=k2+gi.Hence asr—>co, v—>k. T 3.From equation (q)above, which expresses time asafunction ofr with v02=2gR, weseethat talsoapproaches inﬁnity asr—>co. Example 16.38. Abody falls from interstellar space atadistance rofrom thecenter oftheearth. Find: 1.Itsvelocity vasafunction ofthedistance r,where rismeasured from thecenter oftheearth. 2.Itsvelocity when itreaches thesurface oftheearth. 3.The time tasafunction ofthedistance r. Take theearth’s center astheorigin andtheoutward direction aspositive. Solution. Thedifferential equation ofmotion ofthebody isthesame asthat of(c)intheprevious example. Integration ofthisequation and 150 PROBLEMS LEADING T0Fmsr ORDER EQUATIONS Chapter 3 insertion oftheinitial conditions gives 1? T =_2Q. (a) ];=ovdv gR/Jam T2 Itssolution is 2___ 2_1___L(b) v--2gR (r T0), which istheanswer toquestion 1. When r=R,i.e.,when thebody isatthesurface oftheearth, its velocity is,by(b), (c) v=-—,i2gR—2iR3-To Thenegative signisneeded because thebody ismoving toward theearth andtheoutward direction ispositive. This istheanswer toquestion 2. From (c)weseethat if1'0isvery large, i.e.,ifthebody isextremely far away from theearth, then 2gR”/ro isvery close tozero, and|v|isextremely close to,butlessthan \/2gR. This means that abody falling from outer space cannever exceed avelocity equal to\/2gR. Aswesawin(i)of Example 16.36, \/2gR =25,100 miles/hour. Hence, ifairresistance is ignored, abody falling from interstellar space willhave avelocity atthe earth’s surface which differs extremely little from 25,100 miles/hour. Weleave ittoyouasanexercise toshow that theanswer toquestion 3is (6) t=l§%[\/it _.2+L;-Arcsin \/;;( r 2r-r=_____./ _2+_°Ac - R‘/% ror r 2rcos To Hint. Start with (b)andfollow themethod weused inExample 16.36 toﬁndtheanswer toquestion 5.You donotneed tomake thesubstitu- tions (j)ofExample 16.36. Comment 16.39. Insolving thetwoprevious problems, weassumed noairresistance. Since there isairresistance, theescape velocity vo would have tobesuﬁiciently greater than \/2gR =25,100 miles/hour to overcome thisresistance. If,however, thebody emerged from theearth’s atmosphere, which israre 100miles above itssurface,* with avelocity equal toorperhaps very slightly more than 25,100 miles/hour, itwould escape theearth. Conversely, theformula forthevelocity ofabody falling from outer space willgive fairly accurate results until theobject reaches theearth’s atmosphere orabout 100miles from itssurface. Con- ‘Acalculation made from ananalysis ofoneofoursatellite’s orbits shows that the density ofatmosphere at932miles above theearth isonethousand million millionths thedensity ofairatsealevel. Lesson 16A-Exercise 151 sidering thegreat distances involved, 100miles isrelatively insigniﬁcant, butitsimportance istremendous. Itcomplicates thewhole problem of exitandreentry ofsatellites. EXERCISE 16A 1.Verify theaccuracy oftheanswer asgiven inthetext toquestion 3of Example 16.38. Inproblems 2—5, assume noairresistance andthat theobject is near theearth’s surface. 2.Aballisthrown vertically upward from theground with aninitial velocity 'of80ft/sec. (a)Find itsvelocity anddistance equations asfunctions oftime. Take the origin atthepoint where theballisthrown andtheupward direction aspositive. (b)What areitsvelocity andheight attheendof1sec? (c)How longandhowhigh willitrise? (d)When willitreach theground andwith what velocity? (e)Would theresults differ iftheballwere aprojectile weighing 5tons? 3.Aman leans over thesideofabridge anddrops astone. Hisstop watch shows that thestone touched thewater in2.1seconds. How high isthe bridge above thewater? 4-.Aballisgiven adownward velocity of8'ft/sec from aheight of120ftabove theground. (a)When willitreach theground andwith what velocity willitstrike the ground? (b)How much lower must onestand inorder todrop aballandhave itreach theground atthesame time astheﬁrstball‘? (c)With what velocity willthesecond ballstrike theground? (d)What isthesigniﬁcance ofthenegative signof-3seconds obtained in(a)? Hint. Substitutet =-3inyour velocity equation. Then solve thisproblem: Ifaballisthrown upward from theground with aveloc- ityof88ft/sec, howhigh willitgo;atwhat height willitsvelocity be 8ft/sec downward; when willitreach thisheight andvelocity? 5.Aperson, 81ftabove theground, drops anobject. With what velocity must asecond person 180ftabove theground throw anobject straight down in order thatboth objects reach theground atthesame time? Intheproblems below, weight inpounds isequal tomass inslugs times theacceleration ofgravity infeetpersecond persecond, i.e., (16.391) W=mg, m=W/g. 6.Aman weighs 160lbonearth. (a)What ishismass? (b)Theacceleration ofgravity onthesurface ofthemoon isapproximately one-sixth that oftheearth. What willheweigh onthesurface ofthe moon? (c)Find formulas comparable tothevelocity and distance equations (16.19) and(16.2) forthesurface ofthemoon. 152 Pnosums LEADING roFmsr ORDER Eouzmons Chapter 3 7.Aballthrown vertically upward from thesurface oftheearth withavelocity of64ft/sec willreach amaximum height of64ftin2sec(verify it).If thrown with thesame velocity onthesurface ofthemoon, ﬁnd, bymeans oftheformulas developed in6(c),comparable ﬁgures forthemaximum height reached andthetime needed toattain thisheight. 8.Ifaman canhigh-jump 5ftonearth, howhigh willhejump onthemoon andhowmuch longer willhebeintheairascompared with thetime inthe airontheearth? Assume hiscenter ofgravity is2ftfrom thetopofthe bar. Hint. Seeproblem 7. 9.Aman whose weight is160lbisinanelevator which isdescending with an acceleration of2ft/sec2. What ishisweight while riding intheelevator? Hint. Use(16.391) ;remember hismass isconstant, andwhen theelevator accelerates down, heaccelerates up. Inproblems 10-17 and20,21,assume that theforce Roftheairre- sistance isproportional tothe ﬁrst power ofthe velocity, i.e., R=kv,andthat thefalling orrising body isnear theearth’s surface. 10.Abody ofmass misdropped from agreat height. (a)Find itsvelocity andthedistance itfalls asfunctions oftime. Take thepositive direction asdownward andtheorigin atthepoint where thebody isdropped. Hint. In(a)ofExample 16.24 replace m/40 byk. Useyour results tocheck theaccuracy oftheanswers given in(f). (b)What isitsterminal velocity? ll.Solve problem 10ifthebody initially isgiven adownward velocity of v0ft/sec. What isitsterminal velocity? Compare with 10(b) above. Note thattheinitial velocity does notaffect theterminal velocity. 12.Abody weighing 192lbisdropped from agreat height. The proportionality constant Icoftheairresistance is12. (a)Find itsvelocity andthedistance itfallsasafunction oftime. Solve independently. Useresults obtained in10onlyasacheck. (b)What isitsterminal velocity? (c)How fardoes itfallin10sec? What isitsvelocity atthatmoment? 13.Solve problem 12ifthebody isgiven aninitial downward velocity of 170ft/sec instead ofbeing dropped. Solve independently. Usetheresults obtained in11onlyasacheck. 14-.When aparatrooper falls freely from agreat height before opening hischute, histerminal velocity isapproximately 175ft/sec. Assume aparatrooper andhischute together weigh 200lb. (a)Find theproportionality factor lcoftheairresistance. Hint. Usethe formula forT.V. found in10(b). (b)Find hisvelocity andthedistance hefallsasafunction oftime. (0)What ishisvelocity attheendof8%sec,17%sec,26}sec,32%sec? (d)How farhashefallen in32%sec? 15.Assume theparatrooper ofproblem 14opens hischute when hehasreached histerminal velocity of175ft/sec, andthat hischute isdesigned togive himasafelanding speed of16ft/sec. (a)What isthenew value ofIo?Hint. Use theformula forT.V. found in problem 11. (b)Find hisvelocity andthedistance hefallsafter heopens hischute as functions oftime. Lesson l6A—Exercise 153 16. 17 18. 19. 20. 21.(c)What ishisvelocity attheendof1sec,2sec,3sec,4sec,5sec? (d)How farhashefallen in5sec? (e)Doyour answers in(c)and(d)suggest asafeheight atwhich hecan open hischute? (f)Ifheopens hischute ataheight of1040 ft,inhowmany seconds does hereach theground? Aparatrooper jumps from aplane ﬂying horizontally atagreat height- When hefeels that hehasreached asteady velocity (i.e., when hehas reached histerminal velocity), heopens hischute. Assume thissteady veloc- ityis180ft/sec. (a)Find hisvelocity andthedistance hefallsasafunction oftime before thechute opens. Hint. Useformula forT.V. found in10(b) andsolve form/k. (b)Calculate hisvelocity attheendof11}sec,22}sec,33$sec,45sec. (c)How farhashefallen in45sec? Iftheforce oftheairresistance is50lbwhen abody isfalling atavelocity of25ft/sec, what isthevalue oftheproportionality constant koftheair resistance? Forthisk,ﬁndtheterminal velocity ofafalling body weighing 100lb.Ifthebody hasaninitial velocity of20ft/sec, ﬁnditsvelocity and distance equations asfunctions oftime. Abody weighing 96lbbegins tosinkassoon asitisplaced inwater. Two forces actonittooppose itsmotion, anupward force duetothebuoyancy oftheobject andaforce duetotheresistance ofthewater. Assume the buoyant force is12lbandtheresistance ofthewater is6v.Take theorigin onthesurface ofthewater anddownward direction aspositive. (a)Find thevelocity andposition ofthebody asfunctions oftime. (b)Find itsterminal velocity. Thespeciﬁc gravity ofabody isdeﬁned astheratio ofitsweight tothe weight ofanequal volume ofwater. Assume abody isreleased from thesurface ofamedium whose speciﬁc gravity isone-fourth thatofthebody andthatthe resistance offered bythemedium ismv/3. (a)Find thevelocity ofthebody asafunction oftime. Hint. Thespeciﬁc gravity ofthemedium equal to5;thatofthebody, implies thatthemedi- um’s upward buoyant force isone-fourth theweight ofthebody. (b)Find itsterminal velocity. Abody ofmass misshotstraight upfrom theground withaninitial velocity ofvoft/sec. (a)Find thevelocity andposition ofthebody asfunctions oftime. Take thepositive direction upwards andtheorigin ontheground. (b)How highwillthebody riseandwhen willitreach thismaximum height? Anobject weighing 64lbisshot straight upfrom theground with aninitial velocity of96ft/sec. Assume theforce oftheairresistance is4v. (a)Find thevelocity andposition oftheobject asfunctions oftime. (b)How high will thebody rise and when will itreach this maximum height? Check theaccuracy ofyour answers in(a)and(b)with the formulas obtained inproblem 20. (c)When andwith what velocity willitstrike theground‘? Note thatthe down trip takes longer than theuptrip, whereas when there isnoair 154 Pnostmus LEADING roFrasr ORDER Equarrons Chapter 3 resistance both times arethesame. Note alsothat thereturn velocity issmaller than theinitial velocity. (d)Doyoubelieve youwould getthesame answer forthevelocity ofthe falling body when itstrikes theground andforthetime ofthedown tripifyouused theformulas forafalling body asfound inproblem 10, with yhaving thevalue determined in(b)? Tryit. Inproblems 22-30, assume that theforce Rofairresistance ispro- portional tothe second power ofthe velocity, i.e., R=kvz, and that thefalling orrising body isnear theearth’s surface. 22.InExample 16.27, wefound thevelocity vofafalling body asafunction of thedistance fallen. Starting with equation (a)ofthisexample, ﬁndthe velocity anddistance ofafalling body asfunctions oftime. Find itsterminal velocity. Compare with (e)ofExample 16.27. 23.Abody ofmass mfallsfrom agreat height. Itsinitial velocity isvoft/sec. Find: (a)Itsvelocity asafunction ofthedistance fallen. Take thepositive direc- tion downwards andtheorigin atthepoint offall. (b)Itsvelocity asafunction oftime. (c)Itsterminal velocity. Compare with (e)ofExample 16.27 andwith problem 22.Note thattheinitial velocity does notaffect theterminal velocity. 24.Abody ofmass misﬁred vertically upward from theground ataninitial velocity ofvoft/sec. (a)Find itsvelocity asafunction ofitsheight. Take positive direction upwards andorigin onground. (b)Find itsvelocity asafunction oftime. (c)Find itsheight asafunction oftime. (d)When willitreach maximum height? (e)How high willitrise? Nora. These formulas arevalid onlywhile thebodyisrising When itbegins tofall,formulas developed inExample 16.27 andproblem 22must beused. Compare with problem 21where weused oneformula tocalculate thetime foraround trip. 25.With what velocity willthebody ofproblem 24return totheearth andhow long willittake foritsdescent? Hint. Read note inproblem 24.Toﬁnd thevelocity, use(d)ofExample 16.27 with yequal tothevalue ofthe maximum height asfound inproblem 24(e). Note thatthereturning velocity islessthan theinitial velocity vo.Toﬁndthetimeofdescent, usetheformula foryasfound inproblem 22.Here itisnoteasy toseethat thetime of descent islonger than thetime ofascent. 26.Thevelocity ofaparachutist atthemoment hischute opens is160ft/sec. Theforce oftheairresistance ismvz/8. Find: (a)Hissubsequent velocity anddistance asfunctions oftime. Take the origin atthepoint where thechute opens and thepositive direction downwards. (b)Hisvelocity 1secafter hischute opens; 2secafter. (c)Histerminal velocity. (d)I-Iow farhefallsintheﬁrstsecond; inthesecond second. (e)Approximately when hereaches theground ifheis1064 ftabove the earth when hischute opens. Lesson 16A—Exe1-cise 155 27.Abody ofmass misdropped from aplane ﬂying horizontally 1mileabove theearth. Theforce oftheairresistance is2mk2v2. Theterminal velocity ofthebody is100ft/sec. Find: (a)Thevalue oftheconstant k.(Hint. T.V. =\/mg/k; replace loby2mk2.) (b)Thevelocity ofthebody asafunction oftime. (c)Thevelocity ofthebody attheendof3sec. (d)When thebody reaches avelocity of60ft/sec. 28.Abody fallsfrom agreat height. Itsterminal velocity is10ft/sec. (a)Find itsvelocity anddistance equations asfunctions oftime. Hint. T.V. = \/mg7E. Solve fork/m. (b)Find itsvelocity equation asafunction ofdis- tance. 29.Aparatrooper andhischute, which together weigh 192lb,drop from an airplane moving horizontally. Heopens hischute attheendof10sec. Assuming theproportionality constant ofairresistance is1/120 when the chute isclosed and4/3when itisopen, ﬁnd: (a)Hisvelocity asafunction oftime before thechute isopened. (b)Histerminal velocity before thechute isopened. (c)Hisvelocity attheendoftheﬁrst10sec. (d)Hisvelocity asafunction oftime after thechute isopened. (e)Histerminal velocity after thechute isopened. (f)Hisvelocity attheendof15sec,i.e.,hisvelocity 5secafter thechute is opened. Solve independently andthen check your results withtheformulas found in problems 22and23. 30.Amanandhisparachute weigh 192lb.Assume thatasafelanding velocity is16ft/sec andthatairresistance isproportional tothesquare ofthevelocity, equaling %lbforeach square footofcross-sectional area oftheparachute when itismoving at20ft/sec atright angles tothedirection ofmotion. What must thecross-sectional areaofaparachute beinorder thatthepara- trooper land safely? Hint. First ﬁndtheforce oftheairresistance. Then ﬁndksuch that T.V. =16.Then ﬁndthenumber ofsquare feetofpara- chute thatwillmake theforce oftheairresistance equal tokvz. Inproblems 31-39, assume noairresistance and that theobject is farenough from theearth sothat equation (16.35) applies. Usethefol- lowing data: R=4000 miles, g=32ft/secz, \/2gR =6.96 mi/sec. 31.With what velocity must arocket beﬁred inorder toreach aheight of400 miabove theearth; 4000 miabove theearth? Solve independently. Check your results with (h)ofExample 16.36. 32.The“air” 200miabove theearth issothinthatitwillhardly slow aspace vehicle. Itiscalled theF-2region oftheatmosphere. What velocity should arocket have at200miles above theearth, ifallitsfuelisexhausted atthat point, inorder togoanother 3800 mi? 33.Abody isshot toaheight of400miandthen starts tofall. What isits velocity (a)when ithasfallen 200miand(b)when itisatthesurface of theearth? Hint. Use(a)ofExample 16.38 with 1'0=R+400=4400. Note that thevelocity, when itreaches theground, isthesame asthe velocity required topropel it400miles upward. Seeproblem 31. 34.Assume abody falls from rest atadistance of61R, i.e., atadistance of 244,000 mifrom thecenter oftheearth (equivalent tothemoon’s distance from thecenter oftheearth). With what velocity andinhow many hours willitreach theearth? 156 Pnoansms Lmnme roFrasr Onnaa Eqmrrons Chapter 3 Abody isﬁred straight upwith aninitial velocity equal totheescape veloc- ity\/ Find: (a)Thevelocity oftheprojectile asafunction ofthedistance rfrom the center oftheearth. (b)When itwillhave reached 244,000 mi,thedistance ofthemoon from thecenter oftheearth. Hint. Use(e)and(q)ofExample 16.36. Abody isﬁredstraight upwithaninitial velocity vowhose magnitude isless than escape velocity. When willitreach itsmaximum height? Hint. The maximum height thebody willreach isgiven in(h)ofExample 16.36. Express thisvalue ofrinterms ofaandbasdeﬁned in(j).Then use(m) and(n). Nora. This time equation isvalid only forarising particle. If youwish tocompute itsreturn time, youmust usethetime equation given in(d)ofExample 16.38, with r0equal totheheight from which itbegins itsfall. Show thatifvoisvery much lessthan theescape velocity \/2g}iR, then the time fortheobject toreach itsmaximum height, asgiven inproblem 36, isapproximately v0/g. Hint. Replace theArcsinfunction byitsseries ex- pansion, Arc sin2:=:2:+2:3/6 +~-'.Then eliminate 1202and higher powers ofvo.Toseesome justiﬁcation forthiselimination, letvo=~310- mi/sec inthetime equation ofproblem 36. Abody isshot straight upfrom thesurface ofthemoon with aninitial velocity vo.(a)Find thevelocity vofthebody asafunction ofitsdistance r,,,from thecenter ofthemoon. (b)Find theescape velocity ofthebody. Theradius ofthemoon isapproximately 1080 mi;theacceleration ofgravity onthesurface ofthemoon isapproximately one-sixth that oftheearth. Take theoutward direction from themoon aspositive. (a)Prove thatifaparticle were placed approximately nine-tenths ofthe distance Dfrom thecenter oftheearth tothecenter ofthemoon onaline connecting moon toearth, Fig.16.392, theparticle would beatrest,i.e.,the gravitational pulls ofmoon andearth onaparticle placed nine-tenths ofthe distance from thecenter ofearth tothecenter ofthemoon areequal. As- sume themass ofthemoon is1/81 themass oftheearth. Hint. Apply (16.32) toboth earth andmoon. Then equate thetwoforces. : T D-7' Earth Moon I m /4 Neutral point so _Q10 10 D Figure 16.392 (b)Setupthedifferential equation ofmotion ofaparticle shotfrom the surface oftheearth toward themoon, considered ﬁxed, taking intoaccount both theearth’s andmoon's gravitational attractions. Solve theequation Lesson 16A-Answers 157 4-0. 2 3 4. 5 6 7 8 9 10. ll 12 13 14with t=0,v=vo.Take thepositive direction asoutward from theearth, andtheorgin attheearth’s center. (c)Atwhat velocity must theparticle beﬁred inorder toreach the neutral point? Hint. Inthevelocity equation found in(b),youwant v=0,when r=ﬁ;D. Assume R,,.2 =6R2/81, D=61R+R/4, g...=g/6,where R...istheradius ofthemoon andg...istheacceleration dueto theforce ofgravity ofthemoon. Note thesmall effect ofthemoon’s gravita- tional attraction. (d)Atwhat velocity must aprojectile beﬁredinorder toreach themoon? Ifabody were dropped inaholebored through thecenter oftheearth, it would beattracted toward thecenter with aforce directly proportional to thedistance ofthebody from thecenter. (a)With what velocity willitpassthecenter? (b)When willitreach theother endofthehole? NOTE. The motion ofthebody isknown assimple harmonic motion. A fuller discussion ofthismotion canbefound inLesson 28. ANSWERS 16A (a)v=—32t +80,y=—16t2 —|—80t. (b)48ft/sec, 64ft. (c)2%sec,100ft. (d)5sec,—80 ft/sec. (e)No. 70.56 ft. (a)2%sec,88ft/sec. (b)20ft. (c)80ft/sec. (d)If3secago, theballwere thrown upward from theground with a velocity of88ft/sec, itwould reach aheight of121ftandhave avelocity of8ft,/sec asitpassed the120ftpoint onitswaydown. 44ft/sec. 16‘ (a)5slugs. (b)26§lb. (c)v=—?+ c1,y=—1§t2+ c1t+ 62. 384ft,12sec. 15ftifoneassumes hescales thebarhorizontally sothat heraises his center ofgravity 2ftonearth; 6times aslong. 150lb. 2 e)»=%<1—e-""'">. 1/=%:+%<e""""'—1).(b)T.V. =mg/lc. v=% __ e—kl/m) + voe-kl/7!!’ 2y (1_e-kl/In)‘ T.V. =mg/Ic, same asin10(b). (a)v=16(1 —e’2‘), y=16t+ 8(e‘2‘ -1). (b)16ft/sec. (c)y=8(19+ e‘2°) =152ft,v=16(1 —(F20) =16ft/sec. (a)v=16(1-rm)+not-21 =16+154e'2‘, y=16z+77(1-vi").(b)16ft/sec. (c)237ft,16ft/sec. (=1)Iv=$-2(b)v=1750 _e-32¢/115), y=l75t+ 1372i (e-32¢/115 _1). (C)139.7ft/sec,167.9, 173.6,174.6. (d)4791a. 158 Paoauams LEADING 'roFmsr ORDER Eotwrrons Chapter 3 15.(1.)12.5. (6)v=16(1- 8"")+175$”, 1,=16z+1? (1-tr"). (6)37.5ft/sec,18.9,16.4,16.05,16.01. (<1)159.5rt.(e)Over 160ft. (f)Approx. 60sec. _ 1s0’ _16.(1.)v=1so(1-e8"“), y=1so¢+§- (e8"“-1). (b)v(11{) =l80(l ——0'2) =155.6 ft/sec, 176.7, 179.6, 179.9. (c)Approx. 7100 ft. 11.k=2;T.V.=soft/sec;v =so-30e‘"‘”‘° =so-s0e"°"‘-"“; y=50:-1339(1-6-"’5°) =50:-3%§(1-e'°"“‘). 18.(a)v=14(1 -F2‘), y=14[t——§(1—e-2‘)]. (b)T.V. =14ft/sec. 19.(a)v=72(1 ——e"/3). (b)T.V. =72ft/sec. 20.(a)v=%(e_k”"' -—1)—|—v0e_“"", 2 y=<%+%) (1—e"“"") —$1. kv kv(b)y=%l<v0 —%log »t =%log 21.(a)v=16(7e"2‘ —1),y=56(1 —e"2‘) —16t. (b)y=8(6-log7)=32.4ft,t=}log7 =0.97sec. (c)3.5secapprox., -16 ft/sec. (d)Same answer. 21HmI1nt _ 22.v=\/mg/ktanh\/glc/mt =\/nig/kff/.__i1 1e2kqImt+1 m m eV|1kI1nl +e-—‘\f||k/ml y=Ilogcosh\/gk/mt =Tlog?—-7-?-1 T.V. =\/mg/la ft/sec. Fordeﬁnitions ofcoshandtanh, see(18.91) and(18.92). 26.(1.)v2=%(1-¢"2"""") +vo”e"”‘""". Sincethepositive direction is downward, thepositive square rootmust betaken forvwhen thebody isfalling. (b)ﬂv+ =ﬂvo+- Wv—— \/Itvo- (6)T.V.=\/E71? ft/sec. 24.(a)v2=2%(e_2k"I"' ——1)-1-v02e_2k"/"'. Since thepositive direction is upward, thepositive square rootmust betaken forvwhen thebody is rising. (b)t=Vm/kg (Arctanx/k/mg vo—Arctan\/lc/mg v), orv=\/gm/k tan(c—\/kg/m t),Where c=Arctan\/k/gm vo. Theformula forthetime tisvalid only forarising body.asas Lesson 16A—Answers 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36.(0) (d) (e) 1) ﬂ= where c=Arctan\/It/gm v0.Fordeﬁnition ofcosh, see(1891) (a)v (b) (d) (a) (b)v (C) (a)v (b)v (a)v (b) (d)v (e) 600*3y=— t=c\/m/kg. my=Tlogsecc. V"'11/(IW02 +ma)voft/sec. \/m/gk cosh"1 seccorcosh\/gk/mt =secc,log[cos(c—Vlcg/mt)]_ COS C 11+9e““ _ 11—9e‘4‘ 16.49 ft/sec, 16.009 ft/sec. (c)16ft/sec. 29.5ft.,16.1ft. (e)66sec. k=1/25. e0.64z _I =100—-——~-e0.64: +1 74.4ft/sec. (d)2.2sec. 86.41 _1 10t8.I1h (3.20 =10T r e'—|—1 3.21 -3.2: $logcosh3.2t=1%-0log€-——_:i—— - uh/1___e—0.641/_ ezn/E/15 __1y= =151.8ezn/E/15+ 1 T.V. =151.8 ft/sec. (c)147.4 ft/sec. =121.18e16”3 +1_ 1.1se‘°‘"“ -1 12ft/sec. (f)12ft/sec. sqft. 2.10mi/sec, 4.92mi/sec. 4.7mi/sec. (a)1.45mi/sec. (b)2.10mi/sec. 6.9mi/sec, 119hours. (a) t b t=v=Rx/2g/r, (b)50.6hours. 1 a 1r -2bR —-——\/ R+bR2——i|:——Arcsin :1 a 2b\/—b 2 “ gR2 [vov 2gR -—002 (gR —002)] gR gR(2gR _v02)3/2 —|—Arccos 2 /ii UR a/2loo zgflt W2+2Amsin U0(2gR—62> ./fl=16i-,y =161+ 8log(5.5~4.56‘“) 160 Paoannms LEADING 'roFmsr Oannn EQUATIONS Chapter 3 2 2QR». R». - -3g_(9,)9=90+-5- -r--1,where R...1stheradius ofthemoon. (b)vo=\/gR,,,/3 =1.48mi/sec. dv_—gR2 g,,.R,,.239.(b)I»;-————r, +—(D_r),» R22,12,.’ 2,12,.’v2=2g-T+%+vo2 —2gR—%z. (c)vo=\/2gR (0.99) =99percent oftheescape velocity oftheearth when thegravitational pullofthemoon isignored. (d)Thevelocity must have apositive value when itreaches theneutral point. See(c). 4-0.(a)4.9mi/sec. (b)42.5 min. LESSON 16B. Horizontal Motion. Ifabody moves inahorizontal direction, asonatable orplatform, africtional force develops, which operates tostop theprogress ofthemotion. This frictional force isdueto 1.Thegravitational force oftheearth pressing thebody totheplatform. 2.The smoothness orroughness ofthesurface oftheplatform. This quality ofthesurface, i.e.,itsroughness orsmoothness, ischaracterized bymeans ofaletter pt,called thecoeﬂicient offriction ofthesurface. Deﬁnition 16.4. The frictional force ofabody moving onahori- zontal surface is,bydeﬁnition, equal totheproduct ofthecoeﬂicient of friction /1ofthebody and thegravitational force mg, i.e., frictional force =;.i(mg). Comment 16.41. From experience weknow that itrequires agreater force tobegin themovement ofanobject than itdoes tokeep itmoving. There arethus two coefficients offriction, onecalled static friction, which operates atthestart ofthemotion, theother called sliding friction, which operates after themotion hasbegun. Inaddition tothefrictional force, abody also may besubject toa resisting force duetotheairorother medium inwhich itmoves. Example 16.411. Anobject onasledispulled byaforce of10pounds across afrozen pond. Object andsledweigh 64pounds. The coefficients ofstatic andsliding friction arenegligible. However, theforce oftheair resistance istwice thevelocity ofthesled. Ifthesled starts from rest, ﬁnditsvelocity attheendof5seconds andthedistance ithastraveled in that time. What isitsterminal velocity? Solution. Here m=64/32 =2andtheforce oftheairresistance is given as2vpounds. Hence thedifferential equation ofmotion ofthesled Lesson 16B HORIZONTAL Morron 161 is(remember mass Xacceleration ofabody =netforce acting upon it) do dv(3.) 2E—10'-21), Itssolution is (b) 0=5+c1e_'. The initial condition ist=0,v=0.Hence cl=-5,and (b)becomes (c) v=5(1—e“‘). When t=5, (<1)v=5(1—rs)=5(1-0.0067) =5(0.9933) =4.97ft/sec. By(c)theterminal velocity isfound tobe5ft/sec. Toﬁndthedistance :1:traveled intime t,weintegrate (c)toobtain (9) 1”=5(l+6-‘) —|—62- When t=0and 1:=0(taking theorigin atthestarting position), we ﬁnd, from (e),C2=-5. Hence (e)becomes (f) x=5(t+e"‘--1). And when t=5, (g) :1:=5(5+e“5—1)=5(4+0.0067) =20ftapprox. Example 16.42. Aboyweighing 75pounds runs foraslide andreaches itatavelocity of10ft/sec. Ifthecoefficient ofsliding friction 11between hisshoes andtheiceis1/25, howfarwillheslide? Ignore wind resistance. Solution. ByDeﬁnition 16.4, thefrictional force is1/25 -75=3 pounds. Since thisistheonly force which isopposing themotion andthe mass oftheboyis75/32, thedifferential equation ofmotion becomes 75dv dv_ 32 <“> 3—2m""3' s--55' By(16.11), wecanwrite (a)as dv__Q __3_2(b) 1)‘? - 25» vdv— 25dz, where asisthedistance measured from thebeginning oftheslide. Integra- tion of(b)andinsertion oftheinitial andﬁnal conditions results in 0 a: 32(c) /;=10vdv-—%’/;=od:c. 162 Paoanens LEADING 'roFmsr Oanna Equxrroxs Chapter 3 Itssolution is (d) 50=ﬁx, a:=39ftapprox. Example 16.43. Solve theprevious problem ifawind isblowing against theboywith aforce equal tohisvelocity. Solution. The differential equation (a)above must now bemodiﬁed toread [with dv/dt replaced byitsequal v(dv/dx), —see(16.11)], 75d d 32(a) 55v-é=——3——v, ;li_—'_L3=—%dx, 0 1: _n) __Q] /;=l0 <1 U+3dv- 75z=°dx. The solution of(a)is (b) -3log3—-10+3log13=-3-§:c, 2:=13.1ftapprox. Example 16.44. Aboat isbeing towed attherate of18ftsec. At theinstant when thetowing lineiscastoff,aman takes uptheoarsand begins torowwith aforce of20pounds inthedirection ofthemoving boat. Ifman andboat together weigh 480pounds, andtheresistance is equal toivpounds, ﬁndthespeed oftheboat attheendof30seconds. Solution. Thenetforce acting ontheboatattheinstant t=0when thetowing lineiscastoffistheman's force of20pounds, lesstheresisting force of5',-vpounds. The mass ofman andboat is480/32 =15.Hence thedifferential equation ofmotion is d 7 d 7 4(a) 1s§=20-Z», 3?—l—%v=§- Itssolution byLesson 11Bis D=Ce-—7t/60 +579“ Substituting in(b)theinitial conditionst =0,v=18,weﬁndc=46/7. Hence (b)becomes (C) U=37§_e—7t/60 + When t=30, (6) v=#,@e"’/2 +11,11=11.6ft/sec. Lesson 16B—Exe1-cise 163 1. 2. 3. 4-. 5. 6. 7. 8 9EXERCISE 16B Aboypulls asled, onwhich aparcel hasbeen placed, with aconstant force ofFlb.Thesledandparcel together weigh mglb.Thefrictional force of theiceontherunners isnegligible. However, theforce oftheairresistance isktimes thevelocity ofthesled. Ifthesledstarts from rest, ﬁnd: (a)Itsvelocity asafunction oftime. (b)Itsdistance as.afunction oftime. (c)Itsdistance asafunction ofvelocity. (d)Itsterminal velocity. Inproblem 1,assume sledandparcel weigh 96lb,that airresistance is§ times thevelocity andthattheboyismoving thesledataconstant rateof 4.5ft/sec) (i.e., theterminal velocity ofthesledis4.5ft/sec). (a)Find th_econstant force which theboyisapplying tothesled. (b)Find thevelocity anddistance equations asfunctions oftime. (c)How farhasthesledmoved in5min? Solve independently andthen check your answers with formulas found in problem 1. Aboyweighing mglbrunsforaslideandreaches itwithavelocity ofvoft/sec. Thecoefficient ofsliding friction between hisshoes andtheiceisr.(a)Find hisvelocity asafunction ofdistance. Ignore airresistance. (b)How far willheslide? Aboyweighing 80lbrunsforaslideandreaches itwithavelocity of12ft/sec. Thecoefficient ofsliding friction between hisshoes andtheiceis1/20 and thewind blows against himwithaforce equal totwice hisvelocity. (a)Find hisdistance asafunction ofhisvelocity. (b)How farwillheslide? A32,000-ton shipstarting from restbegins tomove because oftheactions ofitspropellers that exert aforward force of120,000 lb.Theforce ofthe water resistance is50000. Find thevelocity oftheshipasafunction oftime anditsterminal velocity. Thebrakes areapplied toacar,traveling onaslippery road, when ithas slowed down toaspeed of6mi/hr =8.8ft/sec. Itslides 80ftbefore coming toastop. Compute thecoefficient ofsliding friction between tiresandstreet. Neglect theforce ofairresistance. Abody weighing 50lbrests onatable whose coeﬁicient ofsliding friction is1/25. Thebody isattached byastring toaweight of14lbthathangs vertically over thetable. Atthemoment thesystem isreleased, the50-lb body is15ftfrom theedgeofthetable. Assume noother forces areoperating. When andwithwhat velocity doesthebody leave thetable? Hint.Thetotal mass ofthesystem is64/32. Theforward thrust ofanairplane duetoitspropellers isFlb.Theairre- sistance iskvz. Find theterminal velocity oftheplane. An8-lbbody starting from restisbeing pulled along asurface, whose co- efficient ofsliding friction is1,byaforce thatisequal totwice thedistance ofthebody from itsstarting point 2:=0.Airresistance isv2/8. Find the velocity ofthebody asafunction ofitsdistance. Hint. Theresulting differ- ential equation islinear inv2. 164 Pnosu-zms Lmnrne T0Fmsr Oannn Eqtwrrons Chapter 3 10.Aman andhisboat weigh 320lb.Theman exerts aforce of16lbonthe oars. Theresistance ofthewater istwice thespeed. Find: (a)Thevelocity oftheboat asafunction oftime. (b)Itsspeed after 5sec. (c)Itsterminal velocity. 11.Amanandhisboat weigh 400lb.Atthemoment themanpicks uphisoars torow, theboat ismoving attherateof22ft/sec. Iftheresistance ofthe water is2vlbandtheoarsexert aconstant force of15lb,ﬁndthevelocity oftheboat asafunction oftime; alsoﬁndtheterminal velocity oftheboat. mswnns 16B 1.(a)v-£0-8-'“""). —kl/on @=%(t+'“T——;*)-m< FF-lav)(c)2:=I—v—7:-log—F——- - (d)T.V. =F/kft/sec. 2.(a)Qlb. (1.)v=so-e_‘I27), 1=30+276-” -27).(C)1228.5ft. 3.(a)02=voz-—2rga:. (b)x=v02/2rg. 4.(a)v-21%? =12-ﬁx. (b)10.1. 5.v=24(1 —e"”40o), T.V. =24ft/sec. 6.0.015. 8.T.V. =VF/k ft/sec. 1.z=2.2sec, v=13.4ft/sec. 9.v’=16(x-2+26-’). 10.(a)v=8(1—e'°-2‘). (b)v=5.1ft/sec. (c)T.V. =8ft/sec. 11.v=}(15—|— 29e'4”25); T.V. =15/2 ft/sec. LESSON 16C. Inclined Motion. Quantities that have both magni- tude anddirection, such asforce, velocity, acceleration, arecalled vector quantities. Itistheusual custom to y Q represent avector quantity byalinewith A anarrowhead atoneend. Themagnitude PL R ofthevector isgiven bythelength ofthe //’ lineand itsdirection bytheinclination /W/ oftheline. InFig. 16.5, PQrepresents a I . O xvector quantity. With PQashypotenuse, weconstruct aright triangle whose sides Figure 16.5 areparallel totheasand yaxes. Then thevectors PRandRQarecalled respec- tively thexcomponent andtheycomponent ofthevector PQ. Their respective magnitudes aregiven by Lesson 16C Incnmsn MOTION 165 (16.51) |PR| =IPQcos0|, |RQ| =IPQsin0|, andtheir respective directions bythedirections ofthearrowheads. Comment 16.511. Avector formerly waswritten with anarrow over ittodistinguish itfrom alinesegment. Thecurrent practice, andtheone weshall adopt, istousebold face type. Avector may also bebroken upinto twoormore components inany directions. InFig. 16.52, wehave broken upthevector PQinto four component vectors PR,RS,ST,TQ. Note thateach vector begins where theother leaves offandthat the Q ﬁnal vector’s arrowhead touches theoriginal vector’s T arrowhead. Ifabody moves along alinewhich isinclined to thehorizontal, theeffective force causing ittomove P 3 downhill isthat component ofthegravitational force acting inadirection parallel tothemotion. The forces R opposing themotion arethefrictional force and the Figure 16.52 wind resistance. Here thefrictional force isequal to theproduct ofthecoefficient offriction pand thecomponent ofthe gravitational force acting inadirection perpendicular tothemotion. Example 16.53. Atoboggan with twopeople onitweighs 520pounds. Itmoves down aslope whose gradient is5/12.Ifthecoefficient ofsliding friction is1/50, andtheforce ofthewind resistance is5times thevelocity, ﬁnd thetime itwilltake thetoboggan toreach thebottom ofa650-ft long incline. What would theterminal velocity beiftheslope were of inﬁnite length? Solution. Letabetheangle ofincline oftheslide. Since itsgra- dient =5/12, tana=5/12. Hence (Fig. 16.54a) sinoz=5/13 andcos a=12/13. Thegravitational force, which isthecombined weight ofsled 520sinoz \+ \1 411/P (0,0) 520cosa a ‘ 12 (<1) (bl Figure 16.54- andtwopeople, equals 520pounds. Therefore themagnitude ofthecom- ponent ofthegravitational force inthedirection oftheincline (Fig. 16.54b) is520sinoz=520(5/ 13)=200pounds. The magnitude ofthe 166 PROBLEMS Lnanmo 'roFmsr ORDER EQUATIONS Chapter 3 component ofthegravitational force perpendicular tothe incline is 520cosa =520(12/13) =480pounds. Since thecoefﬁcient ofsliding friction is1/50, thesliding frictional force is480(l/50) =9.6pounds. Themass ofthebody is520/32. Hence thedifferential equation ofmotion is(remember mass Xacceleration =netforce acting onthebody) (a) §3?§§%=200—9.6——5v=190.4--5v, %+%v=11.7. Itssolution bythemethod ofLesson 11Bis (b) v=38+c1e‘“/13. Theinitial conditions aret=0,v=0.Hence cl=-38, and(b)becomes (c) v=38(1 —em“/13). Integration of(c)gives (d) s=38(t+¥;1e_“/'3) +C2. Ifwetake theorigin atthestarting point ofthetoboggan, then t=0, s=0.Therefore by(d),C2=—123.5. Hence (d)becomes (e) s=3s(z+14%-4'/1“) -123.5. When s=650,weobtain from (e) (f) 20.4=¢+#,=*1@-“/ 13, whose solution ist=20.4 seconds approximately. This ist_hetime it willtake thetoboggan toreach thebottom oftheincline. From (b),theterminal velocity is38ft/sec (lett—>oo). EXERCISE 16C Intheproblems below, take theorigin atthestart ofthemotion and thepositive direction intheinitial direction ofmotion. 1.Atoboggan with fourboys onitweighs 400lb.Itslides down aslope with a30°incline. Find theposition andvelocity ofthetoboggan asfunctions of time ifitstarts from restandthecoefficient ofsliding friction is Ifthe slope is123.2 ftlong, what isthevelocity ofthetoboggan when itreaches thebottom? Neglect airresistance. 2.Abody weighing 400lbisshotupa30°incline with aninitial velocity of 50ft/sec. Thecoeflicient ofsliding friction is-115.(a)Find theposition and velocity ofthebody asfunctions oftime. (b)How long andhowfarwill thebody move before coming torest? Neglect airresistance. 3.Two particles start from restandfrom thesame point onthecircumference ofacircle inavertical plane. Onemoves along avertical diameter; theother along achord ofthecircle (anychord willdo). Iftheonlyforce acting onthe particles isthatofgravity, prove thatboth particles reach thecircumference ofthecircle atthesame time. Lesson l6C—Exe1-cise 167 4. 5. 6. 7. 8. 9. 10.Abody weighing Wlbslides down aslope with an01°incline. Itsinitial velocity isvoft/sec. Thecoefficient ofsliding friction isr.Ignore airre- sistance. (a)Express thevelocity andthedistance ofthebody asfunctions oftime. (b)Express itsvelocity asafunction ofdistance. (c)Express thetime asafunction ofthedistance. Usethese formulas toverify theaccuracy ofyour answers to1. Abody weighing Wlbisshotupaslope withana°incline. Itsinitial velocity isvoft/sec. Thecoefficient ofsliding friction isr.Ignore airresistance. (a)Find theposition andvelocity ofthebody asfunctions oftime. (b)How longandhowfarwillthebody move before coming torest? Usethese formulas toverify theaccuracy ofyour answers to2. Abody weighing Wlb,starting from rest, slides down aslope with ana° incline. Thecoefficient ofsliding friction israndtheforce oftheairre- sistance isk'vlb. (a)Express thevelocity andthedistance ofthebody asfunctions oftime. (b)Express itsdistance asafunction ofthevelocity. (c)What isitsterminal velocity? Aboyandhissledweigh 96lb.Starting from rest,heslides down anincline whose gradient is{11.Thecoefficient ofsliding friction is514-,andtheforce of theairresistance istwice thevelocity. (a)Find thevelocity andthedistance ofthesledasfunctions oftime. (b)What ishisterminal velocity? (c)When willhereach thebottom oftheslope ifitis272ftlong? (d)What ishisvelocity atthebottom? Solve independently. Usetheformulas found inproblem 6toverify the accuracy ofyour answers. Abody weighing Wlbisshot upana°incline with avelocity ofvoft/sec. Thecoefficient ofsliding friction israndtheforce oftheairresistance is lcvlb. (a)Express thevelocity andthedistance ofthebody asfunctions oftime. (b)How longandhowfarwillitgo? Abody weighing 64lbisshot upanincline, whose gradient isi,with a velocity of96ft/sec. Thecoefficient ofsliding friction is1andtheforce of theairresistance is-1150lb. (a)Find thevelocity andthedistance ofthebody asfunctions oftime. (b)How longandhowfarwillitgo? Solve independently. Usetheformulas found inproblem 8toverify the accuracy ofyour answers. Atoboggan with twopeople onitweighs 300lb.Itstarts from restdown a slope, 1mile long, from aheight 200ftabove ahorizontal level. Theco- efficient ofsliding friction is-1-8-5andtheforce ofthewind resistance is proportional tothesquare ofthevelocity. When thevelocity is30ft/sec, thisforce is6lb. (a)Find thevelocity ofthetoboggan asafunction ofthedistance andof thetime. (b)With what velocity willthetoboggan reach thebottom oftheslide? (c)When willitreach thebottom? (d)What would itsterminal velocity beiftheslide were inﬁnite inlength? 168 Pnostsns Lmnmo T0Fmsr Oannn Equarrons Chapter 3 ANSWERS 16C 1.v=15.4t, s=7.7t2, v=61.7ft/sec. 2.(a)v=50—18.77:, s=50¢—9.39%’.(b)t=2.7sec,s=66.6ft. 4.(a)v=vo+g(sina —rcosa)l;s=vol+4; (sina —-rcosa)t2. (b)v2=1:02+2g(sina —rcosa)s. \/v02 +2gs(sin a—rcosa)—vo(c)t= . 'g(sm oz—rcosoz) 5.(a)v=v0—g(sina +rcos a)t,s=vol-—4%(sina +rcosa)t2. (b)t=vo/g(sin oz+rcosa)sec,s=v02/2g(sin a+rcosoz). 6.(a)v=lg(sina —rcos a)(1 —c_"”"'), s=%,(sina ——rcosa)<t-|—%le_“/"' A A kv .(b)s=-7]?(-0 —IlogT) ,where A=mg(s1na ——rcos a). (c)T.V. =I/TV(sina —rcosoz). 7.(a)v=27.2(1 -e"2‘/3), s=27.2(t—|— -3-e'2”3 -—%). (b)T.V. =27.2ft/sec. (c)11.5sec. (d)27.2ft/sec. 8.(a)v=W(sina;|c—rcos a)(e_;,,/,,, __1)+Doe-“/,,., 8= [g(1_,—~/»,_,]+%(1_,-»~~,_ kv <">‘=$1“;(1+ S°°* 8_m_ ,, (1+i__)f,‘k 1.2 E W(sina—|—rcos .1)' 9.Intheformulas: W=64,r=1,k=-115,sina =~§,cosa =§, to=96,m=64/32=2. ewes‘ —12 -0.0014s10. (8.) U=74.1 ?rI) =54840. '—611e +1 (b)68ft/sec. (c)30sec,approx. (d)74.1ft/sec. LESSON 17. Pursuit Curves. Relative Pursuit Curves. LESSON 17A. Pursuit Curves. The path traced byabody which always moves inthedirection ofaﬁxed point orofanother moving object iscalled apursuit curve. Example 17.1. Apilot always keeps thenose ofhisplane pointed toward acity Tduewest ofhisstarting point. Ifhisspeed isvmiles per Lesson 17A Punsurr Cunvns 169 hour andawind isblowing from thesouth attherateofwmiles perhour, ﬁndtheequation oftheplane’s path. Assume that itstarts from aflying ﬁeld which isatadistance amiles from T. Solution (Fig. 17.11). LetP(z,y) betheposition oftheplane atany time t.The vector representing itsspeed hasmagnitude vandispointed toward T.Call 0theangle thisvector makes with thehorizontal linecon- necting theﬂying ﬁeld andT.Thewind vector points duenorth andhasa magnitude w.Thediagonal oftheparallelogram formed bythevectors v Y Actual direction T ofplane +w +—> /Avcos0J‘ 6 1T(0.0) (11.0) Xx Figure 17.11P(m/) vsin0 andwthen represents theactual direction andmagnitude oftheplane’s velocity attime t.Since thisdirection ischanging instantaneously, itmust betangent tothepath oftheplane atP.Itmust therefore betheslope dy/dz: ofthedesired curve. Ourproblem then istoﬁndanequation which expresses dy/da: asafunction ofacandy. The respective components oftheairplane’s velocity inthexand y directions are d:z:__ @____ .(a) 87- vcosﬂ, dt- vsmﬂ. Hence theeffective velocity oftheplane intheydirection, taking the wind’s velocity intoaccount, is d .(b) I5=-vs1n0+w. From Fig. 17.11, weseethat (c) sin0=——2—y—~—» cos0=ix——- Vw+y2 \/<v2'+?/2 170 Pnostams Lsanmo roFmsr ORDER EQUATIONS Chapter 3 Substituting these values intheﬁrstequation of(a)andin(b)weobtain (<1) @=i‘”” .@=-my +1»dt ,/x2+y2 dl ./x2+y2 ' Division ofthesecond equation in(d)bytheﬁrstgives wpi dy -—vy+w\/w’+y’ y_7 x2+y2 (6) dz= —v:z: = :0 ' Forconvenience, welet w(f) lo-51 sothat lcrepresents theratio ofthespeeds ofwind andplane. Hence (e) now becomes (g) wdy=(11—kvw’ +1/2)dr- This equation isofthehomogeneous type discussed inLesson 7.Itcan therefore besolved bythemethod outlined there. Easier perhaps isto make useoftheintegrable combination (10.31). The necessary steps are outlined below without further comment. (Initial conditions aret=0, w=¢1,1/=0.:/’= y/w=0-) <1-d__@ do/> __g<11)L?-ﬁ—”—?’-_ x~/_1+<y‘/@‘>2dx. \/1_ﬁ’;E,- xdw. u/I w - do/x) =_11“’ /./==»./m <1) 1<>r(f;+ \/mi) =-k1<»g§-Let (1<) 1.= Then (j)becomes (1) log(u+\/1—|-u2)= —klog-Er (I11) u+\/1_-W =(3)4. <1» (:)“’*~2(z)"w< <<»> 2(2)?‘=(z>'”"~1» Lesson 17A Ptmsurr Crmvns 171 -1x5*6"] (P) "-§l(;) *2' Replacing ubyitsvalue in(k),weobtain xr>s*(“>1“l(x>‘“”W1 (‘ll y=2a-;=§a '5' Replacing kbyitsvalue in(f),wehave ﬁnally (r) y=g[(§)1-(w/v) _(%)1+<w/1.)], astheequation ofthepath. Comment 17.111. Equation (r)enables ustoobtain interesting con- clusions inregard tothepath oftheplane. Case 1.Wind speed w=plane speed v.Inthiscase(r)becomes (s) y=g(1—%:)» :t2=——2a(y-—-g)» which istheequation ofaparabola, Fig.17.12. Note thattheplane will never reach itsdestination T. YW T=2 a w_(0,5)T-1fIv I I 1 Foo) M) ’ I T(0.0) (0,0) X Figure 17.12 Figure 17.13 Case 2.Wind speed w>plane speed v.Inthis case w/12 >1so x1—<'l0/U) a<1»/1.)-1 that 1—(w/v) <0.Hence asst—>0, [== ]—>co. Weseefrom (r),therefore, that as2:-—>0,y—>oo.Again theplane will never reach T.Arough graph ofitspath isshown inFig. 17.13 with w=2v. 172 Pnonnnms Lmnmo 'roF1Rs'r Onnnn Eqrmrrons Chapter 3 Case 3.Wind speed w<plane speed v.Inthis case w/v <1,so that 1—(w/v) >0.We seefrom (r),therefore that when :1:=0, y=0.Hence theplane willreach thetown T.Arough graph ofitspath isshown inFig. 17.14, with v=2w. Y 2-1v_2 T(0,0) (a,0)X Figure 17.14 Example 17.2. Solve theproblem ofExample 17.1 byusing polar coordinates. Solution (Fig. 17.21). The components ofthewind velocity winthe radial and transverse directions arerespectively [forthederivation of these formulas, seeLesson 34,(-34.2)]. (a) %=wsin9, r%?=wcos0. Y Actual velocity ofPlane wcos9 wsin0 P(1..v)=(r,9) r 0 T(0.0) 01,0) X Figure 11.21 Hence theeffective velocity oftheplane intheradial direction, taking intoaccount thewind’s speed wandtheplane’s speed v,is (b) %= —-v+wsin0. Lesson 17A Punsurr Cunvns 173 Dividing (b)bythesecond equation in(a),weobtain dr__—v+wsin0___ v(c) %- wcos 0 - wsec 0+tan0. Let v(d) k-—E- Then (c)becomes (e) g=(ksec0—|—tan6)d6. Itssolution, byintegration, is (f) logr+logc=lclog(sec0+tan0)—logcos0, which canbewritten as (g) cr=(sec0—|—tan0)”sec0. Att=0,r=a,0=0.Substituting these values in(g),wehave (h) ca=1, c=1/a. Theequation ofthepath is,therefore, after replacing lcbyitsvalue in(d), (i) r=a(sec 0+tan0)_"/“’ sec0, or (j) rcos0=a(sec 0+tan0)“"/"’. Example 17.3. Solve theproblem ofExample 17.1 ifthewind is blowing with avelocity winadirection which makes anangle 0:with the vertical, Fig.17.31. Y Direction of thewind W a IUCOB (I T(0,0) wsinor (a,0) X Figure 17.31 Solution. Weshall give twomethods bywhich thisproblem may be solved. 174 Pnonnrms LEADING T0Frnsr Onnnn Equxrrons Chapter 3 Method 1.Choose theaxessothatthedirection ofthewind becomes the yaxis(Fig. 17.32). Theinitial conditions att=0,therefore, become (a) :t=acosa, y=asina. Direction I ofwind W Y Actual velocity l++ ofplane -> P(z,y) Position ofplane vsin0 at‘=0 a Po(a eosa,asina) ,. 4 asina W vcos0.'1! aoosa r(o,o) x Figure 17.32 Thedifferential equations ofmotion, however, areexactly thesame as those inExample 17.1, namely (b) %€=—vcos0, %:=—vsin0+w. Proceeding justaswedidinExample 17.1, weobtain, (c) log(u+\/1+u'~’)=—Icloga:+logc, lc=w/v. Butnowatt=0,a:=acosa, u=y/x=(asina)/(acosa) =tana. Substituting these values in(c),andsolving forc,weobtain (d) c=(tana+seca)(acosa)". Hence (c)becomes, after replacing ubyitsvalue y/zc, (6)log+, +10;{Ck=log[(11.11.1+sec..)(acos...)"1. Simpliﬁcation of(e)gives (f) a:"_1(y +V22 +y?)=(tanoz—|—seca)(acoscc)", k="w/v. Replacing lcbyitsvalue w/v, weobtain ﬁnally astheequation ofthe plane’s path (g) x‘""""‘(y +\/x=+7*)=can<1+seca)(acos<==)“"'- Lesson 17A—Exe1-cise 175 Weleave ittoyouasanexercise todraw rough graphs ofitspath as wedidinComment 17.11. Remember tanoz,seca,cosaandaarecon- stants. WDirection ofthewindY ,1I F] wcos or P(x»J') ‘_wsinor V vsin0 ii‘ A 1,voos 0 9 x T(0,0) (a,0) X Figure 17.33 Method 2(Fig. 17.33). From Fig. 17.33, weseethat theresultant of theeffective velocities ofwind and plane inthe:1:and ydirections are respectively (h) %?=—vcos0+wsina, %=—vsin0+wcosa. Dividing thesecond equation in(h)bytheﬁrstweobtain G) Q;__—vsin0+wcosa_ d:v_ —vcos0+wsina In(i)replace sin0byitsvalue y/\/x2 +3/2andcos6by:1:/\/1:2 +1/2. Itthen simpliﬁes to . vy ___ vac .(1) <i— +wcoSa)d:t-< \/?:—l-!—;+wsma)dy, which isahomogeneous equation. Remember v,wcosa,andwsinaare constants. Hence itcanbesolved bythemethod ofLesson 7.Theinitial conditions are:1:=a,y=0.Weleave ittoyouasanexercise tocom- plete. EXERCISE 17A 1.Aman swims across ariver 100ftwide, always heading foratreedirectly across from hisstarting point. Hecanswim attherateof3ft/sec. (a)Find theequation ofhispath ifthecurrent iscarrying himdownstream attherateof1ft/sec; 3ft/sec; 4ft/sec. (b)Draw thegraph ofeach equation. 176 Pnosnnus Lnnnmo roFmsr Onnnn Equarrons Chapter 3 Solve independently andcheck your answers with (r)ofExample 17.1. 2. 3. 4. 5. 6. 7. 8. 9. 10.Inproblem 1show thattheman willreach thetreeontheopposite bank if thecurrent is1ft/sec, that hewillreach apoint ontheopposite shore 50ftfrom thetreeifthecurrent is3ft/sec, thathewillnever reach theop- posite shore, ifthecurrent is4ft/sec. Solve problem 1byusing polar coordinates. Solve independently andcheck your answers with (j)ofExample 17.2. Aninsect steps ontheedge ofaturntable ofradius athatisrotating ata constant angular velocity a.Itmoves straight toward thecenter ofthetable ataconstant velocity vo.Findtheequation ofitspathinpolar coordinates, relative toaxesﬁxed inspace. (Hint. If0istheangle through which the turntable hasrotated intime tandristhedistance oftheinsect from the center atthat moment, then d0/dt =oz,1-(d0/dt) =ra.) Draw agraph oftheequation ifa=10ft, a=1r/4radians/sec, vo-=1ft/sec. How many revolutions willthetable have made bythetimetheinsect reaches the center? Assume inproblem 4thattheinsect always moves inadirection parallel to thediameter drawn through thepoint where hesteps onthetable. (a)Find theequation ofitspath relative toaxesﬁxed inspace. (b)What kind ofcurve isit?Hint. Change theequation torectangular coordinates ifthepolar form isunfamiliar toyou. Aboystands atA,Fig.17.34. Bymeans ofastring lftlong, heholds a boat, which isinthewater atB.Hestarts walking inadirection per- pendicular toAB,always keeping thestring taut. Find theequation ofthe boat’s path. Ignore theheight oftheboyabove thehorizontal andassume thatthestring isalways tangent tothepath. Theresulting curve isknown asatractrix. Hint. Theslope ofthetractrix is:tan0=dy/dz. Yno.1) l J‘ I 0 x 412"J/2 A(0.0) X Figure 17.34 Apilot always keeps thenoseofhisplane pointed toward acitywhich is400 miduewest ofhim. Awind isblowing from thesouth attherateof20mi/hr. Hisspeed is300mi/hr.Find theequation ofhispath. Solve independently, byusing both rectangular andpolar coordinates. Check theaccuracy ofyour answers with (r)ofExample 17.1and(j)ofExample 17.2. Apilot always keeps thenose ofhisplane pointed toward acitywhich isa miduenorth ofhim. Hisspeed isvmi/hr andawind isblowing from the west atwmi/hr. Find theequation ofhispath. Solve problem 8ifthecityisamiduesouth ofthepilot. Apilot always keeps thenose ofhisplane pointed toward acitywhich is 300miduewest ofhim. A25-mi/hr wind isblowing inadirection whose Lesson 17B Rnnxrrvn Ptmsurr Crmvn 177 slope is Hisspeed is200mi/hr. Find theequation ofhispath. Solve in- dependently. Useboth methods outlined inthislesson. Check theaccuracy ofyour diﬁerential equations with (b)and(j)ofExample 17.3. 11.Solve problem 10ifthewind isblowing inadirection whose slope is-—§. Usemethod 2ofExample 17.3. Solve independently andthen check the accuracy ofyour differential equation with (j)ofthisexample. 12.Assume inproblem 4thattheinsect moves straight toward alight which is ﬁxed inspace directly above theendofthediameter drawn through the point where itsteps onthetable. Find thedifferential equation ofitspath inpolar coordinates. ANSWERS 17A [(I>2/a (1)4/3] 2 1.y=50 W —-W ;:c=—200(y ——50), KZW"1 *1=5°m *166'3.r=100(1 ——sin0)/(1+ sin0)2;r=100/(1+ sin0); rcos0=100(sec 0—|—tan0)-3/4. 4.r=a—g0(origin atcenter). 1}revolutions. 5.(a)2v0(r sin0)=a(a2 -—r2). (b)Circle: center (0,—v0/a), radius Va2+1:02/0:2. 6.as=110;’-“ll-i W -\/11 —-y’. 14/15 16/15It 1 _ = ._15 7.y—200 — ],rcos0 400(sec 0+tan0) . 1—(wI v) 1+(wI v) 8.2:=g — ]with positive directions totheeastand tothesouth. 9.Same as8with positive directions totheeastandtothenorth. 10.First method: solution isy+\/2:2 -1-y?=2(240)1I8a:7I8. Second method: Differential equation is (200y —2l|\/2:2 +yz)dz—(200: —15\/2:2 +yz)dy=0. ll.Differential equation is (2003; -20\/2:2+y?)dz:—(2002: +15\/2:2+1/2)dy=0. Q__ h -th 1dbth "1 12,d0_ m+vosin(o_¢),were¢/is eangemae yeorigina diameter andalinedrawn from itsend(i.e.,where thelightis)totheinsect’s position. LESSON 17B. Relative Pursuit Curve. Iftheorigin ofacoordi- natesystem isnotﬁxed ontheground butisattached toapursued object, then thepath described byapursuer, moving always inthedirection of theobject heispursuing, iscalled arelative pursuit curve. Itis,in 178 Pnonu-ms LEADING T0Fmsr Omma EQUATIONS Chapter 3 effect, thepath traced bythepursuing body asseen byanobserver in thepursued object. Example 17.4. Aﬁghter plane whose speed isVpischasing abomber plane whose speed isVB.Thenoseoftheﬁghter plane isalways pointed toward thebomber which isﬂying inadirection making anangle Bwith thehorizontal. Find thepath traced bytheﬁghter asplotted byan observer inthebomber. Solution (Fig. 17.41). _Call (0,0) theorigin ofacoordinate system ﬁxed ontheground, and (0,6) theorigin ofacoordinate system which is moving with thebomber plane. The position oftheﬁghter plane atany Y P(x,y)56,?) =position ofﬁghter attime t VF IQ.sin(1I'-9) T4-+ ;v-;vB=7 “'0 —>\@.cos(1r-0) VB VhB 3BID. xs'x=E 1-0 ‘ ZVBcosB Q(5,5)=(xB» ya)=position ofbomber attime t (0,0) X Figure 17.41 time tistherefore given bytwosetsofcoordinates, one(z,y) with respect totheﬁxed ground origin (0,0) andtheother, (28,111) with respect tothe moving origin (0,0). Asmeasured from anobserver ontheground, theeffective velocities of theﬁghter plane inthexandydirections are(seeFig.17.41) (a) %=VFcos(1r—0)=—VF cos0, dy . .E=—VFS1Il (1r-—0)=——Vps1n 0. Iftwoplanes areapproaching each other along astraight line, then to anobserver inoneplane, itwillseem asifhewere standing stillandthe other plane were coming toward him with aspeed equal tothesum of thespeeds ofthetwoplanes. Ifthetwoplanes aremoving away from Lesson 17B RELATIVE Ptmsurr Convs 179 each other along astraight line, then itwillseem toanobserver inone plane asifhewere standing still and theother plane were going away from himwith aspeed equal tothesum ofthespeeds ofthetwoplanes. If,however, theplanes aregoing inthesame direction along astraight line, then toanobserver inapursued plane itwillseem, ifhisspeed is slower than theother, asifhewere standing still and theother plane were coming toward himwith aspeed equal tothedifference ofthespeeds ofthetwoplanes. Ifhisspeed isgreater than theother, then itwillseem tohimasifhewere standing stillandtheother plane were going away from him with aspeed equal tothedifference ofthespeeds ofthetwo planes. Here thehorizontal component ofﬁghter andbomber velocities arein thesame direction. Hence (seeFig. 17.41), toanobserver inthebomber (weassume theﬁghter’s component isgreater than thebomber’s), itisas ifhewere standing still, andtheﬁghter plane were coming toward himin thepositive asdirection with avelocity equal tothedifference oftheir twoxcomponents ofvelocity, i.e.,therate ofchange ofiftoanobserver inthebomber is (b) gl’=—Vpcos0-Vgoosﬁ. Thevertical components ofﬁghter andbomber velocities areinopposite directions and pointed toward each other. Hence, thevelocity ofthe ﬁghter relative toanobserver inthebomber, isinthenegative ydirec- tion andisthesum oftheir twovertical velocities, i.e.,therate ofchange ofZ]is (0) %f=-(V,sin0+VBsin5). Dividing (c)by(b),weobtain (d) @_ Vpsin0+VBsinﬁ dZt— VFO0S0+ V3COSﬂ, which hasthesame form as(i)ofExample 17.3. Remember Vp,VBsin)3, and VBcosBareconstants. Hence itcanbesolved bythemethod sug- gested there. Weleave ittoyouasanexercise tocomplete. Theinitial con- ditions willdepend onthedistance anddirection oftheﬁghter plane from thebomber plane att=0,i.e.,atthemoment when thepursuit began. Example 17.5. Solve theproblem ofExample 17.4 byusing polar coordinates. Solution (Fig. 17.51). Toanobserver inthebomber, when hemoves totheright itappears tohim asifhewere standing stillandtheﬁghter plane were moving totheleft. Hence, toanobserver inthebomber, 180 Pnonusms LEADING roFmsr Onnnn EQUATIONS Chapter 3 when hemoves with avelocity VB,represented bythelower vector in Fig. 17.51, itappears tohimasifhewere standing stillandtheﬁghter is P(r,0) position ofﬁghter atanytime t VBsin(0-B) ) 9-I5 0-ﬂVB VF Direction ofbomber’s ﬂight VBcos(0—B)V ‘ VB0ﬁve°°B(9-B) 9-B gq VBsin(0-B) O Figure 17.51 moving with avelocity VBrepresented bytheupper vector inthisﬁgure. Theradial andtransverse components oftheupper VBare w %=-nmm-M r%=mmw-m Asseen from anobserver inthebomber, therefore, theeffective velocity oftheﬁghter intheradial direction is(remember itisthesum ofthetwo velocities inthesame negative rdirection) (b) %= —VF —-V3cos(0—13), andinthetransverse direction is (Q r§=nmw-m Dividing (b)by(c),weobtain d — V <d>.;.=~:::§:_§:—.:@»<~~@>-Itssolution is (e)logr+logc =—log sin(0—5)—%log [csc(0—B)-—cot(0—3)], which simpliﬁes to Lesson 17B RELATIVE Punsurr CURVE 181 (f) Cr:[csc(0—B)—cot(0—ﬁ)]_V"/VB sin(o-,2) _ 1 [1—cos(0-5)]-"""’B 6) 1Hsin(0——13) sin(0— I[sin<0—ﬁ)l"“"”””’[1 —cos<0—a>1”"”B'Nora. Theinitial conditions aret=0,r=ro,0=00. Comment 17.52. Asviewed byanobserver inthebomber, itisasif hewere standing still, andthepath oftheﬁghter asgiven in(f)(which is theresultant ofboth planes’ motions) were dueentirely totheﬁghter plane’s movements. Comment 17.521. Bymeans of(f),wecandraw arough graph of theﬁghter plane’s path, asseen from theobserver inthebomber. Case 1.IfVB=VF,i.e.,iftheground speed ofboth planes arethe same, then (f)reduces to 1 (g) cr_-1——c0s(0—ﬂ)' which istheequation ofaparabola [see(34.68) andcomment following]. Arough graph ofthepath oftheﬁghter plane asviewed from the bomber isgiven forthiscase inFig. 17.53. Itisevident that theﬁghter willnever reach thebomber. Fighter path VF=W3 (0.0) '3Bomber Figure 17.53 182 Pnonnams LEADING T0Fmsr ORDER Eqozmous Chapter 3 (h) CrCase 2.IfVF>V3,then 1—(VF/VB) <0,and(f)canbewritten as = [Sin (0_ﬂ)](VF/VB)-1 I [1—008(9—l3)lV"'V” Arough graph oftheﬁghter plane asviewed from thebomber isgiven for thiscase inFig. 17.54 (with VF=2V3). l. 2. 3VF=2VB B (0,0) Figure 17.54 EXERCISE 17B Anairplane Aisﬂying with aspeed of200mi/hr inadirection whose slope is i.Aplane B,50miles duenorth ofhim, starts inpursuit. Plane B,whose speed is300mi/hr, always keeps thenoseofhisplane pointed toward A.Find theequation ofB'spath asobserved from A.Userectangular andpolar co- ordinates. Hint. Inpolar coordinates att=0,0=1r/2,tanﬁ =§,r=50. Solve problem 1,inpolar coordinates only, ifAisﬂying inasoutheasterly direction andplane B,attheinstant hestarts inpursuit, is50miles north- eastofA.Hint. Att =0,0=1r/4, B=—1r/4, r=50.Make aﬁgure corresponding to17.51. Bevery careful ofsigns. Solve problem 1,inpolar coordinates only, ifAisﬂying inanortheasterly direction andplane B,attheinstant hestarts inpursuit, is50miles south- eastofA.Hint. Att =0,0=—1r/4, )3=1r/4, r=50.Make aﬁgure corresponding toFig.17.51. Bevery careful ofsigns. Lesson 17M A.Fnow orWATER THROUGH ANORIFICE 183 ANSWERS 17B 1.Differential equation inrectangular coordinates is § 2°\m,,,,,h,=0;=0-=50 11¢a0o§+1@0\/m ’’”' Solution inpolar coordinate is: (5—4cos0 —3sin 0)3/2 _50\/5 (sin0 +cos0)1'2 5(}\/5 (cos0—sin0)1I2 2''_ ' 3'T= '(\/5-—cos0+S1110) (\/5—cos0—S1110) LESSON 17M. MISCELLANEOUS TYPES OF PROBLEMS LEADING TO EQUATIONS OF THE FIRST ORDER A.Flow ofWater Through anOriﬁce. When water ﬂows from a tank through asmall hole initsbottom, ithasbeen proved that therate ofﬂow ofwater isproportional tothearea ofthehole andthesquare root oftheheight ofthewater inthetank. Hence (11.6) %’=-kaﬂ, av=-kaﬁ .11, where Visthenumber ofcubic feetofwater inthetank attime tsec, kisapositive proportionality constant, aisthearea ofthehole insquare feet, histheheight infeetofthewater above thehole attime t. The minus sign isnecessary because thevolume ofwater isdecreasing. IfAisthecross-sectional area ofthewater surface attime t,then dV=Adh. Substituting this value ofdVinthesecond equation of (17.6) andtaking lc=4.8,aﬁgure determined experimentally asvalid under certain conditions, weobtain (17.61) Adh=-4.811% at,Ah“/2dh =-4.s1.~r’d¢, where Aisthecross-sectional area insquare feet ofthewater surface attime tsec, histheheight infeetofthewater level above thehole ororiﬁce attime t, ristheradius infeetoftheoriﬁce. 184 Pnonuans LEADING roFmsr Onnsn Eouxrrous Chapter 3 1. 2. 3 4-. 5. 6.With theaidof(17.61), solve thefollowing problems. Atank, whose cross section measures 3ftX4ft,isﬁlled with water toa height of9ft.Ithasaholeatthebottom ofradius 1in.(a)When willthe tank beempty? (b)When willthewater be5fthigh? Hint. Att=0, h=9andin(17.61) r=11,11 =12.Awater container, whose circular cross section is6ftindiameter and whose height is8ft,isﬁlled withwater. Ithasaholeatthebottom ofradius 1in.When willthetank beempty? Solve problem 2ifthetank rests onsupports above ground sothatits8ft height isnowinahorizontal direction, Fig.17.62, andtheholeofradius 1in. isinitsbottom. U M7.”__.-.___ // II Figure 17.62 Awater tank, intheshape ofaconical funnel, withitsapex atthebottom and vertical axis, is12ftacross thetopand18fthigh. Ithasaholeatitsapex ofradius 1in.When willthetank beempty ifinitially itisﬁlled with water? Awater tank, intheshape ofaparaboloid ofrevolution, measures 6ftin diameter atthetopandis3ftdeep. Aholeatthebottom is1in.indiameter. When willthetank beempty ifinitially itisﬁlled with water? Acylindrical tank is12ftindiameter and9fthigh. Water ﬂows intothe tank attherateof1r/10 ft3/sec. Ithasaholeofradius 1}in.atthebottom. When willthetank befullifinitially itisempty? Hint. Adjust theﬁrst equation in(17.6) toreﬂect theincrease involume duetothewater intake. Remember dV=Adh.After simpliﬁcation, youshould obtain 9 e / __df‘___ = dt 5-0 12__W 4320 ¢=0 l Ifyouhave trouble integrating, seehintgiven inanswer. B.First Order Linear Electric Circuit. The differential equation which results when anapplied electromotive force E,aninductor LL,andaresistor Rareconnected inseries is, Fig. 17.621,R dz .Figure 17_62l (17.63) L-ii+R1=E. Lesson 17M C.Srmnr STATE Fnow orHEAT 185 Foranunderstanding ofthisequation andforthemeaning ofthese terms, read Lesson 30.Theunits weshall adopt are ohms forthecoefficient ofresistance oftheresistor, henries forthecoeﬁicient ofinductance oftheinductor, volts forlthe electromotive force (also written asemf), amperes forthecurrent. 7.Solve thedifferential equation (17.63),ifE=E0andatt=0,thecurrent i=£0.What isthelimiting current ast—+w? 8.Solve (17.63) ifE=E0sinwtandatt=0,i=£0. 9.Aninductance of3henries andaresistance of30ohms areconnected inseries withanemfof150volts. Att =0,i=0.Find thecurrent whent =0.01 sec. 10.Aninductance of2henries andaresistance of20ohms areconnected inseries with anemfof100sin150tvolts. Att=0,'i=0.Find thecurrent when t=0.01 sec. ll.When anemfisdisconnected from acircuit inwhich acurrent isﬂowing, i.e., att=0,E=0,thecurrent iscalled aninduced current. The term Ldi/dt isthen referred toasaninduced electromotive force. Atthe moment anemfisdisconnected from acircuit inwhich there isaresistance of30ohms andaninductance of6henries, thecurrent 1'=15amp. (a)Find thecurrent equation asafunction oftime. (b)When willthecurrent be 7.5amp? C.Steady State Flow ofHeat. When theinner andouter walls of abody, asforexample theinner andouter walls ofahouse orofapipe, are maintained atdiﬂerent constant temperatures, heat will ﬂow from the warmer wall tothecolder one. When each surface parallel toawall has attained aconstant temperature, wesaythat theﬂow ofheat hasreached asteady state. Inasteady state ﬂow ofheat, therefore, each surface parallel toawall, because itstemperature isnow constant, iscalled an isothermal surface. Isothermal surfaces atdifferent distances from an interior wall will, ofcourse, have different temperatures. Inmany cases thetemperature ofanisothermal surface isafunction only ofitsdistance :0from aninterior wall, andtherate ofﬂow ofheat inaunit time across such asurface isproportional both tothearea ofthesurface andtodT/dz, where Tisthetemperature oftheisothermal surface. Hence dT(17.64) Q-—kA E. where Qistherateofﬂow ofcalories* ofheat in1secacross anisothermal surface, k,theproportionality constant, iscalled thethermal conductivity ofthematerial that isbetween thewalls, ‘Acalorie isequal totheamount ofheat required tochange thetemperature of1 gram ofwater 1degree centigrade. 186 Pnontnms LEADING 'roFIRST ORDER EQUATIONS Chapter 3 Aisthearea insquare centimeters ofanisothermal surface, Tisthetemperature incentigrade degrees oftheisothermal surface, xisthedistance incentimeters oftheisothermal surface from an interior wall. The negative sign isused toindicate that heat ﬂows from theinterior wall ofhigher temperature totheexterior wall oflower temperature. With thehelp of(17.64), solve thefollowing problems. 12.Theinner andouter radii ofahollow spherical shellare4cmand9cmrespec- tively. The thermal conductivity ofthematerial between thewalls is 0.75cal/deg-cm-sec. Theinner surface iskept ataconstant temperature of 100°C andtheouter surface at0°C. Find: (a)Therateofheat losspersecond ﬂowing outward through theexterior oftheshell. (b)Thetemperature Tofasurface 5cmfrom thecenter. Hint. In(17.64) A=41rr2, where r,which replaces a:intheformula, isthe radius ofanisothermal surface. Initial conditions are:r=4,T=100; r=9,T=0. Note, from answer, that Qhasaconstant value. Inthis example, itis 2l6O1r cal/sec. Hence, through each isothermal surface, thismuch heat escapes each second. Ifasurface isnearer thecenter ofthesphere, sothat itsareaissmaller than theareaofasurface farther away, more heatperunit areawillescape each second from thenearer surface than from thefarther one. Thetotal lossofheat through each surface is,however, thesame. 13.Asteam pipe ofnegligible thickness hasaninside radius of6cm. Itisin- sulated with 3cmcoating ofmagnesia, whose thermal conductivity is 0.000175. Theinterior ofthepipe ismaintained at100°C andtheouter surface at0°C. Find: (a)Therateofheatlosspersecond ﬂowing outward from each meter length ofpipe. (b)Thetemperature ofanisothermal surface whose radius is8cm. Hint. In(17.64), A=21rr(100), where r,which replaces acintheformula, istheradius ofanisothermal surface. D.Pressure—Atmospheric and Oceanic. Weconsider acolumn ofairofcross-sectional area A,ofheight dhandatadistance hunits from thesurface oftheearth, Fig. 17.65. Forsimplicity, weassume theearth isaplane, theairisatrest, and isunaffected bychanges oflatitude, longitude, andtemperature. The positive direction isupwards. Letpbe thepressure* onthiscolumn ofair,duetotheweight ofalltheairabove it.Hence thetotal force Ponacolumn ofairofcross-sectional area A isP=Ap. Bydifferentiation weobtain (a) dP=Adp, ‘Pressure istheforce acting perpendicularly onaunitareaofsurface. Lesson 17M D.ATMOSPHERIC ANDOCEANIC Pnassunn 187 which gives thechange ofthetotal force Pforachange inthepressure p. Letpbetheweight ofaunit volume ofair.Therefore theweight ofacol- umn ofairofheight dhandcross-sectional area Ais (b) (Adh)p. The decrease intotal force asonegoes from theheight htotheheight h+dhmust beequal totheweight of thecolumn ofairofthickness dh.Hence equating thenegative of(a)with (b),we obtain L - T+ d(17.66) —Adp=pAdh, i=—p, which states that therate ofchange of Surface °f airpressure with height isequal tothe y theearth negative oftheweight ofaunit volume of airatthat height. The units weshall use are: pounds percubic foot forp,pounds Figure 17.65 persquare foot forp,andfeetforheight. Ifpisoceanic pressure andhisthedepth below sealevel, then (17.66) becomes dz»_ (17.67) 3_p, where pistheweight percubic foot ofocean water. With theaidof(17.66) and(17.67), solve thefollowing problems. 14.Assume that theairisanisothermal gas. Therefore byBoyle's law, its pressure anddensity arerelated bytheformula (17.67l) p=lcp. (a)Determine thepressure oftheairasafunction oftheheight habove theearth ifatsealevel theairpressure is14.7lb/sqin.Hint. In(17.66), replace pbyIcp. (b)What isthevalue oftheconstant kin(17.671) ifp=0.081 lb/cu ftat sealevel? (c)What istheairpressure at10,000 ft,at15,000 ft,at70,000 ft,at50mi? (d)Show thatthepressure iszeroonly when hisinﬁnite. (e)Express thedensity ofairasafunction ofheight. Hint. In(17.66) re- place dpbydp/k asobtained from (l7.671). (f)What istheweight oftheairatanaltitude of1000 ft,of5000 ft,of 10,000 ft,of50,000 rt? 15.Ifairexpands adiabatically, i.e., without gaining orlosing heat, then P=kp1.4 (a)Find thepressure oftheairasafunction ofheight. Assume thatatsea level p=14.7lb/sq in.andp=0.081 lb/cu ft.Hint. First ﬁndthe value ofIc.Then in(17.66) replace pby(p/k) 5/7. 188 PROBLEMS LEADING T0FIRST ORDER EQUATIONS Chapter 3 (b)How high istheatmosphere, i.e.,atwhat height isp=0?Hint. In (17.66) replace dpby1.4lcp°*4 dp.Solve forp.Then ﬁndhwhen p=0. 16.Assume thattheweight pofacubic footofseawater, under apressure of plb/ft2 isgiven bytheformula (17.672) p=k(l—|—2-10_8p) lb/fta, andis64lb/ft3 atsealevel. (a)Find thevalue ofk.Hint. Atsealevel p=0. (b)Find thepressure ofseawater asafunction ofitsdepth below sealevel. Hint. In(17.67), replace pbyitsvalue asgiven in(17.672). (c)Find theweight percubic footofseawater asafunction ofdepth. Hint. In(17.67), replace dpby108dp/2k asdetermined from (17.672). (d)What isthepressure anddensity ofseawater at20,000 ftbelow sealevel? 17.Iftheearth isassumed spherical instead ofplanar, show that (17.66) be- comes ti 2 ﬁ =—P — i where Ristheradius oftheearth. Hint. SeeFig.17.69. Thevolume ofa thin spherical shell ofairatadistance 1'units from thecenter oftheearth andthickness dris411-72 dr.Itsweight, therefore, is(41rr2 dr)p. Thetotal d . ’“\ h q Figure 17.69 force Ponthisspherical surface duetotheweight oftheairoutside itis P=4172]). Hence dP=41r(r2 dp—|—27pdr). Set—dP, which isthede- crease inthetotal force Pasonegoes from thedistance rtothedistance r+dr,equal totheweight ofthespherical shell. E.Rope orChain Around aCylinder. When arope orchain, assumed uniform, ﬂexible, and inextensible, iswound around arough cylindrical post, asmall force applied atoneendoftherope orchain can control amuch larger force applied attheother end. Forexample, aman canhold incheck alarge weight bywinding arope attached toit,asufﬁ- cient number oftimes about apole. Forapost, whose axisishorizontal, ithasbeen proved that thediffer- ential equation dT .(17.7) E=6r(cos 0+psin 0)+pT Lesson 17M F.MOTION or11Conrtnx SYSTEM 189 expresses thetension Tintherope orchain atapoint Ponit,when therope orchain isjustontheverge ofslipping, seeFig. 17.71. Inequation (17.7), Tisthepullortension ontherope atanypoint Ponit,inpounds, ristheradius ofthecylinder, infeet, /1isthecoeﬂicient offriction between rope andpost, 6istheweight oftherope orchain inpounds perfoot, 0istheradial angle ofP. With thehelpof(17.7), solve thefollowing problems. 18.Show thatthesolution of(17.7) is (17.72) T=L [(1-,1”)sin0_2,.66$01+¢e"’.1+#2 19.Achain weighs 6lb/ft. Ithangs over acircular cylinder with horizontal axisandradius rft.Oneendofthechain isatA,Fig.17.71.How farmust theother endextend below D,sothatthechain isontheverge ofslipping. Hint. The initial conditions are: 0=0,T=0,and 0=‘I’,T=Z8, where listhelength oftheportion ofthechain overhanging atD. 20.Achain weighs 8lb/ft. Ithangs over acircular cylinder with horizontal axisandradius rft.Oneendofthechain isatB,Fig.17.71, theother end justreaches toD.What istheleast value ofitsothat thechain willnotslip? Hint. B Theinitial conditions are:0=1r/2, T=0, andifthechain isnottoslip,thenat0=1r, T-'=' P(rr0) 21.Achain weighs 1lb/ft. Ithangs over a M circular cylinder with horizontal axisand D radius Qft.Oneendofthechain reaches three quarters around thetoptoC’,Fig. 17.71; theother endextends below D.The I coefficient offriction between chain and cylinder isQ.Ifthechain isontheverge of slipping, ﬁndthelength loftheoverhang. _ 22.Iftheaxisofthecylinder isvertical, then F18“1‘° 17-71 therope’s orchain’s weight which nowacts vertically sothatitdoesnotpress down upon thecylinder, haslittle effective force. Hence, informula (17.7), 6which istheweight perunitlength of rope, may betaken tobezero. Theformula then simpliﬁes toC (17.73) %=pT. With thehelp of(17.73), solve thefollowing problem. Alongshoreman is holding ashipbymeans ofahawser wound around avertical post. The shipispulling ontheropewith aforce of5tons. Ifthecoefficient offriction between rope andpostis§,andtheman exerts aforce of50lbtoholdthe ship, approximately how many turns ofrope isheusing? Hint. Initial condition is0=0,T=50.Wewant 0when T=10,000. F.Motion ofaComplex System. Solve problems 23-27 below, by useofNewton’s lawofmotion F=ma=mv(dv/dy), where Fisthe 190 PROBLEMS Lnanms 'roFmsr Onnsn Equurons Chapter 3 algebraic sum ofalltheforces acting onabody ofmass m,and aisthe acceleration ofthecenter ofgravity ofthebody. 23.A24-ft chain weighing 6lb/ft hangs over africtionless support, which is more than 24ftabove theﬂoor. Initially thechain isheldatrestwith 10ft overhanging ononesideofthesupport and14ftontheother side, Fig.17.74. iy=12 14a. l+ y=0,equilibrium _2 position —y1-¢—\ j‘< @,_¢ Figure 17.74- How long after itsrelease andwith what velocity willthechain leave the support? Hint. When 12ftofchain overhang oneach sideofthesupport, thechain isinequilibrium. Callthisposition y=0.Then ifthedistance of oneendofthechain from equilibrium isy,thedistance oftheother endis-1/. Theeffective force moving thechain istherefore 2;/5. Themass ofthechain is248/32. Initial conditions aret =0,v=0,y=2. 24.Inproblem 23,assume thatthe14-ft overhang justtouches theﬂoor. Com- pute thevelocity with which theother endwillleave thesupport. 25.Achain is12ftlong. Sixfeetofthechain areheldextended onafrictionless ﬂattable which ismore than 12ftabove theground, theother 6fthangs overthetable. When andwith what velocity willtheendofthechain leave thetable after itsrelease? Hint. m=126/32, F=y6,where yisthedis- tance oftheoverhanging partofthechain from theedge ofthetable and6 isitsweight perfoot. Initial conditions aret=0,y=6,v=0. 26.Assume, inproblem 25,thatthetable isonly 4ftabove theground. With what velocity willthechain ofproblem 25leave thetable? 27.Ithasbeen proved thatwhen amass particle slides without friction down a ﬁxed curved path whose equation isy=f(z), itsdiﬁerential equation of motion, with upward direction positive, is (17.75) vclv=—gdy, where gistheacceleration duetogravity, visthevelocity oftheparticle along thecurve, andyisthevertical position oftheparticle attime t.Therefore, v=ds/dt, where sisthedistance theparticle hasmoved along thecurved path. Show that thesolution of(17.75), with initial conditions t=0, v=my=1/o,is2 (17-76) v2=(g) =voz+20(yo ——y)- Lesson 17M G.VARIABLE Mass. Rocxrrr MOTION. 191 Bymeans ofthesubstitution in(17.76) ofds/dt =V1—l—(dy/da:)2 dz/dt, theequation ofthepath y=f(z)andtheinitial condition yo=f(a:0), show that (17.76) becomes (1.77, =' dt 1+lf'(1)l’ Thesolution of(17.77) willgive2:asafunction oft.With :2:known, wecan determine ybymeans ofthegiven equation ofthepath y=f(z). Use(17.77) tosolve thefollowing problem. Aparticle moves along a smooth wire, shaped intheform oftheparabola 2:=—y. Initially itisat theorigin andhasavelocity of4ft/sec. Find theposition oftheparticle at theendof5sec.Hint. f(z) =——:i:2,f(a:Q) =0,f'(x) =-21. G.Variable Mass. Rocket Motion. Inthestraight line motion problems thus farconsidered, themass ofaparticle orofabody remained constant throughout themotion. If,however, themass itself isalsochang- ingwith time, decreasing orincreasing, then Newton's second law of motion nolonger holds andmust bemodiﬁed. Ithasbeen proved, inthe case ofabody ofvariable mass moving inastraight line, that thedifferen- tialequation governing itsmotion isgiven by dv dm where misthemass ofthebody attime t, visthevelocity ofthebody attime t, Fisthealgebraic sum ofalltheforces acting onthebody attime t, dmisthemass joining orleaving thebody inthetime interval dt, uisthevelocity ofdmatthemoment itjoins orleaves thebody, relative toanobserver stationed onthebody. Note that (17.78) differs from Newton's second lawofmotion bythe term udm/dt. Ifamass dmleaves thesystem inthetime interval dt,dm/dt willbea negative quantity; ifitjoins thesystem, dm/dt willbeapositive quantity. With theaidof(17.78), solve thefollowing problems. 28.Arocket, which weighs 32M lbandcontains fuelweighing 32mg lb,ispro- pelled straight upfrom thesurface oftheearth byburning 3210lboffuelper second andexpelling itbackwards ataconstant velocity ofAft/sec relative toanobserver ontherocket. Assume thattheonlyforce acting ontherocket isthatofgravity. Find thevelocity oftherocket andthedistance ittravels asfunctions oftime. Take positive direction upward. Hint. In(17.78), the variable mass mattime tism=M+mo—kt;therefore dm/dt =-—k. Therelative velocity uofdmis—A. Theforce ofgravity attime tisF= —(M +mg—kt)g. Answers are _____ ___ _ It M—|-mo(17.79) U— gt A103 (1 -?+ moI): 0§I<T 1 192 (17.8) 29. (17. (17. 30. 31. 32. 33. 34.PROBLEMS LEADING 'roFmsr ORDER EQUATIONS Chapter 3 y= At——§gt2+%(M+mo—kt)log(1— M+mo5 L . O_t< k Nora. Iftherocket moves infreespace sothat itisnotsubject tothe gravitational force oftheearth, then F=Oin(17.78) andg=0in(17.79) and(17.8). (a)Show that, when therocket’s fuelofproblem 28isexhausted, ithas reached atheoretical height of A 2AM M ;"“~%<@:°>+ .1%....)-Hint.Themass moofthefuelwillbeexhausted intime t=mo/k. Sub- stitute thisvalue in(17.8). (b)Show thatitsvelocity atthatmoment is ___9m_ L. 82) v— k Alog (M+ mo) Arocket ofmass M,containing fuelofmass mo,fallstotheearth from a great height. Itburns anamount Icofitsmass persecond andejects itdown- ward withaconstant velocity relative toanobserver ontherocket ofAft/sec. Find thedistance itfallsintime t.Take positive downward direction. Hint. Seeproblem 28. Arocket anditsfuelhave mass mo. Atthemoment itstarts toburn an amount kofitsmass persecond, itismoving with avelocity vo.Thefuel isejected backwards with justenough velocity, sothat theejected fuelis motionless inspace. Find thesubsequent velocity anddistance equations oftherocket asfunctions oftime. Hint. Fortheejected fueltobemotion- lessrelative toanobserver ontheearth, thebackward velocity ofthefuel must equal theforward velocity oftherocket; remember, thefuelatthe instant ofejection hasthesame forward velocity vastherocket itself. Relative toanobserver ontherocket, however, itwillseem tohimasifhe were standing stillandtheejected fuelmoving away from him atthe rateof—vft/sec, where —l—vft/sec ishisownvelocity inthepositive direc- tion. Therefore uin(17.78) is—v. The variable mass m=mo—kt anddm/dt =—k. Solve problem 31iftherocket were moving infreespace. Hint. F=0 in(17.78). Abody moves inastraight lineinfreespace with avelocity ofvoft/sec. Initially itsmass ismoandasitmoves itadds toitsmass lcslugs persecond. Find itsvelocity anddistance equations asfunctions oftime. Hint. In (17.78), F=0andassuming theadded mass dmisstationary inspace, then itsvelocity relative toanobserver onthebody is—v,where visthevelocity ofthebody. Seeproblem 31. Aspherical raindrop fallsunder theinﬂuence ofgravity. Itsmass increases bytheaddition ofstationary moisture particles ataratewhich ispropor- tional toitssurface area. Initially itsradius 1'=1'0.Find itsacceleration, velocity, anddistance equations asfunctions oftime. Take positive direction downward. Show thatifinitially r0=0,theacceleration hastheconstant Lesson 17M H.ROTATION orALIQUID 193 value g/4. Forhints, seeanswer section. First trytosolve without making useofthese hints. 35.Achain unwinds from acoilheldatrest. Itfallsstraight down under the inﬂuence ofgravity, which istheonlyacting force. Initially lftofthechain areunwound. Find itsvelocity asafunction oftime. Take positive direction downward. Forhints seeanswer section. 36.Solve problem 35,ifthechain must ﬁrstslide along africtionless plane in- clined atanangle 0with thehorizontal before dropping straight down. Assume thatthecoilisheldatrestatadistance lftfrom oneendoftheplane andthatinitially oneendofthechain isattheendoftheplane. Forhints, seeanswer section. H.Rotation oftheLiquid inaCylinder. 87.Avessel ofwater isrotated about avertical axiswith aconstant angular velocity w.Show thatwhen thewater ismotionless relative tothevessel, the surface ofthewater assumes theform ofaparaboloid ofrevolution. Find theequation ofthecurve made byavertical cross section through theaxisof thecylinder. Hint. SeeFig.17.9. Twoforces actonaparticle ofwater atP: Y T+ : I Tangent tothesurface i mwzxXaxis 1" .<mgNormal tothesurface Figure 17.9 onedownward duetotheweight mgoftheparticle; theother mwzz dueto thecentrifugal force* ofrevolution. Since theparticle ofwater ismotionless, theresultant ofthese twoforces must beperpendicular tothesurface ofthe water atP;ifitwere not,then theresultant would itself have acomponent offorce inadirection tangent tothecurve andthus cause theparticle of water atPtomove. Youcantherefore equate tan0=dy/dz withmwza:/mg. 38.Assume thattherotating vessel ofproblem 37isacylinder containing aliquid whose weight perunitvolume isp.Ifthepressure ontheaxisofthevessel is pg,show thatthepressure atthesurface oftheliquid atadistance rfrom theaxisisgiven bythedifferential equation dp/dr =pw2T/g. Then show thatitssolution isp=po+(pw21‘2/2g). Hint. Asinproblem 37,usethe factthatanelement ofvolume isinequilibrium under thepressures onits surfaces andcentrifugal force. ‘For adeﬁnition ofcentrifugal force seeLesson 30M, A. 194- Paonnnns Lmnmo 'roFmsr Oansn EQUATIONS Chapter 3 39. 1. 2. 3. 6. 7. 8. 9. 10. 13 14. 15. 16 19. 20 21 22 23. 24-. 25. 26 27. 30 31 32 33.12 =Assume thecylinder ofproblem 38contains agaswhose weight perunit volume ispandwhich obeys Boyle's lawp=kp,where pispressure and kaconstant. Show that p=poe“"’/2*". Hint. Replace pbyp/kinthe formula given inproblem 38. ANSWERS 17M. (b)12(s—~/5)/1. min.4.30.5_r_nin. 32\/6/1 min. 512\/3min.Make thesubstitution u=12-\/h;65min. 1"=%<1-rm") +i0e_R”L; 1'=E0/R.(a)3_6/‘I’ min. 18\/2min. . E . ._ ._1=;5— [Rsinwt—wLcoswt—l-wLe MIL] +ioeRM’. 0.476 amp. ll.(a)i=15e-5‘. (b)0.14sec. 0.299 amp. 12.(a)Q=21601’ cal/sec. (b)64°C. (a)8.611’ cal/sec. (b)29°C. (a)p=l4.7e""' lb/sq in.=2117e-"" lb/sq ft. (b)0.000038. (c)10.0lb/sq in.,8.3lb/sq in.,1.0lb/sq in.,0.00060 lb/sq in.(e) p=0_081e-0.000038h_ (f)0.078 lb/cu ft,0.067, 0.055, 0.012. (a)p2/7 =p02/7 —§k'5/7h, where po=14.7lb/sq in.=2116.8 lb/sq ft andk=2116.s(0.0s1)-7/5.(b)h=.}(p°/po) =.}(211e.s/0.081) =91,500 ft=17§mi.(9.)k=64. (b)p=5-107(¢1='8h/1°” -1). (O)p=calm/1°‘.(d)1,296,500 lb/ftz, 65.7 lb/fta. l=ﬂiﬂ (1+e"'). Note that foraﬁxed [1,Zisafunction only ofr. 2;;=(1——;i2)e"‘/2, 11.=0.7324. l=0.63 ft. 0=15.9radians =slightly more than glurns ofrope about thepost. v=—§\/ 210ft/sec;t =\/§log(6—|— 35)sec. 18.1ft/sec. __ 17.0ft/sec; \/§log(2+\/3)sec. 15.3 ft/sec. 2:=20ft,1/=-400 ft,or2:=-20 ft,y=-400 ft. 2-At-5014+ mg-—kt)log(1 -ii-).M+mo It M+mou=tot 0§t<kl _ 2lcm0v0 —gmoz 21¢('"° I“)+2k(m0 -kt)' 9 22 Zkmovo -—gmoz Itm(2m0kt—kt)— l0g 1-—%t- movo/(mo -—kt). y=$108 %,' mo»./<m.+ kc.y=$107;f)= y= |)= Lesson 17M—Answers 195 34. 35. 36. 37.First show, seeExercise 15D, 12,thatr=ro-l—kt/6,where ristheradius of theraindrop attime t,kisaproportionality constant and6isthemass per cubic footofwater. Hence dr=kdt/6.Thevelocity uoftheparticle dm is—vwhere visthevelocity oftheraindrop attime t,since relative toan observer moving with theraindrop, itisasifhewere stationary andthe particle were moving tohim, inthenegative upward direction, with a velocity equal tohisownvelocity atthemoment themoisture particle at- taches itself totheraindrop. Thevariable mass mattime tism=4-n36/3; dm/dt =k4-1'12. Answers are Acceleration a=- arc»asQ:|#§/_\U-1+ Velocityv =—— r—-— » 2 4 g +_2.5). Replacing rbyro+lct/6intheabove equations willgive acceleration, velocity, anddistance equations asfunctions oftime. Themass ofchain attime tis(l+y)6,where 6isthemass ofchain perunit length andl+yisthedistance fallen intime t.Therefore dm/dt = 6dy/dt =6v.Thevelocity ofapiece dmofthechain asitleaves thespool, relative toaperson moving with thechain is—v,where visthevelocity of thechain attime t.Change dv/dt in(17.78) tovdv/dy. Answer is311)‘ 7?’4 T0 (.3)Distance y= ﬂHwf=%w+nP4ﬁ Seeproblem 35.Acting force nowis[(lsin0)6—l-y6]g, where yisthevertical distance thechain hasfallen intime t.Answer is »W+w=§wwMw+n+fw+mi 1/=40212/2g, iftheorigin istaken atthelowest point oftheparabola. Chapter 4 Linear Differential Equations ofOrder Greater Than One Introductory Remarks. InLesson 11,weintroduced theimportant linear differential equation oftheﬁrst order. The linear differential equa- tionofhigher order which weshall discuss inthischapter haseven greater importance. Motions ofpendulums, ofelastic springs, offalling bodies, theﬂow ofelectric currents, andmany more such types ofproblems are intimately related tothesolution ofalinear differential equation oforder greater than one. Deﬁnition 18.1. Alinear differential equation oforder nisan equation which canbewritten intheform (1811) f»(w)y("’ +f»-1(@=)y(”‘” +---+f1($)y' +fo(w)y =Q01), where fo(x), f,(:c), ---,f,,(:r:), andQ(:c) areeach continuous functions of:1: deﬁned onacommon interval Iandf,,(x) #0inI.* Note that inalinear differential equation oforder n,yandeach ofits derivatives have exponent one. Itcannot have, forexample, terms such asy2or(y’)1/2 or[y(")]3. Deﬁnition 18.12. IfQ(z) ¢0onI,(18.11) iscalled anonhomo- geneous linear diﬁerential equation oforder n.If,in(18.11), Q(z) E 0onI,theresulting equation (13-13) f»(@>)1/l") +f.._1(r)y("—” +'"-1-fi(1=)y' +fo($)?/ =0 iscalled ahomogeneous linear differential equation oforder n. ‘The notation “f,.(z:) E0inI”(oronIorover I)means f,.(x) =0forevery a:inI (oronIorover I)andisread as“f.,(:c) isidentically zero inI(oronIorover I)." Thesymbol f.,(:n) ,=.€0inImeans f,.(:v) isnotequal tozeroforevery a:inI(oronIor over I),although itmay equal zeroforsome zinI.Itisread as“f,,(a:) isnotidentically zeroinI(oronIorover I).” 196 Lesson 18A Courmzx Numasas 197 Remark. Donotconfuse theterm homogeneous asused inLesson 7 with theabove useoftheterm. Foraclearer understanding ofthesolutions oflinear differential equa- tions oforder nyoushould befamiliar with: 1.Complex numbers andcomplex functions. 2.Themeaning ofthelinear independence ofasetoffunctions. Weshall, therefore, brieﬂy discuss these topics inthisandthenextlesson. LESSON 18. Complex Numbers and Complex Functions. LESSON 18A. Complex Numbers. Weshall notattempt toenter intoalengthy discussion ofthetheory ofthenumber system, butwill indicate only brieﬂy theimportant facts weshall need. Thereal num- bersystem consists of: 1.The rational numbers; examples arethepositive integers, zero, the negative integers, fractions formed with integers such as§,——=§,and1. 2.Theirrational numbers; examples are\/2, \'/5, 1r,ande. Deﬁnition 18.2. Apure imaginary number istheproduct ofa realnumber andanumber iwhich isdeﬁned bytherelation i2=—1. Examples ofpure imaginary numbers are3i, -—5i, \/2i,andxi/iri. Deﬁnition 18.21. Acomplex number isonewhich canbewritten intheform a—|—bi,where aandbarerealnumbers. Itisevident that thecomplex numbers include thereal numbers since every realnumber canbewritten asa+Oi.They alsoinclude thepure imaginary numbers, since every pure imaginary number canbewritten as0+bi. Deﬁnition 18.22. The ainthecomplex number z=a+biiscalled thereal part ofz;thebtheimaginary part ofz. Note that b,theimaginary part ofthecomplex number, isitself real. Remark. The complex numbers developed historically because ofthe necessity ofsolving equations ofthetype :v2+1=0, :v2—|-2:0-l—2=0. Ifwerequire xtobereal, then these equations have nosolutions; if,how- ever, xmay becomplex, then thesolutions arerespectively 1:=:l:iand av=—l=1:i.Infact, ifweadmit complex numbers, wecanmake the following very important assertion. Itissoimportant that ithasbeen labeled the“Fundamental Theorem ofAlgebra.” 198 HIGHER ORDER LINEAR DIFFERENTIAL Eouurons Chapter 4 Theorem 18.23. Every equation ofthefarm (18.24) anx" +a,,_1:z:"'1 +---+ala:+ao=0,an96O, where thea’sarecomplex numbers hasatleast onerootandnotmore than n distinct roots. Itsleftsidecanbewritten as (18-25) an(w—r1)(w-12)---(w——rt), where ther’sarecomplex numbers which need notbedistinct. Comment 18.26. Although Theorem 18.23 tellsusthat (18.24) hasn roots, itdoes nottellushow toﬁndthem. Ifthecoelﬁcients arereal numbers, then youmay have learned Horner’s orNewton’s method for ﬁnding approximate realroots. Oryoucansetyequal totheleftsideof (18.24) anddraw acareful graph. The approximate values ofthecoordi- nates ofthepoints where thecurve crosses thexaxis willalso give the realroots of(18.24). Toﬁndtheimaginary roots of(18.24) isamore difﬁcult matter. How- ever, there aremethods available forapproximating such roots.* Deﬁnition 18.3. Ifz=a+bi,then theconjugate ofz,written as2,is§= a—— bi. Toform theconjugate ofacomplex number, change thesignofthe coefficient ofi.Forexample ifz=2+3i,2=2—3i;ifz=2-—3i, Z=2+3i. Deﬁnition 18.31. Ifz=a+bi,then theabsolute value ofz, written as|z|,is =\/a2 +b2. Forexample, ifz=2+3i,then Imaginaryaxis |z|=\/2"+32—-=\/ﬁ;ifz=2— 3i, then =\/22 —|—(-—3)§ = _2+3i, ,2+3i \/T8. (Norm. Byourdeﬁnition |z| isalways anon-negative number.) If,inarectangular coordinate system, welabel thexaxisasthe (om) Realaxis realaxisandtheyaxisastheim- aginary axis, wemay represent any complex number z=a+bigraphi- _2_3i. '2_3i cally. Weplotaalong therealaxis andbalong theimaginary axis. In Fig. 18.32, wehave plotted the Figure 18.32 numbers 2+3i,2—3i,-2+3i, -2-—-3i. Bythegraphical method ofrepresenting complex numbers, itiseasy to establish relationships between acomplex number and thepolar coordi- ‘W.E.Milne, Numerical Calculus, Princeton University Press, 1949. Lesson 18A Conrnnx Ntnmnns 199 nates ofitspoint. InFig.18.33, wehave represented graphically thecom- plexnumber z=:c+iyandthepolar coordinate (r,0) ofitspoint. If 22z=x+iyE(r,9) |zl=r=1/x +y J)‘ L (°>°) I Figure 18.33 youexamine thisﬁgure carefully, youwillhave notrouble establishing thefollowing relationships. (18.4) |z|=r=\/iii? (18.41) st=rcos0, (18.42) y=rsin0. Deﬁnition 18.5. Ifin (18.51) z=re+yi, (called therectangular form of5),wereplace xby(18.41), andyby (18.42), weobtain (18.52) z=1'cos0+irsin0=r(cos 0+isin 0), (called thepolar form ofz). Deﬁnition 18.53. Thepolar angle 0inFig.18.33 iscalled theArgu- ment ofz,written asArgz.Ingeneral, Argzisdeﬁned tobethesmallest positive angle satisfying thetwoequalities __ll, '=l/.. (18.54) cos0--'2' sin0I2‘ Formulas (18.4) to(18.42) plus(18.52) and(18.54) enable ustochange acomplex number from itsrectangular form toitspolar form andvice versa. Remark. Note thatwehave written argzwith acapital Atodesig- natethesmallest orprincipal value of0.Ifwritten with asmall a,then argz=Argz:2n1r,n=1,2,3... Example 18.55. Find Argzandthepolar form ofz,iftherectangular form ofzis (a) z=l——i. 200 HIGHER ORDER LINEAR Drrrnnnmnu. Equarrons Chapter 4 Solution. Comparing (a)with (18.51), weseethat :1:=1and y= —-1. Therefore by(18.4) (b) |z|=1'=\/12+(-1)2 =\/5. By(18.54) (c) cos0=%» sin0==%- Hence 0isafourth quadrant angle. ByDeﬁnition 18.53, therefore, (d) Arg2= Finally by(18.52), andmaking useof(b)and(d),wehave (e) z=\/2(cosL,Z—r+isin'%r), which isthepolar form ofz. Example 18.56. Find therectangular form ofzifitspolar form is (a) z=\/2(cos%r+isin%)- Solution. Comparing (a)with (18.52), weseethat (b) r=\/5, 0=g- Hence by(18.41) and(18.42), (c) x=\/2cos1r=i§ and y=\/2_sinZ-r=l/§~3 2 3 2 By(18.51), therectangular form ofzistherefore (d) Z=t(\/2 +\/55% LESSON 18B. Algebra ofComplex Numbers. Complex numbers would notbeofmuch useifrules were notavailable bywhich wecould add, subtract, multiply, and divide them. The rules which have been laid down forthese operations follow theordinary rules ofalgebra with thenumber i2replaced byitsagreed value —-1. Ifzl=a+biand 22=c+diaretwocomplex numbers, then bydeﬁnition, (13-6) l1+Z2=(ll+bi)+(6+di)=(<1+6)+(b+(1)1) (18.61) 21—22=(a+ bi)—-(c+di) =(a—c)+(b—d)i, Lesson 18C Com=1.r:x FUNCTIONS 201 (18.62) 2122 =(a+bi)(c +di)=ac-1-adi+bci+bdi2 =('16—bd)+(ad+bc)i, 2__a-4—bi c— __(ac-I-bd)+(bc——ad)' (18.63) ;§—c+d,-c_ ._ c,+d, ‘ —ad 2 2 =@+m?*'~ °+d"°- Example 18.64. If21=3+5iand22=1-—3i,compute 2,+22, Z1"-Z2,Z152, 11/22- Solutions. 2,+22= (3+5i)+(1—3i)= 4+2¢. 2,-22= (3+5i) -(1-3.") =2+8i'. 2,2,=(3+5i)(1-31')=(3+15)+(—9+5)i=18—41'. 2l_3+5i_l+3i_(3-l5)+(9+5)i___12_,_14i 22__1—-3i1+3i_ 1+9 _ 1010'8‘5% LESSON 18C. Exponential, Trigonometric, and Hyperbolic Func- tions ofComplex Numbers. If2:represents areal number, i.e.,if2; cantakeononlyrealvalues, thentheMaclaurin series expansions (which youstudied inthecalculus) fore‘,sin:2:andcoszcare 2 3 (18.7) e‘=1+%+%—|—%+---, ——oo<:c<oo, 3 5 (18.71) sin:z:=a:—gL,+€—,—---, —oo<:z:<oo, 2 4 (18.72) cos:c=1—%+%—---, —-oo<:c<oo. Each ofthese series isvalid forallvalues of1,i.e.,each series converges forallx.If2represents acomplex number, i.e.,if2cantake oncomplex values, wedeﬁne 2 3 <18-73) e‘=1+%+§+§j+---. 3 5 (18.74) sinz=z——%—|—%i——---, (.) cos2—_—§—,—|—,T—---. 1875 —1*2‘4 Ithasbeen proved that each ofthese series alsoconverges forallz. 202 HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS Chapter 4 Example 18.76. Find theseries expansion forcosi. Solution. By(18.75) with 2=iand i2=-1,wehave (a). 1 1 1C0S't=1-I-5!‘-1-Ii-I-Ki-1-"' Note that cositurns outtobearealnumber. Bymeans of(18.73) to(18.75), wecanprove thefollowing identities. These aretheonesyouwillmeet most frequently, andtheonesweshall need. (18.8) (18.81) (18.82) (18.83) (18.84) (18.85) (18.86)e°=1, ezlezg =ezl+z,’ e"‘=cos2+isin2, e"‘=cosz —isin2, . 1 _s1n2 =Z(e”—e"), cos2=§(e“+e““), e‘250foranyvalue of2. Proofs of(18.8) to(18.86). Wegivebelow anoutline oftheproofs of (18.8) to(18.86). Rigorous proofs would bebeyond thescope ofthistext. Proof of(18.8). In(18.73) replace 2byzero. Proof of(18.81). By(18.73) 2 3 en: 2 3 e‘"=1+%+%+%+--- Since each series alsoconverges absolutely, wemay multiply them to obtain e“e” =1+(Z1+22)+ +2122+ ifL22 __z1z22 Q3)+(31+ 21+ 2!+81+ =1+(Z111122) +(Z12-!z2)2 +(Z1-gl-12):, +___ =e=1+=2 Lesson l8—Exercise 203 Proof of(18.82). Replace 2byizin(18.73). There results, with 1’=-1, - i2 11222 i323 11424 ‘"=1+"1‘1+3F+'8T+'U+"' . Z2 Z23 Z4 _ 22 2‘ ) 23 )+1_z__.§T+... =cosz +isin2[by(18.74) and(18.75)]. Proof of(18.83). Replace 2by—izin(18.73) andproceed asabove. Orreplace 2by-2in(18.82) andthen note by(18.74) and(18.75) that sin(—z) =-—sin2; cos(-2) =cos2. Proof of(18.84). Subtract (18.83) from (18.82) andsolve forsin2. Proof of(18.85). Add(18.82) and(18.83) andsolve forcos2. Proof of(18.86). By(18.81) and(18.8) e2e—: =e:+(-—-8) =e0=1‘ Since theproduct e'e" =1,e‘cannot have thevalue zeroforany2. Thecombinations es_e-s and es_‘_e-s 2 2 appear sofrequently inproblems that, forconvenience, symbols have been introduced torepresent them. These are (18.9) 5111112= (18.91) cosh2= sinh2e‘—e_'(18.92) tanhz -Ts,” -ei——,+e_'- These three functions arecalled respectively thehyperbolic sine, the hyperbolic cosine, andthehyperbolic tangent of2. EXERCISE 18 1.Find theconjugate ofeach ofthefollowing complex numbers. (a)1+1'. (b)3.(c)—3i. (d)5—6i.(e)-2. (f)2i.(g)-3—4i. 2.Find theabsolute value ofeach ofthecomplex numbers in1above. 204 HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS Chapter 4 3 4-. 5 6 7 8Find Arg2andthepolar form of2ofeachofthecomplex numbers in1above. Find therectangular form of2,ifitspolar form is: (a)2(cos 2+ isin - (e)\/2(cosix—|—isinQ1). (b)3(cos %—|—isin - (f)\/3(cos2:1-1-isinZ1’). (c)4(cos ‘I’—|—isin1r). (8)\/3(cos§-Ir-1-isingr). (<1)5(cosg+11111 - Given 21=-2+i,22=3-2i. Find: (a)21+22. (b)21-22. (c)2122. (d)21/22. If21and22aretwocomplex numbers, prove: (a)21+22=21+ 22- (<1)2122=2122- (b)-21 =-21. (d)21/22 -21/22, 229-40. With thehelpof6,prove thatif21,22,23arecomplex numbers and Z=Z112+Z1(=2 +Z3)+Z1-1-22-is-l-21/22, 22#0, then E=323%-55+ Z)+2_1+5 -5+5/E. No'rE. Ifyouhave proved 7,then youhave proved thatif2isobtained byadding, subtracting, multiplying, anddividing complex numbers 21,22, 23,---,then2canbeobtained byperforming thesame arithmetic operations onH,5,Z,~-~- With thehelpof7,prove thatifrisarootof (a) ate"+a.._1w"_1 +---+an+do=0, (b) 9. 10 1 2 3.then risarootof _ _ -1 _ _a,,:c"-1-a,,_1a:" —|—----1-a12—|—an=0. With thehelp of8,prove that ifthecoeﬂicients in8(a)arerealandrisa rootof8(a), then 1‘isalsoarootof8(a). NOTE. Ifyouhave succeeded in proving 9,then youhave proved thefollowing very important theorem. If thecoeﬂicients in8(a) arereal, thenitsimaginary roots must occur incon- jugate pairs. Prove thattheproduct oftwoconjugate complex numbers isapositive real number. ANSWERS 18 (.1)1-1. (11)3.(C)81'. (<1)5+6i(e)-2 (f)-2.‘.(g)-8+4.". (11)\/8. (11)3.(c)8.(d)\/H. (.1)2.(f)2.(g)5. (a)%»\/2(cos-:~+isin:E)- Lesson 19A LINEAR Iunsrannsuca orFUNCTIONS 205 (b)O,3(cos 0°+isin 0°). (c)3%»3(cos 3§+ isingg)- (d)s09°4s’, \/(T(cos309°4s' +isin309°4s'). (e)1r,2(cos 1r+isin1r). 1' ‘I’ .. ‘I’ (f)-2-,2(cos 5+ 181115) - (g)233°8', 5(cos 233°8' —|—isin233°8'). 4.(a)\/5+1'\/5. (b)-g+i=§'\/§. (0)-4. (<1)51". (e)-1-i.(r)—\ﬁi.(2)1}\/5-ig. 5.(a)1—i. (b)—5—|—311. (c)-4+711. (d)-335 —ills. LESSON 19. Linear Independence ofFunctions. The Linear Differential Equation ofOrder n. LESSON 19A. Linear Independence ofFunctions. Deﬁnition 19.1. Asetoffunctions f1(:e), f2(x), ---,f,,(a:), each de- ﬁned onacommon interval I,iscalled linearly dependent onI,if there exists asetofconstants cl,C2,---,en,notallzero, such that (19-11) ¢1f1($) +¢2f2(1»‘) +~+¢».fn($) =0, forevery xinI.Ifnosuch setofconstants cl,c2,---,c,,exists, then theset offunctions iscalled linearly independent. Deﬁnition 19.12. Theleftsideof(19.11) iscalled alinear combina- tion ofthesetoffunctions f1(:::), f2(:|:), ---,f,,(:c). Example 19.13. Foreach ofthefollowing setsoffunctions, determine whether itislinearly dependent orindependent. 1.1:,-—2:c,——3:v,4:::, I:—oo <iv<<73. 2.x",:v", p¢q, :c>0, :v<0. 3.e"‘,e“, p9-6q, I:—oo <as<oo. 4.e“,0,sina:, 1, I:—oo <:v< 00. Solution. First weobserve that allthefunctions ineach setare deﬁned ontheir respective intervals. Second weform foreach setthe linear combination called forin(19.11). The linear combination ofthesetoffunctions in1,equated tozero, is (11) 61¢+c2(—2=v) +ca(—3Iv) +c4(4Iv) =0- Wemust now askandanswer thisquestion. Does asetofc’sexist, not allzero, which willmake (a)atrue equation forevery :2:intheinterval 206 HIGHER Onnan LINEAR Drrrsamrruln Eouxrrons Chapter 4 —-oo <1:<oo?Rewriting (a)as (1)) (61—262-"363-l"464)"? =0» weseeinunediately there areaninﬁnite number ofsetsofc’swhich will satisfy (b),andtherefore (a),forevery xinI.Forexample, (<=) ¢1=2, ¢2=1. ¢a=0. ¢4=0; c1=5, c;=3, c3=1, c4=1, aretwosuch sets. Hence byDeﬁnition 19.1thesetoffunctions in1is linearly dependent. Thelinear combination ofthesetoffunctions in2,equated tozero, is 6113’ +62$‘, = Ifweassume ac"and2;’arelinearly dependent functions inI,then by Deﬁnition 19.1, there must exist constants c1and02both notzerosuch that (d)isanidentity inx.Since clandC2arenotboth zero, wemay choose oneofthem, sayc1;é0.Dividing (d)by01x‘, weobtain (e) x""=—£"3: paéq.61 Thevalue oftheleftsideof(e)varies with each :0intheinterval :1:>0, :1:<0.Theright side, however, isaconstant forﬁxed values ofcland e2.Hence toassume that1:1’,an’arelinearly dependent leads toacontra- diction. They must therefore belinearly independent. Theproof thatthefunctions in3,e",e",paéqarelinearly independent ispractically identical with theproof thatac’andav”,p¢qarelinearly independent. Ithastherefore been lefttoyouasanexercise; seeExercise 19,1. Thelinear combination ofthesetoffunctions in4,equated tozero, is (f) c1e’+c20+c3sina:+c4-1=0. Choose cl=03=c4=0andCg=anynumber notzero. Then (f)will read (g) 0-e’+c-2-0+0-sin:c+0-1=0, which isatrueequation forevery xinI.Hence thegiven setislinearly dependent. Comment 19.14. Whenever asetoffunctions contains zeroforone ofitsmembers, thesetmust belinearly dependent. Allyouhave todois tochoose forevery cthevalue zero, except theonewhich isthecoefficient ofzero. Lesson 19B Tar: Lmmn DIFFERENTIAL Eouxrron orOanrzn n207 Ifasetoffunctions were selected atrandom, onecould work along time trying toﬁndasetofconstants cl,C2,---,c,,,notallzero, which would make (19.11) true. However, afailure toﬁndsuch asetwould notof itself permit onetoassert thatthegiven setoffunctions wasindependent, since thepossible values assignable tothesetofc’sareinﬁnite innumber. Itisconceivable that such asetofconstants exist, noteasily discoverable, forwhich (19.11) holds. Fortunately there aretests available todeter- mine whether asetoffunctions islinearly dependent orindependent. A discussion ofthese tests willbefound inLesson 63BandExercise 63,5. You may bewondering what linear dependence andlinear independence ofasetoffunctions hastodowith solving alinear differential equation of order n.Itturns outthat ahomogeneous linear differential equation has asmany linearly independent solutions astheorder ofitsequation. (For itsproof, seeTheorems 19.3 and65.4.) Ifthehomogeneous linear differ- ential equation, therefore, isoforder n,wemust ﬁndnotonly nsolutions butmust alsobesure that these nsolutions arelinearly independent. And asweshall show, thelinear combination ofthese nsolutions willbe itstrue general solution. Forinstance, ifwesolved afourth order homo- geneous linear differential equation andthought thefourfunctions :z:,—2:v, -317, 4:2:inExample 19.13—1 were itsfour distinct anddifferent solutions, wewould bevery much mistaken. Allofthem canbeincluded intheone solution y=cx.Hence wewould have tohunt forthree more solutions. Ontheother hand, thefunctions :1;andx2,which weproved were linearly independent (seeExample 19.13—2 with p=1,q=2),could betwodis- tinct solutions ofahomogeneous linear differential equation oforder two. Inthat event, thelinear combination clx—l-02x2 would beitsgeneral solution. Hence insolving annthorder homogeneous linear differential equation, wemust notonly ﬁnd nsolutions, butmust also show that they are linearly independent. LESSON 19B. The Linear Differential Equation ofOrder n. Wehave already deﬁned in18.1thelinear differential equation oforder n. Inconnection with thisequation, there aretwoextremely important theorems. The proof oftheﬁrst one, Theorem 19.2 below, istoocom- plicated tobegiven atthisstage andwould only lead ustoofaraﬁeld. Inorder, therefore, nottodelay thestudy ofmethods ofsolving nth order linear differential equations more than necessary, wehave post- poned itsproof toLesson 65.Theproof ofthesecond one, however, Theorem 19.3below, isgiven inthislesson. Theorem 19.2. Iff0(x), f1(:2:), ---,f,,(:c) andQ(:c) areeach continuous functizms of:1:onacommon interval I,andf,,(a:) sf0when xisinI,then 208 Hronnn Osman LINEAR DIFFERENTIAL Eouxrrons Chapter 4 thelinear diﬁerential equation (19-21) f»(w)y‘"’ -l-fn-1(3)!/("'” +'''-l-f1(1v)y' +fo(w)y =Q(w) hasoneandonly onesolution, (19-22) 1/=1/(iv), satisfying thesetofinitial conditions (19-23) 1/($0) =1/0, 1/($0) =1/1, "', 1/"_D(1Po) =I/n-1, where 1:0isinI,andyo,y1,---,y,,_1 areconstants. This theorem isanexistence andauniqueness theorem. Itisanexistence theorem because itgives theconditions under which asolution of(19.21) satisfying (19.23) must exist. Itisalso auniqueness theorem because it gives conditions under which thesolution of(19.21) satisfying (19.23) is unique. Asremarked previously, theproof ofthisveryimportant theorem hasbeen postponed toLesson 65;seeTheorem 65.2. Right nowweshall content ourselves with theproofs ofthree important properties oflinear differential equations that weshall need foraclearer understanding oftheremaining lessons ofthischapter. These properties areincorporated inthefollowing theorem. Theorem 19.3. Iffo(:c), f1(:z:), ---,f,,(:c) andQ(z) areeach continuous functions of:1:onacommon interval Iandf,,(x) 96Owhen xisinI,then: 1.Thehomogeneous linear diﬁerential equation (19-31) f1-(16)?/l") +fn-1(w)y‘"_” +---+f1(=v)1/' +fo(¢)!/ =0 hasnlinearly independent solutions y1(:v), y2(a:), ---,y,,(:c). 2.Thelinear combination ofthese nsolutions (19-32) y.(w)=on/1(w) +en/2(r) +---+cny-.(r). where c1,C2,---,0,,isasetofnarbitrary constants, isalsoasolution of (19.31). Itisann-parameter family ofsolutions of(19.31). (Weshall explain later thesigniﬁcance ofthesubscript cin‘y,.) 3.Thefunction (19-33) y(w)=y¢(w)+yp(w), where yc(a:) isdeﬁned in(19.32) andy,,(x) isaparticular solution ofthe nonhomogeneous linear diﬁerential equation corresponding to(19.31), namely (19-34) f..(w)y("’ +f»_1(w)y‘"_” +---+f1(w)1/’ +fo(@)!/ =Q(¢). isann-parameter family ofsolutions of(19.34). Lesson 19B Tm’-: Lmma DIFFERENTIAL Equxrron orOsman n209 Proof of1.Weshall, forthepresent, limit theproof of1tothespecial casewhere thecoefﬁcients f0(x), ---,f,,(:c) areconstants. Theproof will consist inactually showing, inlessons 20to22which follow, howtoﬁnd these nlinearly independent solutions. Theproof, forthecasewhere the coefficients arenotconstants, hasbeendeferred toLesson 65,Theorem 65.4. Proof of2.Byhypothesis each function yl,y2,---,y,,isasolution of(19.31). Hence each satisﬁes (19.31). This means, byDeﬁnition 3.4 andtheRemark atthebottom ofpage 22,that (3) f»($)1/1(") +fn—1($)l/1(n_1) +'''-1-f1($)?/1' —|—f0(1?)!/1 =0, fn(111)3/2“) +1»-1(3)?/2(”_1) -1-'''-1-f1(1?)?l2' -l"fo($)1/2 =0, f..(w)y.."" +f.._1(w)y.."“” +---+f1(w)1/..' +fo(w)y.. =0- Multiply theﬁrstequation of(a)byc1,thesecond byc2,---,thelastby c,,,where 01,C2,---,c,,arenarbitrary constants, and add them. The result is (b) f..(1)[¢1y1‘”’ +cm“) +---+¢..y..""l+f“_1(x)[c1y1(n—1) +c2y2(n--1) +___+cnyn(n--1)] +___ +fo(w)l¢1y1 +cm+---+61:1/nl=0- This lastequation canbewritten as* (C) f..(w)[¢1y1 +cm+---+ch/l‘”’ '1‘fn-1(5'3)l¢1!/1 -l"'''+cnl/nl(n_l) '1‘‘‘‘ +f0(17)lc1I/I '1'‘''+cnl/nl =0- By(19.32) each quantity inside thebrackets isyc.Hence (c)is (d) fn($)1/c(n) '1'fn-1(x)yc(n_1) +''''l'f0($)llc =0» which saysthatyasatisﬁes (19.31). Itistherefore asolution. Since itis alinear combination ofnindependent solutions andcontains nparameters, itisann-parameter family ofsolutions. Proof of3.Byhypothesis, y,,isaparticular solution of(19.34). Hence, byDeﬁnition 3.4, (8)f»(fv)yp(") +f»-i(1=)1/15"") +---+f1($)!/,1’ 'l'fo($)9p =Q01)- 'For example, u"<=>+-/'<==>-‘%+“%‘3-=W =<11+o"- 210 Hrcnan Onnmz LINEAR Drrranrznruu. EQUATIONS Chapter 4- Adding (d)and(e)weobtain (1) f»(@>)(!/c +yp)(") +fn_1(fv)(y¢ +yp)("'” '1'''‘'l‘f0(37)(l/c "1'yr)= This lastequation says that (ye+y,,)satisﬁes (19.34) andistherefore a solution. Further, since y=y,+y,,contains nparameters, itisan n-parameter family ofsolutions of(19.34). Definition 19.4. The solution y¢(:r) in(19.32) ofthehomogeneous equation (19.31) iscalled thecomplementary function of(19.34). Hence theuseofthesubscript ciny,(x). Remark. The subscript piny,,(x) of(19.33) isused todistinguish it from they,part ofthen-parameter family ofsolutions. Comment 19.41. Weshall prove inLesson 65,seeTheorem 65.5, that thefunction y¢(x) of(19.32), which isann-parameter family ofsolu- tions of(19.31), isinfactitstrue general solution inaccordance with our Deﬁnition 4.7. Every particular solution of(19.31) canbeobtained from itbyproperly choosing thenarbitrary constants. Wetherefore shall refer tothisn-parameter family ofsolutions asageneral solution. Weinfer fur- ther from thistheorem that ifoneperson hasobtained thenindependent solutions yl,1/2,---,y,,(whose linear combination isthegeneral solution) byusing onemethod, then itisnotpossible forasecond person using an- other method toﬁnd ageneral solution which isessentially different from it.The second person’s solution willbeobtainable from theﬁrst by aproper choice oftheconstants cl,C2,---,c,,in(19.32). Asimilar statement canbemade forthefunction y(z) in(19.33). The proof willbefound inTheorem 65.6. Weshall therefore refer toitasthe general solution of(19.34). EXERCISE 19 1.Prove that ifp96q,thefunctions e"ande"arelinearly independent. Hint. Follow themethod used inproving thelinear independence ofthe functions inExample l9.13—2. 2.Prove thatthefunctions e"andxe“arelinearly independent. Hint. Make useofExample 19.13—2 with p=0,q=1,andthe-fact that 2'#0for allz. 3.Prove thatthefunctions sin2:,0,cosa:arelinearly dependent. 4.Prove thatthefunctions 3e2‘and—2e2‘ arelinearly dependent. Itisextremely important thatyouprove thestatements inExercises 5to 7below. Follow themethod used toprove statements 2and3ofTheorem 19.3. 5.Ify,isasolution of (19-5) f-.(¢)1/"') +---+f1(1)y' +fo(1)y =Q(I). then Ay,isasolution of(19.5) withQ(z) replaced byAQ(:c). Lesson 20 LINEAR Eouxrron wrrn CONSTANT COEFFICIENTS 211 6.Principle ofSuperposition. (Also seeComment 24.25.) Ifgmisasolution of(19.5) with Q(z) replaced byQ|(x) andy,,,isasolution of(19.5) with Q(z) replaced byQz(a:), then y,=1/,1+yp,isasolution of fr-(r):l/"" +---+fi(1¢)y' +fo(1)y =Q1(I) +Q2(I)- 7.Ify,,(a:) =u(a:)—|—1Iv(x) isasolution of (19-51) f..(¢)y"" +---+f1(1)1/' +fo(I)1/ =13(1)+119(1), where fo(a:), ---,f,,(z) arerealfunctions of2:,then (a)therealpartofy,,,i.e.,u(x), isasolution of f»(1)y"') +~--+fi(I)z/’ +fo(1)y =R(1), (b)theimaginary partofy,,,i.e.,v(x), isasolution of nmww~~+nmr+nmy=so Hint. Two complex numbers areequal ifandonly iftheir realparts are equal andtheir imaginary parts areequal. 8.Assume that y,=:2isasolution ofy”+ y’-2y=2(1+ 2:—:2). Use5 above toﬁnda particular solution ofy"+y’-—2y=6(1—|—:2:——2:2). Answer: y,,=32:2. Verify thecorrectness ofthisresult. 9.Assume thaty,1=1+0:isasolution of 1/”—y’+1/=1, andy,,,=e2‘isasolution of 1/”—1/’+y=3e"- Use 6above toﬁnd aparticular solution ofy”—y’+y=:0:—|—3e2". Answer: y,,=1+1:+e2’.Verify thecorrectness ofthisresult. 10.Assume that y,,=(-115cos2:—3%sin2:)+i(315sin:2:+13;;cosx)isasolu- tion ofy”—-3y’—|—2y=e“=cos:0+isin2:.Use7above toﬁnd a particular solution of(a)y"—-3y’+2y=cos2:,(b)y"—-3y’+2y= sin1.Ans. (a)y(z) =T15cosa:—335sin2:.(b)y(z)=315sinx+335cos2:. Verify thecorrectness ofeach ofthese results. ll.Prove thattwofunctions arelinearly dependent ifoneisaconstant multiple oftheother. 12.Assume f1,fz,f3arethree linearly independent functions. Show that the addition tothesetofoneofthese functions, sayf1,makes thenewsetlinearly dependent. 13.Prove that asetoffunctions f1,f2,---,_f,,,islinearly dependent iftwo functions ofthesetarethesame. 14.Prove thatasetoffunctions, f1,f2, ---,f,,,islinearly dependent, ifasubset, i.e.,ifapartoftheset,islinearly dependent. LESSON 20. Solution oftheHomogeneous Linear Dilferential Equation ofOrder nwith Constant Coefﬁcients. LESSON 20A. General Form ofItsSolutions. Inactual practice, equations ofthetype (19.31), where thecoeﬁicients arefunctions of:1: with norestrictions placed ontheir simplicity orcomplexity, donot usually have solutions expressible interms ofelementary functions. And 212 HIGHER Onmsn Lmmn DIFFERENTIAL EQUATIONS Chapter 4 even when they do,itisingeneral extremely diﬂicult toﬁnd them. If, however, each coeﬁicient in(19.31) isaconstant, then solutions interms ofelementary functions canbereadily obtained. Forthenext fewlessons, therefore, weshall concentrate onsolving thedifferential equation (20.1) a,,y(") -1-a,,_1y("_n +----1-a1y' -1-aoy=0, where a0,a1,---,anareconstants andan960. Without going into thequestion ofmotivation, letusguess that a possible solution of(20.1) hastheform (20.11) y=em”. Wenow askourselves thisquestion. Forwhat value ofmwill(20.11) be asolution of(20.1)? ByDeﬁnition 3.4,itmust beavalue forwhich n n1 (20.12) an8%;em—|—a,,_1 £1: em+---—|—a1(%em’+aoem =0. m__ m:Since thekthderivative ofe‘-mke ,wemay rewrite (20.12) as (20.13) a,,m"e"“’ +a,,_1m"'"‘e"“ -1----—|—a1me"“‘ -1-aoem =0. By(18.86), em‘960forallmandx.Wetherefore candivide (20.13) by ittoobtain (20.14) a,,m" +a,,._1m"“1 +----1-alm —|—an=0. Weatlasthave theanswer toourquestion. Each value ofmforwhich (20.14) istrue willmake y=em‘asolution of(20.1). But (20.14) isanalgebraic equation inmofdegree n,andtherefore, by thefundamental theorem ofalgebra (seeTheorem 18.23), ithasatleast oneand notmore than ndistinct roots. Letuscallthese nroots ml, mg,---,mn,where them’sneed notallbedistinct. Then each function (20-15) U1=e"“‘, 92=8"”. "'. 11..=e"‘"" isasolution of(20.1). Deﬁnition 20.16. Equation (20.14) iscalled the characteristic equation of(20.1). Norm. The characteristic equation (20.14) iseasily obtainable from (20.1). Replace ybymandtheorder ofthederivative byanumerically equal exponent. Insolving thecharacteristic equation (20.14), thefollowing three pos- sibilities may occur. 1.Allitsroots aredistinct andreal. 2.Allitsroots arerealbutsome ofitsroots repeat. 3.Allitsroots areimaginary. Lesson 20B Roors REAL ANDDrsrmcr 213 Weshall discuss each oftheabove three possibilities separately. Other possibilities may also occur as,forexample, when allroots aredistinct butsome arereal and some areimaginary. Itwill bemade apparent why such other possible combinations donotrequire special consideration. Remark. Wehave already commented, seeComment 18.26, inregard tothediﬂiculty, ingeneral, ofﬁnding thenroots ofthecharacteristic equation (20.14). Ifanequation ofthistype, ofdegree greater than two, were written atrandom, theprobability isvery high that itwould have irrational orcomplex roots which would bedifficult andlaborious toﬁnd. Since this isadifferential equations text and notanalgebra text, the examples used forillustration and exercises have been carefully chosen sothat their roots canbereadily found. Inpractical problems, however, ﬁnding theroots oftheresulting characteristic equation may notbeand usually will notbeaneasy task. Wecannot emphasize this point too strongly. LESSON 20B. Roots oftheCharacteristic Equation (20.14) Real and Distinct. Ifthenroots ml,mg,---,m,,ofthecharacteristic equation (20.14) aredistinct, then thensolutions of(20.1), namely, I/1 =emlxr l/2=emit: '''2yn =em“: arelinearly independent functions. (For proof when n=2,seeExercise 19,1. Forproof when n>2,seeExample 64.2.) Hence byTheorem 19.3 [seeinparticular (19.32)] andComment 19.41, (20.21) y.=c1e’”1‘ -1-c2e"'*" -1----+c,,e""" isthegeneral solution of(20.1). Example 20.22. Find thegeneral solution of (a) y”'+2y”—y’—2y=0- Solution. ByDeﬁnition 20.16, thecharacteristic equation of(a)is (b) m3+2m2-m—~2=0, whose roots are (c) ml=1, mg=——l, m3=-2. Hence by(20.21) thegeneral solution of(a)is (d) ye=cle”+c2e_” +c3e'2“’. Example 20.23. Find theparticular solution y(z) of (fl) y"—3y’+2y=0, forwhich y.(0) =1,y¢'(0) =0. 214 Hrennn ORDER LINEAR Drrrnnnmrnt EQUATIONS Chapter 4 Solution. ByDeﬁnition 20.16, thecharacteristic equation of(a)is (b) m2—3m+2=0, whose roots are (c) m1=1, m2=2. Hence by(20.21) thegeneral solution of(a)is (<1) 21¢=vie”+6262”- Bydifferentiation of(d),weobtain (6) 2/.’=ere‘+26262‘- Substituting thegiven initial conditions zr=0,y,,=1,y,’=0in,(d) and(e),there results (fl 1=61+62, 0='C1+262. Solving (f)simultaneously forcland C2,weﬁnd cl=2,c2=--1. Substituting these values in(d),weobtain therequired particular solution (g) y=28’—6“- LESSON 20C. Roots ofCharacteristic Equation (20.14) Real but Some Multiple. Iftwoormore roots ofthecharacteristic equation (20.14) arealike, then thefunctions (20.15) formed with each ofthese n roots arenotlinearly independent. Forexample, thecharacteristic equa- tion of (=1) y”—4y’+4y=0 IS (b) m2—4m+4=O, which hasthedouble rootm=2.Itiseasy toshow thatthetwofunc- tions yl=e2‘and1/2=e“arelinearly dependent. Form thelinear combination (c) c1e2" +c2e“ =0 andtakecl=1,c2=—1. Hence thegeneral solution of(a)could not bey=clez’ -1-c202‘. (Remember, ageneral solution ofasecond order linear differential equation isalinear combination oftwo linearly inde- pendent solutions.) Actually wecanwrite thesolution as (<1) 1/=01¢"+62¢“=(61+cm" =Ce”, Lesson 20C Roors Ran. amSour: Mourrrnn 215 from which weseethat wereally have only onesolution andnottwo. Wemust therefore search forasecond solution, independent ofea’. Togeneralize matters forthesecond order linear differential equation, weconﬁne ourattention totheequation (20.3) y”—2ay’ +a2y=0, whose characteristic equation (20.31) m2—2am+a2=0 hasthedouble rootm=a.Let (20.32) y,,=ue“, where uisafunction of2:.Wenowaskourselves theusual question. What must ulooklikefor(20.32) tobeasolution of(20.3)? Weknow, byDeﬁnition 3.4,that(20.32) willbeasolution of(20.3) if (20.33) (ue“)” —2a(ue“")' +a2(ue“') =0. Performing theindicated differentiations in(20.33), weobtain (20.34) e“’(u" +2au’+azu-—2au’ -—2a2u +a2u) =0, which simpliﬁes to (20.35) e“u” =0. By(18.86), e“950.Hence, (20.35) willbetrueifandonlyif (20.36) u”=0. Integration of(20.36) twice gives (20.37) u=cl+C213. Wenowhave theanswer toourquestion. If,in(20.32), uhasthevalue (20.37), then (20.38) y,=(cl+c2:c)e“ willbeasolution (20.3). Itwillbethegeneral solution of(20.3) provided thetwofunctions e“andxe“ arelinearly independent. The proof that they areindeed linearly independent waslefttoyouasanexercise; see Exercise 19,2. Hence byTheorem 19.3 and Comment 19.41, (20.38) isthegeneral solution of(20.3). Ingeneral itcanbeshown that ifthecharacteristic equation (20.14) hasaroot m=a,which repeats ntimes, then thegeneral solution of (20.1) is (20.4) y.=(cl-1-C2113-1-032:2 -1----+c,,a:"'"1)e“‘. 216 Hronnn Onnnn LINEAR Drrrnnnrrrmn Eqtwrrons Chapter 4 Andif,forexample, thecharacteristic equation (20.14) canbewritten in theform (20.41) m’(m —a)3(m +b)‘(m +c)=0, which implies thatitsroot are m=0twice, m=athree times, m=—bfourtimes, m=—conce, then thegeneral solution ofitsrelated differential equation is (20-42) lie=61+62$+(Ca+64$—|—65152)?“ +(ct+61%+car”+¢.x“‘>e"" +c.0e'°’- Observe thatwith each n-fold rootp,e”ismultiplied byalinear com- bination ofpowers of:1:beginning with :z:°andending with :c"_‘. Example 20.43. Find thegeneral solution of (9-) y“’—32”+21/= 0- Solution. Thecharacteristic equation of(a)is (b) m4—3m’+2m=0, whose roots are (c) m=0, m=1, m=1, m=——2. Since theroot1appears twice, thegeneral solution of(a),by(20.42), is (d) llc=01+(C2+63906’ +¢4e"’- Example 20.44. Find theparticular solution of (a) y"~—2y’+y=0, forwhich y,,(0) =1,y,,’(0) =0. Solution. Thecharacteristic equation of(a)is (b) m2—2m+1=0, whose roots arem=1twice. Hence thegeneral solution of(a),by (20.38), is (9) 1/<==(61-1-621$)?- Differentiation of(c)gives (d) ye’=(61+02+c2w)¢'- Lesson 20D Rooms Imornsny 217 Toﬁnd theparticular solution forwhich 1:=0,ye=1,y,’=0,we substitute these values in(c)and(d). The result is (e) 1=cl: 0=C1+C2. Thesimultaneous solution of(e)forclandC2gives cl=1,C2=—l. Substituting these values in(c),weobtain therequired particular solution (f) y=(1——x)e‘. LESSON 20D. Some orAllRoots oftheCharacteristic Equation (20.14) Imaginary. Iftheconstant coefficients inthecharacteristic equation (20.14) arereal, then (seeExercise 18,9) anyimaginary roots it may have must occur inconjugate pairs. Hence if0:+iﬁisoneroot, another root must be0:—iﬁ.Assume now that a—|—iﬁanda—-iﬁare twoimaginary roots ofthecharacteristic equation ofasecond order linear differential equation. Then itsgeneral solution by(20.21) is (205) ye=cl/e(a+ip)z _|_c2Ie(a—-1'5): =clreazeipz +C2/eaze-—igz =eaz(c1rei5z +C2/e-i/3:). By(18.82) and(18.83) (20.51) eiﬁ‘=cosﬁx+isin ﬁx; e_i""’ =cosﬁx--isin ﬁx. Substituting (20.51) inthelastequation of(20.5) and simplifying the result gives (20.52) y¢=e“[(c1' +cg’)cosﬁx—|—i(c1’ —-02')sinﬁx]. Wenow replace theconstant cl’—|—C2,byanew constant clandthecon- stant i(c1' —C2’)byanewconstant 02.Then (20.52) becomes (20.53) ye=e°“(c1 cosﬁx+c2sinﬁx), which isasecond form ofthegeneral solution (20.5). Athird form ofwriting thegeneral solution (20.5), more useful for practical purposes, isobtained asfollows. Write theequality =V61 +022 COSﬂ23+ SiIlB1l7)' 01 62 61 2(20.54) clcosﬁx+02sinﬁx ’( 218 HIGHER ORDER LINEAR Drrrnnnrrrru Eqourons Chapter 4 -zcf+62 91 <52 Figure 20.55 Weseefrom Fig.20.55 that (20.56) sina=-i—, cosas=-—°*—---\/ci’+622 V612+622 The substitution ofthese values intheright sideof(20.54) gives (20.57) clcosﬁx+c2sinﬁx=Val? +c2"(sin6cosﬁx+cos6sinﬁx) =V012 +022sin(ﬁx+6). Prove asanexercise that ifwehad interchanged thepositions ofcl andC2inFig.20.55, equation (20.54) would have become (20.58) clcosﬁx+c2sinﬁx=V012 +022(cos6cosﬁx+sin6sinﬁx) =V012 +622cos(ﬁx-—6). Replacing in(20.57) andin(20.58) anewconstant cfortheconstant \/cl? +027,wemay write thegeneral solution (20.53) ineither ofthe forms (20.59) yc=cc“sin(ﬁx+6)orya=cc“cos(ﬁx—6). Norm. The6intheﬁrstequation isnotthesame asthe6inthesecond equation. Hence inthecase ofcomplex roots, thegeneral solution ofalinear differential equation oforder two, whose characteristic equation hasthe conjugate roots a+ﬁianda-—ﬁi,canbewritten inanyofthefollowing forms: (206) (a)ye=cle(a+1'p)z +c2e<¢-ape, (b)yc=e""(c1 cosﬁx+C2sinﬁx), (0)ya=w“Bin(ﬁx+6), (d)y,=ce""cos(ﬁx—6). The signiﬁcance ofthetwoarbitrary constants cand 6which appear in (20.6) (c)and(d)willbediscussed inLesson 28. Lesson 20D Roors IMAGINARY 219 Example 20.61. Find thegeneral solution of (a) y”-—2y’+2y=0. Solution. The characteristic equation of(a)is (b) m2-2m+2=0, whose roots are1=1:i.Hence in(20.6), a=1,ﬁ=1.Thegeneral solution of(a),therefore, canbewritten inanyofthefollowing forms: (0) ye=c1e(1+i)x +c2e(1—i)x, y,,=e‘(c1 cosx—|—c2sinx), ya=cc”sin(x+6), y,=cezcos(x-—6). Ifthelinear differential equation (20-7) 414?/(4) +(la?/” +1121/” -l"<11!/' -l"any=0,1145*0, hasaconjugate pair ofrepeated imaginary roots, i.e.,if0:-1-iﬁ and a—iﬁeach occurs twice asaroot, then by(20.38) thegeneral solution of(20.7) is (20-71) 11¢=(61+czw)e("‘+"”" +(cs+c4w)¢(°‘_“”‘- The equivalent forms are (20.72) y,=e""[(c1 +C218)cosﬁx+(c,-,+c4x)sinﬁx], 11¢=¢°"l¢1 Bill(19%+51)+62$511*(I3-73+52)], yc=e°"[c1 cos(ﬁx—61)+62$cos(ﬁx-—62)]. Example 20.73. Find thegeneral solution of (a) y“’+2y"+1=0- Solution. Thecharacteristic equation of(a)is (b) m*+2m=+1=0; (m’+1)’=0, whose roots arem==|=itwice. Hence in(20.71) and (20.72) a=0, ﬁ=1.The general solution of(a)therefore canbewritten inanyofthe following forms: (0) ye=(61+cz=v)@" +(ca+c4w)e"", 3/»:=(‘F1—|—02$)COSI+(ca+04$)Sin1*» ye=01sin(x+61)-1-C213sin(x+62), y,,=c1cos(x—6,)—|—62$cos(x—-62). 220 Hronsn Ommn LINEAR Drrrmmnrmn Eqtwrrons Chapter 4 Example 20.74. Find thegeneral solution of (a) y/n (b) m-—12y” +22y’ -—20y=0 Solution. Thecharacteristic equation of(a)is 3—8m2+22m—20=0 whose roots are2,3=|=11.The general solution of(a),therefore, canbe written inanyofthefollowing forms: C1622 +c2e(3+€)x +c3e(3-6):, cle” +e3‘(c2 cosx+c3sinx), ole“ +c263’ cos(x——6), y,=01c“ +c-Zea‘ sin(x+6).(c) ‘$191? EXERCISE 20 Find thegeneral solution ofeach ofthefollowing equations. 99°F?==<=<e@5:: 9.y“’+4z/ + 10.y“)-azy=0,a>0.II III I-l-2y =0. 5.6y" —11y'+4y =0. --31/+2y=0. 6.y”+2y'—1/=0. —y=0. 7.y"'+y"—10y'—6y =0.8 ylll I/+ 4y! 0 +yn_6”! =0_ III I/ ll.y”—2ky'—2y=012.y+4lcy —-12lc2y =0.13_um=0. 14.y”+4y’+4y=0. 20.y(4)+2y”' —11y" —12y'+ 36y=0. 21. 22_y(4) 23.y” 24-.y" 25.1/(4)36y(4) .._ +511QIQ‘Q31 =+_(:'§:wsu+€<=,_._¥:5°y-4y’—2y=0. 17:y satisﬁes thegiven initial conditions. 31-yx=0,,1/(1)=2,32-y+411+4y=0-1/(0)33.y"—21/—|— 5y=0 34.y"~4y’+20y=35' 3y”! + 5y// + yl i l.y 2.y 3.yc1—|—C26Q9‘ -2: ole‘+cgez‘. cie’+age“I1. 1,1/(0)=I1/(1)= y(0) 2.1/(0)1/(1/2) =0,1/’ ANSWERS 20 99"?=e<e=e15. 3y://+ 5y” _|_ 16 ym n u I 4y'+ 5y=0. 26.y"— =0. 27.1/(4) =0. 28.y"'-|- =0. 29.y“)+4y"=0. +6y=0- 30-11“)+211"’+y’ Foreach ofthefollowing equations ﬁndaparticular 1. 4. (1/2) =0.1/(0)=0.u’(0)—61/+-2ay+ 18.y“)+3y"' 19.y(4)-—2y"— —4y =. Q»-I Q‘Q.9,g~g‘.—y=0. 81/= =0.— 0. II 99 +lF ~‘é°£_@_.+'i£=?F =0. solution which =1. =11 f/”(0) =_1' 01+age“ +c3e"3’. clex/2 +C2641/a_ cle(-1+./E): +c2e<-1-\/§)=_ Lesson 21A Msrnon orUnnsrenmnsn COEFFICIENTS 221 I-Ii-I!"'.°2°9°7"'=e=e<e=:=: 12.y 13.y 14-.y= 15.y 20.y 21.y= 22.y 23.y 24-.y 25.y= 26.y 27.y= 28.y 29.y 30.y= 31.y 32.y= 33.y 34-.y 35.y=c103‘ +cze('2+\/-5)’ +c3e(-2-\/-2')‘. c1+cze‘+c3e2‘ +c4e'2'. cle’+c2e"’ +c3e(-2+‘/5)‘ +c4e(_2-‘/-5)’. c1e\/5‘ +c2e"\/5’ +C3cos\/Zx+c4sin\/Ex. mo+-/7-T172)» +¢2e<r=—./Iﬁﬁn, c1e'6"" —|—6262'“. 16.y c1—|—C222+032:2+042:3. 17.y (c1—|—cgx)e_2’. 18.y c1e”/3 +(62+c3x)e". 19.y= (¢1+m)e" +(Ba+c41)e*3‘-he-1+ 6261/2 +C3651/6 +64¢-1/a_ (61+621962‘ -l"(vs+¢41)¢'2‘~ e‘(c1 cos2x+62sin22:). 1:131, 1:1§1, ,2((/5 \/§> 2 2(c1+cgx+c3x2)e2". (c1+cgx)e". c1+02$—|—c3x2+c4e'3‘. 01+c2:c+ c3e‘/§"+ c4e"\/5‘. cle 2 +626 2 =e’ c3cos—x+c4sin——x- c1cosx/§x+ cgsinx/§x+ c3cos\/2x+ c4sin\/2x. e2‘(c; cos4x—|—C2sin4x). (01+6227)cos\/2x+(c3—|—04:2)sin\/2x. c1e"2" +e‘[c2cos\/5x+c3sin\/5x]. c1+62$+c3cos2::+04sin2x. 01+ 62sinx+c3cos2:+04::sinx+c5xcosx. 3—x. (1+3x)e'2’. e’(2cos2x—|—sin2x). {e2"" sin4x. 9 2:/3 <x -4: 16° +416e' LESSON 21. Solution oftheNonhomogeneous Linear Differential Equation ofOrder nwith Constant Coefficients. LESSON 21A. Solution bytheMethod ofUndetermined Coeﬂi- cients. ByTheorem 19.3 and Comment 19.41, the'general solution of thedifferential equation (21.1) an/‘"’ +a.._1u‘"“” +---+aiy’+any=Q(w), where a,,#0andQ(x) é0inaninterval I,is (21.11) l/(3) =l/c(x) ‘l’l/11(33): where y¢(x), thecomplementary function, isthegeneral solution ofthe related homogeneous equation of(21.1) andy,,(x) isaparticular solution 222 Hronsa Oanaa LINEAR DIFFERENTIAL EQUATIONS Chapter 4 of(21.1). InLesson 20,weshowed how toﬁnd ye.There remains the problem ofﬁnding y,,. Theprocedure weareabout todescribe forﬁnding y,,iscalled the method ofundetermined coefficients. Itcanbeused only ifQ(x) consists ofasumofterms each ofwhich hasaﬁnite number oflinearly independent derivatives. This restriction implies thatQ(x) canonly con- tainterms such asa,x",e“,sinax,cosax,andcombinations ofsuch terms, where aisaconstant andkisapositive integer. SeeExercise 21,2. Forexample, thesuccessive derivatives ofsin2xare 2cos2x,-4sin2x,-8cos2x,etc. However, only thesetconsisting ofsin2xand 2cos2xislinearly inde- pendent. Theaddition ofanysucceeding derivative makes thesetlinearly dependent. Verify it.Thelinearly independent derivativcsof x3are 3x2,6x,6. Theaddition tothissetofthenext derivative, which iszero, makes the setlinearly dependent. SeeComment 19.14. However thefunction x", forexample, hasaninﬁnite number oflinearly independent derivatives. Toﬁndy,,bythemethod ofundetermined coefficients, itisnecessary tocompare theterms ofQ(x) in(21.1) with those ofthecomplementary function ye.Inmaking thiscomparison, anumber ofdifferent possibilities mayoccur, each ofwhich weconsider separately inthecases below. Case 1.NotermofQ(x) in(21.1) isthesame asatermof31,.Inthis case, aparticular solution y,of(21.1) willbealinear combination ofthe terms inQ(x) andallitslinearly independent derivatives. Example 21.2. Find thegeneral solution of (11) 1/"+4y’+4y=4x’+6e‘- The complementary function of(a)is (bl 1'/c=(¢1"l"62505-21- (Verify it.)Since Q(x), which istheright sideof(a),hasnoterm incommon with y.this case applies. Aparticular solution y,will therefore bea linear combination ofQ(x) and allitslinearly independent derivatives. These are,ignoring constant coefficients, x2,x,1,e‘.Hence thetrial solution y,must bealinear combination ofthese functions, namely (c) y,=Axz —|—Bx+C+De’, where A,B,C,Daretobedetermined. Successive derivatives of(c)are (d) y,’=2Ax +B—|—De‘, (e) yp”=2A+De‘. Lesson 21A METHOD orUNDETERMINED CoEr'r1c1EN'rs 223 Aswehave repeatedly remarked, (c)willbeasolution of(a)ifthesub- stitution of(c),(d),and (e)in(a)willmake itanidentity inx.Hence (c)willbeasolution of(a)if (f)2A+De‘+4(2Ax +B+De’) —|—4(Ax2 +Bx+C+De‘) E4x2—|—6e’. Simpliﬁcation of(f)gives (g)4Ax2 +(8A+4B)x +(2A+4B+4C)+9De’ E4x2+6e‘. Wenow askourselves thequestion: What values shall weassign toA,B, C,and Dtomake (g)anidentity inx?Weproved inExample 19.13, that x,x2arelinearly independent functions. The proof canbeextended toshow that x°,x,x2arealsolinearly independent functions. Hence the answer toourquestion is:values which willmake each coefficient oflike powers ofxzero. This means that thefollowing equalities must hold. (h) 4A= 8A—|-4B= 2A+4B+4C= 9D= 9°PPJ‘ Solving (h)simultaneously, there results (i) A=1, B=—2, C=%, D=§. Substituting these values in(c),weobtain (i) 1/p= $2-2w+%+§@’, which isa.particular solution of(a).Hence by(21.11), thegeneral solu- tion of(a)is(b)+(j),namely, (k) y=(cl+c2x)e_2" +x2-—2x+-3+§e‘. Example 21,21. Find thegeneral solution of (a) y”—3y’+2y=2xe3’ —|—3sinx. Solution. The complementary function of(a)is (b) 1/.=61¢’+we"- (Verify it.)Since Q(x), which istheright sideof(a),hasnoterm incom- mon with yc,aparticular solution ypwillbealinear combination ofQ(x) andallitslinearly independent derivatives. These are,ignoring constant 224 HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS Chapter 4 coefficients, xesz, ea‘,sinx,cosx.Therefore thetrial solution y,must be oftheform (c) y,,=Axe“ +Be” +Csinx+Dcosx. Successive derivatives of(c)are (d) y,,'=3Axe3’ +Ac“ —|—3Be3” +Ccosx——Dsinx, (e) y,”=9Axe3‘ +6Ae3‘ -1-9Be3’ —Csin x—-Dcosx. Thefunction deﬁned by(c)willbeasolution of(a)ifthesubstitution of (c),(d),and (e)in(a)willresult inanidentity inx.Making these sub- stitutions andsimplifying theresulting expression, weobtain (f)2Axe3’ +(3.4+2B)e3" +(C+3D)sinx -1-(D-—3C)cosxE2xe3' —|—3sinx. Equation (f)willbeanidentity inxifthecoefficients ofliketerms on each sideoftheequal signhave thesame value. Hence wemust have (g) 2A=2. 3A+2B=0, C+3D=3, —3C+D=0. Solving (g)simultaneously, there results (h) A=1, B=—§-, C=Ti‘5, D=19;;. Substituting these values in(c),weobtain (i) yp=xe3” —Z-e3‘ -1--1%;sinx-1--19;;cosx. Hence by(21.11) thegeneral solution of(a)is(b)-1-(i),namely (j) y=cle‘+czez‘ +xea’—§e3‘+-ﬁgsinx+1§6cos x. Case 2.Q(x) in(21.1) contains aterm which, ignoring constant coeﬁi- cients, isx'°times aterm u(x) ofyo,where lciszero orapositive integer. Inthiscase aparticular solution y,,of(21.1) willbealinear combination ofx'°+1u(x) andallitslinearly independent derivatives (ignoring constant coefﬁcients). Ifinaddition Q(x) contains terms which belong toCase 1, then theproper terms called forbythiscase must beincluded inyp. Example 21.3. Find thegeneral solution of (a) y"—3y’+2y=2x2+3e“. Lesson 21A METHOD orUNDETERMINED COEFFICIENTS 225 Solution. Thecomplementary function of(a)is (b) 2/.=ere’+62¢”- (Verify it.)Comparing Q(x), which istheright side of(a),with (b),we seethat Q(x) contains theterm e2”Which, ignoring constant coefficients, isx°times thesame term inye.Hence forthisterm, y,must contain a linear combination ofx°'He2" andallitslinearly independent derivatives. Q(x) alsohastheterm x2which belongs toCase 1.Forthisterm, there- fore, y,must include alinear combination ofitandallitslinearly inde- pendent derivatives. Informing thelinear combination ofthese functions andtheir linearly independent derivatives, wemay omit thefunction e2” since italready appears inyc;seeExercise 21,1. Hence thetrial solution y,must beoftheform (c) y,=Ax” +Bx+C+Dxez’. Successive derivatives of(c)are (d) y,,’=2Ax +B-1-2Dxe2‘ -1-De“, (e) yp”=2A+4Dxe2‘ -1-4De2’. Substituting (c),(d),(e)in(a)andsimplifying, weseethat(c)willbea solution of(a)if (f)2Ax2+(212-6.-in+(2A-3B+20)+De“E2x2+38“. Equation (f)willbeanidentity inxifthecoeﬂicients ofliketerms on each sideoftheequal signhave thesame value. Hence wemust have (g) 2A=2, 2B—6A=0, 2A—3B+2C=0, D=3. From (g)weﬁnd (h) A=1, B=3, C=Z-, D=3. Substituting these values in(c),there results (i) y,,=x2+3x+1}—|—3xe2’. Combining thissolution with (b),weobtain forthegeneral solution of(a) (j) y=x2+3x+-Q+3xe2' +cle‘+c2e2'. Example 21.31. Find ageneral solution of (a) y”—3y’+2y=xez‘ —|—sinx. 226 HIGHER ORDER LINEAR DIFFERENTIAL Eotmrrons Chapter 4 Solution. Thecomplementary function of(a)is (b) yo=c,e"+c2e2‘. Comparing Q(x) which istheright sideof(a),with (b),weseethat Q(x) contains aterm xez‘ which, ignoring constant coefficients, isxtimes a term ehiny¢.Forthisterm, therefore, y,,must bealinear combination ofx1'He2’ ==x2e2' and allitsindependent derivatives. Inaddition we notc Q(x) contains aterm sinxwhich belongs toCase 1.Forthisterm, therefore, y,,must include alinear combination ofitanditsindependent derivatives. Informing thelinear combination ofallthese functions and their independent derivatives, wemay omit thefunction e2’since it already appears iny,.Hence y,must beoftheform (c) y,,=Axzez‘ +Bxe2‘ +Csinx—|—Dcosx. Successive derivatives of(c)are (d) y,,’=2Ax2e2’ —|—2Axe2‘ +2Bxe2‘ +Be“ +C’cosx—Dsinx, (e)y,,”-.=4Ax2e2‘ +8Axe2’ +2.46" +4.3”" +43¢” —Csinx —Dcosx. Substituting (c),(d),(e)in(a)andsimplifying theresult, weseethat(c) willbeasolution of(a)if (f)2AM“ +(2A+B)e2“’+(0+3D)sinx —|—(D-—3C)cosxExez” —|—sinx. Equating thecoefficients ofliketerms oneach sideoftheequal sign, we ﬁndthat (g) 2A=l, 2A+B=0, C+3D=1, D—3C=0. From (g),weobtain G1) A-=§, B=—1, C=-116, D=-P5. Substituting these values in(c),there results (i) y,,=§x2e2’ —xe“ +315sinx+-195cosx. Thegeneral solution of(a)istherefore thesum of(b)and(i). Example 21.32. Find ageneral solution of (a) y”+y=sinsx. Solution. Thecomplementary function of(a)is (b) y,=clsinx+c2cosx. Lesson 21A METHOD orUNDETERMINED COEFFICIENTS 227 By(18.84), ix_ -ix 3 3ix _ —3ix ix_ -ix (°) sinax= =e-si +3(este) =-1sin3x+2sinx. Comgaring Q(x), which istheright sideof(c),with (b),weseethatQ(x) conta'ns aterm, which, ignoring constant coefficients, isx°times aterm sinxiny,.Hence thetrialsolution y,must beoftheform (d) y,,=Asin 3x+Bcos3x+Cx sinx+ Dxcos x. Successive derivatives of(d)are (e) y,’=3Acos3x —3Bsin 3x+Cxcosx+Csinx —Dxsinx +Dcosx. y,"=—9A sin3x—9Bcos3x—Cxsinx+2Ccosx —Dxcosx—2Dsinx. Substituting (c),(d),(e)in(a),weobtain (f)——8A sin3x—8Bcos3x+2Ccosx——2Dsinx =—1sin3x +§sinx. Equating coeﬁcients ofliketerms oneach side oftheequal sign, there results (s) —sA=-1, B=0, c=0, D=--3. From (g),wehave (h) A=31;, D=——§. Substituting these values in(d),weobtain (i) y,=-315sin3x—-§xcosx. Ageneral solution of(a)istherefore thesumof(b)and(i). Case 3.This case isapplicable only ifboth ofthefollowing condi- tions arefulﬁlled. A.The characteristic equation ofthegiven differential equation (21.1) hasanrmultiple root. B.Q(x) contains aterm which, ignoring constant coefﬁcients, isx"times aterm u(x) inye,where u(x) wasobtained from thermultiple root. Inthis case, aparticular solution y,will bealinear combination of x"+'u(x) andallitslinearly independent derivatives. Ifinaddition Q(x) 228 HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS Chapter 4- contains terms which belong toCases 1and 2,then theproper terms called forbythese cases must alsobeadded toy,,. Example 21.4. Find thegeneral solution of (a) y”+4y’+4y=3xe'2”. Solution. Thecomplementary function of(a)is (b) yc=c1e"2” +c2xe‘2’. Weobserve ﬁrstthat thecharacteristic equation of(a),namely m”+ 4m+4=0,hasamultiple root, m=-2. Secondly weobserve that Q(x) which istheright side of(a),contains theterm xe_2‘ which isx times theterm e_2" inyc(oralternately xe‘2" ofQ(x) isx°times theterm xe_2" inyc),andthat thisterm iny,came from amultiple root. Hence, bytheabove remarks under Case 3,r=2,k=1,andr—|—k=3,(or alternately r=2,k=0andr+k=2).Therefore, y,must bealinear combination ofx3e”2’ and allitslinearly independent derivatives [or alternately x2(xe_2‘), which yields thesame x3e'2', anditsderivatives]. Informing thislinear combination, wemay omit thefunctions e_2" and xe"2” since they already appear inyc.Hence y,,must beoftheform (0) y,=Ax3e"2"' +Bx2e"2‘. Thesuccessive derivatives of(c)are (d) yp’=—2Ax3e_2‘ +3Ax2e‘2’ -—2Bx2e_2" +2Bxe'2", (e) yp”=4Ax3e_2“ ——12Ax2e'2" -1-6Axe'2“’ +4Bx2e_2‘ —8Bxe_2’ +2Be'2". Substituting (c),(d),(e)in(a)andsimplifying theresult, weseethat (c) willbeasolution of(a)if / (f) 6Axe'2“ +2Be"2' E3xe"2’. I Equating coefficients ofliketerms oneach sideoftheequal signf, there results (g) A=§, B=0. Substituting these values in(c),weobtain <11) i.=as--*=. Thegeneral solution of(a)is,therefore, thesum of(b)and(h),namely (i) y=§x3e'2‘ —|—c1e_2’ -1-c2xe'2”. Lesson 21A METHOD orUNDETERMINED CoErrIcIENrs 229 Example 21.41. Find thegeneral solution of (a) y”+4y’+41/=367“- Solution. The complementary function of(a)is (b) ye=c1e_2‘ +c2xe_2". The characteristic equation of(a), namely m2+4m—|—4=0,hasa multiple root m=-2. The function Q(x), which istheright side of (a),contains theterm e"2“’which isx°times aterm iny,(oralternately x‘1times theterm xe_2’ inya). Since thisterm iny.came from amul- tiple root, thisCase 3applies. Hence bytheremarks under Case 3, r= 2,lc= 0,and r—|-k= 2(oralternately r= 2,k= ——1 and r—|—lc=1).Therefore y,must bealinear combination ofx2e"2‘ and allitslinearly independent derivatives [oralternately x(xe_2‘), which yields thesame x2e'2“’]. Informing this linear combination, wemay omit theterms e_2“ andxe_2‘ since they already appear iny¢.Hence y, must beoftheform (c) y,_,=Ax2e'2‘, y,’=2Axe"2‘ —2Ax2e_2’, y,,”=2Ae_2" —8Axe_2‘ +4Ax2e'2“. Substituting (c)in(a)andsimplifying theresult, weseethaty,willbea solution of(a)if (d) 2Ae_2” =3672", A= Substituting thisvalue ofAintheﬁrstequation of(c),weobtain (9) yr=25526-21- The general solution of(a)istherefore thesum of(b)and(e). Comment 21.42. Thefunction which wehave labeled y,,,since itdoes notcontain arbitrary constants, hasbeen correctly called, byourDeﬁni- tion 4.66, aparticular solution of(21.1). There are,ofcourse, inﬁnitely many other particular solutions ofthedifferential equation, oneforeach setofvalues ofthearbitrary constants intheyapartofthegeneral solu- tion y=ya+yp.These constants aredetermined intheusual way by inserting theinitial conditions inthegeneral solution y.Donotconfuse, therefore, aparticular solution obtained bythemethod ofundetermined coefficients and theparticular solution which will satisfy given initial conditions. Seeexample below. Example 21.43. Find theparticular solution of (a) y”——3y’+2y=6e”‘ forwhich y(0) =1,y’(0) =2. 230 HIGHER ORDER LINEAR DIrrEREN'rIAI. EqUA'rIoNs Chapter 4 Solution. Bythemethods outlined previously, verify that (b) yo=c1e'+c2e2’, y,,=e". However thisy,,isnottheparticular solution which satisﬁes theinitial conditions. Toﬁndit,wemust ﬁrst write thegeneral solution, andthen substitute theinitial conditions initandinitsderivative. Thegeneral solution of(a)anditsderivative are,by(b), (c) y=cle‘+czen +e"’, y’=cle‘+2c;e2‘ —e_'. Substituting in(c)theinitial conditions x=0,y=1,y’=2,weobtain (d) 1=c,+c2+1, 2=C1+262 * 1, whose solutions arecl=—-3,c2=3.Hence theparticular solution of (a)which satisﬁes thegiven initial conditions is,by(c)andthese values of61,62. (e) y=3e2“ ——3e’+e". LESSON 21B. Solution bytheUse ofComplex Variables. There isanother wayofsolving certain types ofnonhomogeneous linear equa- tions with constant coefﬁcients. Ifin (21-5) a..y‘"’ +a.._1y‘”"" +---+my’+110?!=Q(x). anrs0. thea’sarereal, Q(x) acomplex-valued function (i.e., afunction which cantake oncomplex values) andy,_,(x) isasolution of(21.5), then (see Exercise 19,7): 1.The real part ofy,isasolution of(21.5) with Q(x) replaced byits realpart. 2.The imaginary part ofy,isasolution of(21.5) with Q(x) replaced by itsimaginary part. Remark. Statements 1and2above would stillbevalid ifthecoeffi- cients in(21.5) were real, continuous functions ofxinstead ofconstants. Example 21.51. Find aparticular solution of (a) y”—3y’+2y=sinx. Solution. Instead ofsolving (a),letussolve thedifferential equation (b) y"-3y’+2y=6"- By(18.82), cl‘=cosx+isinx.Itsimaginary partis,therefore, sinx. Hence by2above, theimaginary part ofaparticular solution y,of(b) Lesson 2l—Exercise 231 willbeasolution of(a). Aparticular solution of(b),using themethod. of undetermined coefﬁcients is[take forthetrialsolution y,,=(A+Bz')e“’], (<=) up=fee"+riﬂe“ _ =-I16(cosa: +isinx) +ii;(cosx —|—isinx) =11600331 —-fgsinx +z'(—115-sinx —|—ﬁcosx). The imaginary part ofthesolution y,,is-ﬁn-sina: —|—ﬁcos :0.Hence a particular solution of(a)is (d) yp=ﬁsinx +1%-cos 2:. Question. What would aparticular solution of(a)be,ifinitsinan were replaced bycos1:?[Ans. Therealpart ofy,,,namely y,=-115cos:1: —3%sin2:.] EXERCISE 21 1.Prove that anyterm which isinthecomplementary function y,need not beincluded inthetrial solution yp.(Hint. Show that thecoeﬁicients of thisterm willalways addtozero.) 2.Prove thatifF(a:) isafunction withaﬁnite number oflinearly independent derivatives, i.e.,ifF(")(a:), F("-"(1), ---,F’(1),F(a:)arelinearly independ- entfunctions, where nisaﬁnite number, thenF(:c)consists onlyofsuchterms asa,:v",e",sinax,cosax,andcombinations ofsuch terms, where aisa constant andkisapositive integer. Hint. Setthelinear combination ofthese functions equal tozero, i.e.,set C',.F""(1) +C..-1F""”(1)+ ---+6'iF'(1)+ 6'oF($) =0, where theC"sarenotallzero, andthen show, byLesson 20,thattheonly functions F(:2:) thatcansatisfy thisequation arethose stated. Find thegeneral solution ofeach ofthefollowing equations. " y’—|—2y=4. 10.y"—2y’-—8y=92:0’+10c". +2y=12e‘. ll.y”—-3y’=2e2’sin2:. 2y=e“. 12.y(‘*)—-2y"+y=:2:-sin2:. 2y=sin2:. 13.y” =$2+22:. y=cos2:. 14.y =:4:+sin22:. =8+6e'+ 2sin2:. 15.y =4a:sin:0. =:2. 16.y =zsin 29:. 15':=e=:=.¢~s-we»‘Q:::i::@:§:Q:‘¢=Q=§. ++++++++++(JOQQQ-7cA3¢4-79-7CI~7CI~7~=\g’_=e\_':<e_<@_=e\<=\<=\+4:-F§++++<e<§ro :::z:+++++re‘§<e<e\<=\17.y =x2e-‘. 18.y 31/’—|—y=2e" —:c2e_‘. 2y=e_2‘—|— 2:2. 21.y -—6y=1+e2‘. 20.y"—-3y'+ 2y=we-‘. 22.y+y=sinx+ e“. 23.y"'-—3y”+ 3y’—-y=e‘. 24.y"+y=sin”1.Hint. sinza: =Q—-Qcos 22:. 25.y"’-y’=e2’sinz2:. 26.y(5>+2y’”—|—y’=2:2:+sin:c—|—cos1:.Hint. Solve y(5)+2y”’—|—y’= 2:2:+e“;seeExample 21.51. 27.y"+y=sin22:sin2:.Hint. sin22:sin:2:=Qcos:0:—1}cos32:. 232 H101-nan Onnnn Lmmn Drrrsnnmmu. Equxrrons Chapter 4- Foreach ofthefollowing equations, ﬁnd aparticular solution which satisﬁes thegiven initial conditions. 28-:11"—51/’—61/=83‘,1/(0)=2,1/(0) =1-29.y"__y,_2y=5sina:, y(0) =l,y’(0) =—1. 30-1/:1’—211”+y’=2e"+21.11(0)=0,y'(0)=0,y"(0)=0-31. 32. 33. 9°t~'.°‘9‘?S° 9. 10. ll. 12. 13. 14-. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24-. 25.y+91; =8cos 1:,y(1r/2) =-1,y’(1r/2) =1. 2:—3)y(0) = =3. y"——5y’+6yyn 3 @‘§@'§‘§‘§ U U U U U U U U U U U U U U U U U—:1/’+2ye’(2 , 1,y’(0)F‘,1/(0)=1,2/(0)=-—1- =c1e'2’ +age" +2. =c1e_2‘ —|—age“ +20’. =c1e'2" —|—age" —|—f;;(e“ —3ie"‘). =c1e-2’ +028-’ —|—-11-5(sin 2:-—3cos2:). =cle-2‘ +c2e"" —|—-{O-(3 sinx+cos:0). =c1e'2’ +626-” +4—|—e‘+§(sin 2:—3cos1).ANSWERS 21 =e c1cos——:z:+cgsin—a: +2: —2z.~~~<oow =me“ +cge"2‘ —-we‘—2e‘I 2:: =c1+024:3’ —6?(3sinz +cosaz). =(c1+cz:c)e' +(03+c4:c)e_' +2:—-2:1 - 3 =61+626-‘-F —-Fl RN“; _ 1 .=c1+cge ’+——-—:r—ﬁ(2sin2a:+cos2x).M cl —|— ‘n2:—:2:(a:cosa: ——sin2:). = cos:2: 62s1 =c1cos2:2:—|—C2sin2:1:~—1&6(22:cos2::—sin22:). 4 1 =c1e_' +cease" —|—%- 3—:: -z —z 2—z 33=c1e +cgze +032:e—|—W (20—$2). 2-2 -2 3 7 —z=c1e2+c2e +%——;+Z—a:e2. =c;e2‘ +age’+7}§(6:ce“ +5e“‘). 2: _, ,, 1 2 =c1+czx+¢a1 +. l_=c1cosa:+c2s1na: —-;cosx+§e 1. I —e( 3)c =01+ c2e'+ 03e- +u /f@|-ma“5»-. os2:c=c1cos:r+czsina:+—+——(5—~ -+9cos22:—7sin22:)62 520 Lesson 22B Tan Mm-non orVARIATION orPnamsrsns 233 26.y=c1+c2sina:—|—c3cosx+c4:csin:r+c5:ccosz+:c2 2 +3%(cos1:—sin1:). 27.y.= c1cosa:+c2sina:+Zsina:+%- 28.y=‘H-es’ +§-3-e_‘ —~§5e3". 29.y=Qez’+-Q-e" —-3-sinx +Qcos x. 30.y=:2:2—|—4x+4—|— ($2—-4)e‘. 31.y=cosx+ §cos3z+ sin3:2:. 32.y=e2‘+ me’. 33.6y=—10e2‘ +150‘+0". LESSON 22. Solution oftheNonhomogeneous Linear Diﬁerential Equation bytheMethod of Variation ofParameters. LESSON 22A. Introductory Remarks. Intheprevious lessons of thischapter, weshowed how tosolve thelinear differential equation (22-1) 111.1/‘"’ +an_1y"‘"" +---+my’+aoy=Q(w), <1»¢0, where: 1.The coeﬂicients areconstants. 2.Q(x) isafunction which hasaﬁnite number oflinearly independent derivatives. You may bewondering whether either orboth ofthese restrictions may beremoved. Inregard totheﬁrstrestriction there arevery fewtypes oflinear equa- tions with nonconstant coeﬂicients whose solutions canbeexpressed in terms ofelementary functions andforwhich standard methods ofobtain- ingthem, ifthey doexist, areavailable. InLesson 23,weshall describe amethod bywhich ageneral solution ofasecond order linear differential equation with nonconstant coefﬁcients canbefound provided onesolution isknown. Again therefore, theequation must beofaspecial type sothat theneeded onesolution canbediscovered. Asforthesecond restriction, itispossible tosolve (22.1) even when Q(:c) hasaninﬁnite number oflinearly independent derivatives. The method used isknown bythename of“variation ofparameters" andis discussed below. LESSON 22B. The Method ofVariation ofParameters. Forcon- venience andclarity, werestrict ourattention tothesecond order linear equation with constant coefficients, (22-2) way”+my’+any=Q(x), <12as0, 234 HIGHER Onosn LINEAR DIFFERENTIAL Equxrrons Chapter 4 where Q(x) isacontinuous function ofxonaninterval Iandis#0onI. Ifthetwo linearly independent solutions oftherelated homogeneous equation (22-21) 22?!" +1111/’ +¢1oU=0 areknown, then itispossible toﬁndaparticular solution of(22.2) bya method called variation ofparameters, even when Q(x) contains terms whose linearly independent derivatives areinﬁnite innumber. Inde- scribing thismethod, weassume therefore, thatyouwould have notrouble inﬁnding thetwo linearly independent solutions y;and ygof(22.21). With them weform theequation (22-22) y.-(w)=u1(w)y1(w) +uz(w)y2(w). where ulandU2areunknown functions of2:which aretobedetermined. Thesuccessive derivatives of(22.22) are (22-23) Up’=“1U1' +111'!/1 +u2'U2 +"21/2' =("ii/1' +"2!/2') +(u1'U1 +"2'U2), (22-24) Up”=(uiUi" +"2U2”) +(u1'U1' +142'!/2') +("1'U1 +’"2(U2)(- Substituting theabove values ofyp,y,,’,andy,,"in(22.2), weseethat y, willbea.solution of(22.2) if (22.25) a-.»(u1y1” +"22/2”) +a2(u1’y1’ +u='ya') +¢l2("1'U1 +142'?/2)’ +¢11(u11/1' +H22/2') +¢l1(u1'U1 +"2'U2) 'l'ao("1U1 +142?/2) =Q(x)- This equation canbewritten as (22-26) "1(<12U1” +¢l1U1' +(lo!/1) +u2(<l2U2” -l"21?/2' +1102/2) +a2("1'U1' -l"112'!/2') -l"¢l2(u1'U1 +‘M2’!/2)’ —|—'11("1'U1 +142'?/2) =Q(1’?)- Since ylandygareassumed tobesolutions of(22.21), thequantities in theﬁrsttwoparentheses in(22.26) equal zero. Theremaining three terms willequal Q(x) ifwechoose uland‘M2such that (22-27) 141'?/1 +122'?/2 =0, 111'!/1' -isV2’?/2' =QL) '112 The pair ofequations in(22.27) canbesolved forul’andU2’interms of theother flmctions bytheordinary algebraic methods with which you Lesson 22B Tan Msrnon orVARIATION orPnnam-rrsns 235 arefamiliar. Or,ifyouareacquainted with determinants (seeLessons 31 and63),thesolutions of(22.27) are 0U2 y1 0 ‘E 1/2’ 1/1'Q92 <22-28> "1'=""%y";-» u-'=—r.—@i§'—" U1' U2’ U1’ U2’ These equations (22.28) willalways give solutions foru1'and142’provided thedenominator determinant #0.Weshall prove inLesson 64that, if yland ygarelinearly independent solutions of(22.21), then this de- nominator isnever zero. Integration of(22.28) willenable ustodetermine ulandU2.The sub- stitution ofthese values in(22.22) willgive aparticular solution 31,,of (22.2). Comment 22.29. Since weseek aparticular solution y,_,,constants of integration may beomitted when integrating u,’andM2,. Comment 22.291. Ifthenonhomogeneous linear differential equation isoforder n>2,thenitcanbeshown that (22-3) Up="1?/1 +"2!/2 +'''-l"My-. willbeaparticular solution oftheequation, where yl,1/2,--',3/narethe nindependent solutions ofitsrelated homogeneous equation, and ul’, Mg’, ---,u,.’arethefunctions obtained bysolving simultaneously the following setofequations: (22-31) u1'U1 +u2'U2 +'''+"1/U1» =0, u1'U1' -l"112'!/2' +'''—|—un'Un' =0, V1’!/1(n_1) +142'!/2("_1) +'''+14»/Un(n_1) =% '1| In(22.31), anisthecoefficient ofy(")inthegiven differential equation. Again weremark that since weseek aparticular solution y,,,arbitrary constants may beomitted when integrating u1', 11,2’, ---,u,.’toﬁnd “ls “'2: '''1u1l- Comment 22.32. The method ofvariation ofparameters canalsobe used when Q(x) hasaﬁnite number oflinearly independent derivatives. Intheabove description ofthismethod, theonly requirement placed on Q(x) isthat itbeacontinuous function of:0.You willﬁndhowever, that 236 Hrom-an ORDER LINEAR DIFFERENTIAL Eouxrrons Chapter 4- ifQ(:c) hasaﬁnite number oflinearly independent derivatives, themethod ofundetermined coefﬁcients explained inLesson 21willusually beeasier touse. Toshow you, however, that themethod ofvariation ofparameters willalso work inthiscase, wehave included intheexamples below one which wassolved previously bythemethod ofundetermined coefficients. Comment 22.33. Theproof wegave above toarrive at(22.27) would alsohave been valid iftheconstant coefﬁcients in(22.2) and(22.21) were replaced bycontinuous functions of2:.The method ofvariation ofpa- rameters canbeused, therefore, toﬁndaparticular solution oftheequation (22-34) f2($)U” +fi(¢)y’ +fo($)U =Q(x), provided weknow two independent solutions yland yzoftherelated homogeneous equation (22-35) f2(¢)y" +f1(1)1/' +fo($)U =0- Example 22.4. Find thegeneral solution of (a) y"—3y’—|—2y=sine"’. (N0'rE. This equation cannot besolved bythemethod ofundetermined coeﬂicients explained inLesson 21.Here Q(x) =sine_‘,which hasan inﬁnite number oflinearly independent derivatives.) Solution. The roots ofthecharacteristic equation of(a)arem=1, m=2.Hence thecomplementary function of(a)is (b) yc=cle’+c2e2‘. The twolinearly independent solutions oftherelated homogeneous equa- tionof(a)aretherefore (c) yl=e’and yg=e2‘. Substituting these values andtheir derivatives in(22.27), weobtain, with a2=1,(remember a2isthecoefficient ofy") (d) u1’e‘ +u2’e2" =O, u1’e‘ +u2’(2e2‘) =sine_’. Solvin dforul’anduz’there results 1 (e) u1'=—e_" sine_‘, u2'=e'2‘ sine". Therefore (f) ul=/sin e_’(—e"‘) dx, U2=—-/e_' sine"’(-—e_”) dx. Lesson 22B Tns METHOD orVARIATION orPmxnsrnns 237 Hence (intheintegrands, letu=e_‘,du=—e"" dz) (g) ul=—cos e'"”, M2=—sin e_"+e_’cose_’. Substituting (0)and(g)in(22.22), weobtain (h) y,=——(cos e"‘)e” +(e_’cose_”—sine_’)e2" =—e2“’ sine""’. Combining (b)and(h)gives (i) y=c,e’+c2e2“ —e2‘sine"’, which isthegeneral solution of(a). Example 22.41. Find thegeneral solution of (a) y”+4y’+4y=3:ce_2". (NOTE. Wehave already solved thisexample bythemethod ofundeter- mined coefficients. SeeExample 21.4.) Solution. The complementary function of(a)is (b) 1/.=¢1e"’ +care"? Therefore thetwoindependent solutions oftherelated homogeneous equation of(a)are (<=) 1/1=F“, U2=MT”- Substituting these values andtheir derivatives in(22.27), weobtain (<1) u1’e_2" +u2'(Ie_2‘) =0, u1’(—2e_2’) +u2'(——2:ce'2" +e_2‘) =3xe_2'. Solving (d)forul’andU2’,there results (e) ul’=—-3:02, u2'=3:0. Hence, (f) ul=—x3, T42=§:c2. Substituting (c)and(f)in(22.22), wehave 2 (s) up=—w“e‘”‘ +3%066"”) =%@‘2'r“- Combining (b)and(g)weobtain forthegeneral solution of(a) (11) U=616'" +6216'“ +tfhfva, which isthesame astheonewefound previously inExample 21.4. 238 Hronsn Onnsn LINEAR DIFFERENTIAL EQUATIONS Chapter 4 Remark. Themethod ofvariation ofparameters hasoneadvantage overthemethod ofundetermined coefficients. Inthevariation ofparam- etermethod, there isnoneed toconcern oneself with thedifferent cases encountered inthemethod ofundetermined coefficients. Intheabove example, thecharacteristic equation hasarepeated root andQ(x) con- tains atenn a:e‘2", which is2:times theterm F2"ofyc. Example 22.42. Find thegeneral solution of (a) y”+y=tanx, —g<x<g- (NOTE. This equation cannot besolved bythemethod ofLesson 21. Here Q(x) =tan:1:which hasaninﬁnite number oflinearly independent derivatives.) Solution. Thecomplementary function of(a)is (b) y,=clcos2:+C2sinrt. Thetwoindependent solutions oftherelated homogeneous equation of (a)aretherefore (c) yl=cos1:, ya=sin2:. Substituting these values andtheir derivatives in(22.27), weobtain (d) u,’cosx+Mg’sin2:=0, u1’(——sin 2:)+u-2’cosx =tanx. Solving (d)foru,’andug’,there results , sinzx 1—cos2x , .(e)u;=W= —-W-= —-seca:+cosx, 1&2=smx. Hence (f) ul=—log (secx+tanx)+sinx,u2=—cos :0. Substituting (c)and(f)in(22.22), wehave (g) y,=-—cosxlog (seca: +tanx) +sinxcosx —sinxcosa: =—cosxlog (secx +tanx), —g<x< Combining (b)and(g),weobtain forthegeneral solution of(a), (h)y=c1cos:t+c2sinx ——cosa:log(secx—|—tanx), —-7% <:0< Lesson 22B Tna Mrrrnon orVARIATION orPnnmnraas 239 InFig.22.43, wehave plotted agraph, using polar coordinates, with x asthepolar angle andyastheradius vector, ofaparticular solution of (a)obtained bysetting c1=Cg=0in(h). 1| ~e-s\\v/wee Figure 22.4-3 Example 22.44. If (=1) 1/1=Ivand2/2=if‘ aretwosolutions ofthedifferential equation (b) @221"+my’—y=0. ﬁndthegeneral solution of (c) xzy” +xy’—y=ac,ac¢0. Solution. (See Comment 22.33.) Substituting thegiven twosolu- tions andtheir derivatives in(22.27), itbecomes (<1) W1’+Flu;-’ =0. u1'—-:v_2u2’ =3= Solutions of(d)foru1'andug’,byanymethod youwish tochoose, are 1(6) ’ll1'=%! ’I.t2'=—€' Hence 1 2 (f) u1=§log:t, ug=—%- Substituting (a)and(f)in(22.22) gives (2) y.-=-;l0sw—§- 240 Hroasn Onm-:1: LINEAR DIFFERENTIAL EQUATIONS Chapter 4 Combining (g)with (a),weobtain forthegeneral solution of(c), = -1 ‘E _‘E(11) 1/c1’w+c2w +2l0s=¢ 4- which simpliﬁes to (i) y=c11+c21_1 +:5log1. EXERCISE 22 Usethemethod ofvariation ofparameters toﬁndthegeneral solution ofeachofthefollowing equations. 9°Z*‘.°‘9‘?§'°!°l"'@=eQ==~:===e=e::zz:z::+++++++<=éo&a=:@=e@=:_\I-ll-ll-|l-||-|?‘9‘!°l".°!°Qve~==:<=<ez::==:+++++Mv:éore==Qr§\~+ =sec1. 1/=e“log1. =cot1. =csc1. =sec31. =tan”1. — =sin21. +1/=e"‘/1. =sinz1. =sec1csc1. —|—2y=12e’. I—y'—|- y=e‘log1. —|—y=x2e-1. 15.y'—3y’+2y=cose". =41sin1. Usethemethod ofvariation ofparameters toﬁndthegeneral solution ofeach ofthefollowing equations. Solutions fortherelated homogeneous equation areshown alongside eachequation. 16.12y" —1y'—|— y=1;y1= 1,yg=1log1. 2 2 17-:u"—;y'+;1/=1=l0s1; z/1=m/z==v2-1 18-12:1/”+w1/'—4:1/=13; an=12.1/z=;- 19.12y"+ 1y’—-y=12c"; y1=1,yg=1‘1. 20.212y"+ 31y’ —-y=1'1; yl=11/2, yg=1“1. ANSWERS 22 =c1cos1+ c2sin1+ 1sin1+ cosztlog c0s1. =c1cos1+62sin1+sin1log(csc1—cot1). =c1cos1+ c2sin1+ §tan1sin1. =cle‘+age“ -4}sinz1- =c1cos1—|— c2sin1—|— %cos21+ 1}. =c1e"2’ —|—626-‘ —|—2e‘. 4-1 7.y=c1e"' +c21e_' +£1%— -?‘E"'!‘§°!°L"=e<==:=:<e== =c1cos1+ czsina: —-12cos1-l— 1sin1. ={12e"‘(2 log1 —-3)+(01—|—c21)e"‘. =c1cos1+ czsin1+ sinztlogsina: —-1cos1. =c1cos1—|— c2sin1+ sin1log(sec:t+ tan1) —-2. n-In-IF9299‘QQ°§@ Lesson 23 LINEAR Eqrwrrou wrrn NONCONSTANT COEFFICIENTS 24-1 12.y=e_'(c1 +cg1+1log1). 13.y=c1cos1+ cgsinx+ sin1log(csc1 -cot1) — cos1log(sec1+tan1). 14.y=(c1+cg1)e" +12e"(§ log1—i). 15.y=c1e'+6262’ —e2‘cose“. 16.y=c11—|—cg1log1+2(log1)2. 17.y=c11—|—cg12+<}13log1 —$13. 3 18.y=c112—|—cg1'2—|— 19.y=011+cg1'1 +e"’(1 +1'1). 20.y=0111/2 +cg1‘1 -§1-1 log1. LESSON 23. Solution oftheLinear Differential Equation with Nonconstant Coefficients. Reduction of Order Method. LESSON 23A. Introductory Remarks. Weareatlastready toex- amine thegeneral linear differential equation (23-1) f..(w)2/‘"’ +f.._1(w)y"“" +---+f1(w)y’ +fo(r)y =Q(x), anditsrelated homogeneous equation (23-11) f»(w)!/”) +f.._1(¢¢)y‘”'”'+ ---+f1(fv)y’ +fo(2>)U =0. where fo(1), f1(1), ---,f,.(1), Q(x) areeach continuous functions of1on acommon interval Iandf,,(:c) 9'5Owhen 1isinI. Wehave already remarked that inmost cases thesolutions of(23.1) willnotbeexpressible interms ofelementary functions. However, even when they are, nostandard method isknown ofﬁnding them, asisthe case when thecoefficients in(23.1) or(23.11) areconstants, unless the coefficient functions f,,(:t) areofavery special type; see,forexample, Exercise 23,18. Foranunrestricted nthorder equation (23.11) that hasa solution expressible interms ofelementary functions, thebest you can hope forbytheuseofastandard method istoﬁndoneindependent solu- tion, iftheother n—1independent solutions areknown. Andfor(23.1), thebestyoucanobtain from astandard method istoﬁndoneindependent solution of(23.11) andaparticular solution of(23.1), again provided the other n-—1independent solutions of(23.11) areknown. You cansee,therefore, that even inshowing you astandard method forﬁnding ageneral solution ofonly thesecond order equation (23-12) f2(w)y” +f1(w)y’ +fo(rv)y =Q(x), itisessential that thefunctions f0(1), f1(1), f2(:c) beofsuch acharacter that theneeded ﬁrst solution ofitsrelated homogeneous equation canbe discovered. 242Hiessn Onnrm Lmma Drrranarrrmt EQUATIONS Chapter 4- Comment 23.13. Thefunction yE0always satisﬁes (23.11). Since thissolution isofnovalue, ithasbeen appropriately called thetrivial solution. LESSON 23B. Solution ofthe Linear Differential Equation with Nonconstant Coefficients bythe Reduction ofOrder Method. As remarked inLesson 23A, weassume thatwehave been abletoﬁndanon- trivial solution ylofthehomogeneous equation (23-14) fz(=v)1/" +f1(1=)y' +fo(1=)y =0- Themethod bywhich weshall obtain asecond independent solution of (23.14), aswellasaparticular solution oftherelated nonhomogeneous equation (23-15) f2(1=)1/" +f1(w)?/’ +fo(23)U =Q(x). iscalled thereduction oforder method. Let1/2(1) beasecond solution of(23.14) andassume thatitwillhave theform (23-2) 1/2(2) =1/1(2)I14(1)dw- where u(1)isanunknown function of1which istobedetermined. The derivatives of(23.2) are (23-21) 2/g’(w) =2/11¢+2/1’I11(1)<11, (23-22) U2”($) =U11-1' +U1’?-1 +I/1'" +U1"/2(2) 11$- Substituting theabove values foryz,yg’,andyg”in(23.14), wesee,by Deﬁnition 3.4,that ygwillbeasolution of(23.14) if (23-23) f2($) [1/111' +2?/1'" +U1”/11(2) div] -1-f1(1) [U111 +I1/1'/14(2) div]+fo(1=) [U1(2>)_/14(3) div]=0- Wecanrewrite thisexpression as (23-24) (fz(w)1/1” +f1($)y1’ +fo(rv)2/1) I11(1)11¢ -1-f2($)U11-1' +l2f2(23)U1' +f1($)U1l"- =0- Since wehave assumed that ylisasolution of(23.14), thequantity in theﬁrstparenthesis of(23.24) iszero. Hence (23.24) reduces to (23-25) f2($)U1"' +l2f2(21)U1' -l-f1($)U1lu =0- Lesson 23B REDUCTION orORDER Mmnon 243 Multiplying (23.25) bydx/[uf2(x)y1], weobtain du 2d?/1 —f1($)2.26 — i =—-———— d.(3) “+1/1 rm’” Integration of(23.26) gives (23.27) logu+2logyl=—-I/l%§% dx, 1<>@<uy.=> =-/}"§—"§ dz,22! _1&4, u?/12 =gI/2(1) 7 _J‘f1($)dz u= e fa(1) /y12_ Substituting thisvalue ofuin(23.2), weobtain forthesecond solution of (23.12) _J'(l4, 6 12(3) (23-23) 1/2=y1[T da: Weshall prove inLesson 64,seeExample 64.22, thatthissecond solution ygislinearly independent ofyl. Comment 23.29. Wedonotwish toimply that theintegral in(23.28) willyield anelementary function. Inmost cases itwillnot, since inmost cases, werepeat, thegiven differential equation does nothave a.solution which canbeexpressed interms ofelementary functions. Further the solution ylneed notitself beanelementary function, although inthe examples below andintheexercises, wehave carefully chosen diﬁerential equations which have atleast onesolution expressible interms ofthe elementary functions. Example 23.291. Find thegeneral solution of (a) wzy”+11/’—11=0,w#0. given that y=1:isasolution of(a). Solution. Comparing (a)with(23.14), weseethatfg(x)=:0’, _f1(a:) =x.Hence by(23.28) with yl=2:,weobtain _-lid: —log: yz=x’/5—x.‘%dx=x gaf"d$-“=17 £12-adi; _Z:z:_2___1:“! -2 2 24-4 Hromsn Onman LINEAR DIFFERENTIAL EQUATIONS Chapter 4- Hence thegeneral solution of(a)is (c) y=clx+c2:c'1. Comment 23.292. Although formulas areuseful, itisnotnecessary tomemorize (23.28). Weshall now solve thesame Example 23.29 by using thesubstitution (23.2). With yl==ac,(23.2) becomes (d) 3/2=ac/u(a:) dx. Differentiating (d)twice, weobtain (e) I/2’=mu+/u(:z:) dx, 1/2”=xu’—|—2u. Substituting (d)and(e)in(a),wehave (f) :c2(:z:u' +2u)—|—xixu —|—/u(z) dz]—-:1:/u dz=0, which simpliﬁes totheseparable equation (g) xu’+3u=0, :v¢O. Itssolution is (h) u=x‘3. Substituting (h)in(d),there results 2 1 (i) y2=x/x_3dz=xi—2=?_—2- Hence thegeneral solution of(a)is,asfound previously, (i) y=61¢+62¢"- The same substitution (23.2), namely (23-3) y(z)=y1(w)Iu(rv) dw. inthenonhomogeneous equation (23.15) will yield notonly asecond independent solution of(23.14) butalsoaparticular solution of(23.15). As before, wedifferentiate (23.3) twice andsubstitute thevalues ofy,y’,and y”in(23.15). Theleftsideoftheresulting equation willbeexactly the same as(23.25) found previously. Itsright side, however, willhave Q(x) initinstead ofzero. Hence inplace of(23.25), wewould obtain (2331) f2($)?/1"’ +l2f2($)1/1' —|—f1($)!/1]" =QC”), Lesson 23B Raoucrxou orOnnsn Mmnon 24-5 anequation which isnow linear inuandtherefore solvable bythemethod ofLesson 11B. The substitution ofthisvalue ofuin(23.3), willmake y(z) asolution of(23.15). Comment 23.311. Two integrations will beinvolved inthis pro- cedure; onewhen solving foruin(23.31), theother when integrating uin (23.3). There will, therefore, betwoconstants ofintegration. Ifwein- clude these twoconstants, weshall obtain notonly asecond independent solution y2(:v) of(23.14) andaparticular solution y,,(:c) of(23.15), but alsotheﬁrstsolution y1(x) with which westarted. Hence bythismethod thesubstitution (23.3) willgive thegeneral solution of(23.15). Example 23.32. Given that y=:1:isasolution of a:2y"+xy’—y= O,:c#50, ﬁndthegeneral solution of (a) xzy” +xy’—y=ac,x¢0. Solution. By(23.3), with onesolution y1(:c) =2:,wehave (b) y(z)=a:/u(:c) dx. Itsderivatives are (0) y'(=v)=rm+4/iudw. y"(¢) =W’+u+u. Substituting (b)and(c)in(a)gives (d) a:2(xu’+2u)+x(:vu+/udx)—:c/ud:c=x, which simpliﬁes to (e) mu’+3u=:17‘, u’—|—3u=172. This equation islinear inu.ByLesson 11B, itsintegrating factor is eI3“’_“"‘ =e1°“a =x3.Hence, by(11.19), (f) uz3=/l:z:d:v=§+c1', x_l I-3u=7-+c1x - 246 HIGHER ORDER LINEAR DIFFERENTIAL Eouurons Chapter 4 Substituting thisvalue ofuin(b),weobtain x-1 _3 Z-2 (g)1/(w)=w‘[(—§—+¢1’w )dw=rv[%1<>gw-v1'—§-+02] -1 =glogx —-c1’1;?+C227, which isequivalent to (h) y(z) =c1:::_1 +C313+3log:0. Comment 23.33. We could, ifwehad wished, have obtained (e) directly from (23.31). Comparing (a)with (23.15), weseethat _f2(:z:) =1:2, f1(:c) =atandQ(x) =2:.They,inthisformula isthegiven solution as. Substituting these values in(23.31) willyield (e).Verify it. EXERCISE 23 Usethereduction oforder method toﬁndthegeneral solution ofeach ofthefollowing equations. One solution ofthehomogeneous equation isshown alongside each equation. 1.12y"-a:y'+y =0,y1= x. 2 2 2-1/"—;z/+51/=0. 1/1=w. 3.(2:c2—|— 1)y" —42:1/+ 4y=0,yl=2:. 4-u”+($2—r);1/’-—(w——1):u=0.1/1=1- 1 5-1/”+(;—-;)y’ -11=0,1/1= 1:2. 6-W1/"+ 3w’—y=0.211=1””- 7-1/"+[f(w)—111/’—f(I)u =0.U1=e‘- 8-1/"+rf(1)z/' —f(r)y =0.u1=$- 9-12y"—11/+11 =11,2/1=1- 2 210.y"—-£1/'+;y =xlogz, y1= :0. ll.2:21;"-|— a:y'—41/=2:3, g1=2:2. 12.:c2y”—|— 2:y’—y=1:20", y1=2:. 13.2221/" +31y’—-y=1.yl=11/2.x 14-.x2y” —-2y=22:, y1=2:2. 15-u"+($2—1)!/’-—121/=0.an=e‘- l6.xzy” -—2y=22:2, y1=2:2. 17.a:2y"+ my’—y=1,yl=2:. 18.Thedifferential equation (23-4) aa(I—Io)3y”'+ l12(f¢-9>o)2y"+ <l1(=v-110)!/'+ 1101/=Q(x). I#Io- Lesson 23—Exe1-cise 24-7 where ao,a1,(lg,a3,xoareconstants, andQ(:c) isacontinuous function ofx, isknown asanEuler equation. Prove thatthesubstitution 2:-—1:0=e",u=log(:c —xo), willtransform (23.4) intoalinear equation with constant coefficients. Hint. dy_dydz_ dy du_dxdu _(Z Z0)dz, on-,._,,,-a,du2 du 0dz? <131/ @1221 du aday m“’*m+”a =<“"“°) as Usethesubstitution given inproblem 18above tosolve equations 19-21. 19.(:0:—3)’;/’+ (x—-3);/'+ y=x,2:943. 20.:v"’y"+ my’=0,2:as0. 21.$31/"' +2:z:2y” —2:y’—|—y=2:,:1:950. 22.Prove thatthesubstitutions y_eI|!1(:t)dz’ yl=yleluldz, and y" =y12efy1dz+ y1leIy1d:: willtransform thelinear equation a2(1)1/” +01(1):/’ +ao(-v)y =0 intotheRiccati equation (seealsoExercise 11,25) y1’=—%$—%y1—y12- Hence ify;isasolution oftheRiccati equation, y=ef"141isasolution of thelinear equation. 23.Usethesubstitutions given in22totransform thelinear equation 1:11"—- y'-—2:311=0intoaRiccati equation. Find, bytrial, asolution ofthe Riccati equation; then ﬁndasolution ofthelinear equation. With one solution ofthelinear equation known, ﬁnditsgeneral solution. 24.Prove thatthesubstitution y=—]ﬁ%» f2(1>)¢0, ,,-__;,--+,-(i),— fzu 1'22/“'2 willtransform theRiccati equation, 11'=fo(==) +f1(w)y +f2(1)1/2, f2(r) 9"0. intothesecond order linear equation f2(=v)u" -lf2'(=v) +f1(w)f2(1)lu' +fo($)lf2($)l2u =0- 248 HIGHER ORDER LINEAR DIFFERENTIAL Equu-roNs Chapter 4 25.Asecond order linear diﬁerential equation (23-5) f2(1)1/" +f1(1)1/'+fo(¢)y =Q(1). issaidtobeexact ifitcanbewritten as <23-51) %[M<1> %+Non]=co). i.e.,ifitsleftsidecanbewritten asthederivative ofaﬁrstorder linear expression. Anecessary andsufficient condition thattheequation beexact is that 2 (23.52) ‘175;_gs-1+yoE0. Iftheequation isexact, then M(x)andN(1)of(23.51) aregiven by (23-53) M(1) =f2(1) and N(1)=f1(I) ""f2'(I)- Show thatthefollowing equation isexact andsolve. ($2-—21):/’+4(w-1):/'+21/=62‘- 2: Ans. (212—2:c)y =8?+01:2:—|—C2. 26.Afunction h(:c)iscalled anintegrating factor ofaninexact differential equation, ifafter multiplication oftheequation byit,theresulting equation isexact. Anecessary andsuﬁicient condition that h(:c)beanintegrating factor ofthelinear differential equation (23-6) fz(1):u” +fi(r)1/"’fo(w)i/ =Q(1) isthatitbeasolution ofthedifferential equation , .1’ a(23-61) 3-;(fzh) -3;(f1h)+foh =0- Find anintegrating factor ofthefollowing equation, then multiply the equation bytheintegrating factor anduse(23.52) toprove thattheresulting equation isexact. 1:31;" —(2a:3 —62:2);/' —3(a:3 +212:2—2:c)y =0. Ans. h(2:) =e‘;alsoh(:c) =F3‘. ANSWERS 23 FF!"~==:=e=c1:c+022:log2:. =c1:2:+022:2. =c1:c—|—62(2$2 ——1). 0:3“+noH“ 4-y=C1Z+C2Z/>eTd$. —zzI4 5.y=01x2+022:2/S aria dz. Lesson 23-Answers 6.y=012:1/2+ 6gZ_1. 7.y=c1e’+cze'-/e_’_'m”)“ dz. 6-fz/(=>a= B.y=c1:2:—i—c2n/-7--- 9.y=c1:2:+62$logs: +3(log2:)2. 10.y=012:+62222+1}x3log2:—112:3. ll.y=012:2+cz:t_2 + 12.y=012:+cg:-1 —|—e"“’(1 +2'1). 13.y=c121/2+ 022:-1 —-§a:'1 log:0. 14-.y=c1x2+ cga:-1 -—1:. 15.y=cle’—|—czez/‘e_”_(”3'3) dz. 16.y=012:2—|—c2a:'1 +§:c2log2:. 17.y=cm:—cg:c‘1 ~—1. dzy ..19- y—-6+3, x 3 y=c1cos[log(2:-3)]+czsinllog (x—3)]-|- 5+5- d2y20. 8;; =0, y=c1loga:-1-02. d3 .1’ d .. _21.???’-55%-—%+y=e, y=c1:c+cg:clogx+c3a: 1+—Z(loga: 23.Riccati equation: yl’=22+ Z11—1/12;ll1=1,y=etuz. . 7/2 -:3/2 General solution: y=c1e' —|—C26 .) Chapter 5 Operators andLaplace Transforms Inthenext andsucceeding lessons weshall prove certain theorems by means ofmathematical induction. Wedigress momentarily, therefore, to explain themeaning ofa“proof bymathematical induction.” Suppose wewish toprove astatement about thepositive integers as, forexample, that thesum oftheﬁrst noddintegers isn2,i.e.,suppose wewish toprove (a) 1+3+5+7+~--+(2n——1)=n2. Bydirect substitution wecanverify that (a)istrue when n=1,n=2, n=3,forthen (a)reduces totherespective identities 1=12,1+3=22, 1—|—3—}—5=32.Since itwould beimpossible tocontinue inthismanner tocheck theaccuracy of(a)forevery n(there arejusttoomany ofthem), wereason asfollows. Ifbyassuming (a)isvalid when n=k,where Itis aninteger, wecanthen prove that (a)isvalid when n=k—|—1,itwill follow that (a)willbetrue forevery n.Forthen thevalidity of(a)when n=3will insure itsvalidity when n=4.The validity of(a)when n=4willinsure itsvalidity when n=5,etc., adinﬁnitum. And since wehave already shown bydirect substitution that (a)isvalid when n=3,itfollows that(a)isvalid when n=4,etc.,adinﬁnitum. Weshall now usethisreasoning toprove that (a)istrue forevery n. Wehave already shown that (a)istruewhen n=1,n=2,andn=3. Wenow assume that (a)istrue when n=k.Hence weassume that (b) 1+3+5+7+~--—}—(2lc—1)=k2 isatrue equality. Letusaddtoboth sides of(b)thenext oddnumber in theseries, namely (2k—1+2)=2k+1.Equation (b)then becomes (c)1+3+5+7+---+(2k—1)+(2k+1) =k2+2k+1=(k+1)2. Since wehave assumed (b)istrue, itfollows that (c)istrue. But ifin 250 Lesson 24-A DEFINrrIoN 0FANOPERATOR. LINEAR PROPERTY 251 (a),weletn=k+1,weobtain (<1) 1+3+5+7+---+(2k+l)=(k+1)? which agrees with (c). Therefore, if(b)istrue soalso is(d). Hence we have shown that if(a)istrue when n=k,itistrue when n=k—|—1. But (a)istruewhen n=3.Itistherefore true when n=4.Since (a)istruewhen n=4,itistherefore truewhen n=5,etc., adinﬁnitum. Aproof which uses theabove method ofreasoning iscalled aproof by induction. Comment 24.1. Every proof byinduction must consist ofthese two parts. First you must show that theassertion istrue when n=1(or forwhatever other letter appears intheformula). Second youmust show that iftheassertion istrue when n=k,where kisaninteger, itwill then betrue when n=lc+1. LESSON 24-. Differential and Polynomial Operators. LESSON 24A. Deﬁnition ofan Operator. Linear Property of Polynomial Operators. Anoperator isamathematical device which converts onefunction into another. Forexample, theoperation ofdiffer- entiation isanoperator since itconverts adifferentiable function f(z)into anew function f’(x). The operation ofintegrating I;f(t)dtisalso an operator. Itconverts anintegrable function f(t)into anewfunction F(a:). Because thederivative atonetime wasknown asadifferential coeﬂi- cient, theletter D,which wenow introduce todenote theoperation of differentiation, iscalled adifferential operator. Hence, ifyisannth order differentiable function, then (24-11) D°y=y. Dy=y’. D211=1/”.---,D"y=y‘"’- (Other letters may alsobeused inplace ofy.)Forexample, ify(z) =$3, then D°y =2:3, Dy=dy/dx =3x2, D2y =day/dxz =6x, Day = d3y/da:3 =6,D‘y =d4y/dz‘ =0;ifr(0) =sin0+02,then D°r= sin0—|—02,Dr=dr/d0 =cos0+20;D2r =d2r/d02 =—sin 0+2;if :r(t)=t2,then D°:z: =£2,Dz=dx/dt =2t,D22: ==dzx/dt2 =2,D311:=0. Byforming alinear combination ofdifferential operators oforders 0to n,weobtain theexpression (24.12) P(D) =a0+a,D +a2D2 +---+a,,D", an¢0, where a0,a1,---,anareconstants. Because oftheresemblance ofP(D) toapolynomial, weshall refer to itasapolynomial operator oforder n.Itsmeaning isgiven inthe following deﬁnition. 252 Ornmvrons ANDLumen TRANSFORMS Chapter 5 Deﬁnition 24.13. LetP(D) bethepolynomial operator (24.12) of order nandletybeannthorder differentiable function. Then wedeﬁne P(D)y tomean (24.14) P(D)y =(a,,D" '+---+a1D —|—ao)y =a,,D”y +---+a1Dy +aoy. By(24.11), wecanalsowrite (24.14) as (24.15) P(D)y =a,,y‘">+ +a1y’+ aoy,a,,¢0. Hence, by(24.15), wecanwrite thelinear equation with constant coefﬁ- cients, (24.16) a,,y(") +a,,_1y‘"_” —|—---—|—aly’ +aoy=Q(x), an960, as (24-17) P(D)y =Q(w), where P(D) isthepolynomial operator (24.12). Theorem 24.2. IfP(D) isthepolynomial operator (24.12) andyl,yz aretwonthorder diﬁerentiable fmwtions, then (24-21) P(D)(¢1y1 +622/2) =¢iP(D)y1 +¢zP(D)1/2, where clandC2areconstants. Proof. Intheleftsideof(24.21) replace P(D) byitsvalue asgiven in(24.12). There results (a) (<1»D" +an_1D"_‘ +---+‘MD+<lo)(¢1y1 +@2112)- ByDeﬁnition 24.13, wecanwrite (a)as (b)<1nD"(¢1!/1 +021/2) +<l»—1D"_1(¢1Z/1 +C21/2) +''' +¢l1D(¢1111 —|—621/2) +¢lo(¢1?/1 +621/2)» By(24.11) andtheproperty ofderivatives, (0) D"(¢1y1 +cm)=¢1y""+¢2y""=611)";/1 +¢2D'°y2, foreach It=0,1,2,---,n.Hence (b)becomes (<1)¢1»(¢1D"y1 +6217"?/2) +¢ln_1(¢1D"_1!/1 +¢2D"_11/2) +'''+¢l1(¢1D2/1 +62D?/2) +fl0(¢1y1 +62112) =¢1(a,.D" +a1._1D"" +---+111D+(lo)?/1 +c2(anD" +an_1D"" +---+111D+a<>)y2 =61P(D)?/1 +62P(D)?/2» Lesson 24-A DEFINITION orANOPERATOR. LINEAR PROPERTY 253 which isthesame astheright sideof(24.21). Wehave thus shown that theleftsideof(24.21) isequal toitsright side. Byrepeated application of(24.21), itcanbeproved that (24-22) P(D)(611/1 +621/2 +'''—|—6"!/1») =¢1P(D)@/1 +c2P(D)1/2 +---+c»P(D)z/». where cl,C2,...,c,,areconstants, andeach ofyl,yz,---,y,.isannth order diﬁerentiable function. Example 24.221. UseDeﬁnition 24.13 toevaluate (a) (D2 -—3D+5)(2a:3 +en+sinas). Solution. Here P(D) =D2—3D+5andthefunction yof(24.14) is2x3+e2’+sin2:.Therefore, by(24.14) and(24.11), (b)(D2-31)+5)(2:v3 +e2’+sinx) =D2(2:v3 +e2’-1-sin:0)—-3D(2a:3 +e2‘—|—sinz) —|—5(2:c3 —|—ea‘+sin2:) =12:0+4e2"-—sinas—18:02 —6e2‘—3cosa: +102:3 +502’ +5sinas =102:3 -—18:02 +12:0+3e2” -1-4sin:1:—3cosx. Example 24.222. Use(24.22) toevaluate (a) (D2 —3D+5)(2a:3 +e2’—|—sinze). NOTE. This example isthesame asExample 24.221 above. Solution. By(24.22), with P(D) =D2—3D+5, (b) (D2 —3D+5)(2x3 +e2’—|—sinac) =(D2-3D+s)(2@*) +(D2-31>+s)e’= +(D2 -—3D+ 5)sina: =2(6:e-92:2+52:3)+<46“-st“+56") +(——sin:c —3c0s:c +5sinx) =102:3 ——18x2 +12:2:—|—3e2‘ —|—4sin:1:—3cos2:, thesame result obtained previously. Deﬁnition 24.23. Anoperator which has theproperty (24.21) is called alinear operator. Hence thepolynomial operator (24.12) is linear. 254 Ornmvroas ANDLAPLACE TRANSFORMS Chapter 5 Comment 24.24. With theaidofthelinear property ofthepoly- nomial operator P(D) of(24.12) wecaneasily prove thefollowing two assertions, proved previously inTheorem 19.3. 1.Ifyl,g2,---,y,,arensolutions ofthehomogeneous linear equation P(D)y =0,theny,=c1y1+cg;/2+---+c,,y,.isalsoasolution. 2.Ify,isasolution ofthehomogeneous linear equation P(D)y =0,and y,isaparticular solution ofthenonhomogeneous equation P(D)y = Q(x), then y=y,+y,isasolution ofP(D)y =Q(x). Proof of1.Since y1,yg,---,y,,areeach solutions ofP(D)y =0,we have (3') =or ='0:'''1 = Hence also (b) c1P(D)y1 =0, c2P(D)y2 =0,---,c,,P(D)y,, =0, where cl,Cg,---,c,,areconstants. Adding allequations in(b)andmaking useof(24.22), weobtain (C) P(D)(¢1y1 +em+---+6.2/1.)=0, which implies that y.=clyl+Cg’!/2+---+c,.y,, isasolution of P(D)y =0. Proof of2.Byhypothesis yeisasolution ofP(D)y =0,andy,isa solution ofP(D)y =Q(x). Therefore (<1) P(D)!/¢ =0and P(D)!/p =Q(w)- Adding thetwoequations in(d)andmaking useof(24.21), weobtain (6) P(D)(y¢ +yp)=Q(x), which implies that y=ye+y,isasolution ofP(D)y =Q(x). Comment 24.25. Principle ofSuperposition. Inplace ofthe linear differential equation (a) P(D)?/=Q1+Q2+"'+Qn, where P(D) isapolynomial operator (24.12), letuswrite thenequations (b) P(D)y =Q1, P(D)y =Q2,-'-,P(D)y =Q»- Letylp,y2,,,---,gm,berespective particular solutions ofthenequations of(b). Therefore (C) P(D)y1p =Q1. P(D)!/21» =Q2,---,P(D)y»p =Q..- Lesson 24B ALGEBRAIC Paorsnrms orPOLYNOMIAL OPERATORS 255 Adding alltheequations in(c)andmaking useof(24.22), there results P(D)(1/19+?/2p+"'+l/n9)=Q1+Q2+"'+Qna which implies that (e) I/P=l/19+?/2p+"'+ynp isasolution of(a). Wehave thusshown thataparticular solution y,of(a)canbeobtained bysumming theparticular solutions 1|/1,,312,, ---,gm,ofthenequations of(b). Theprinciple used inthismethod ofobtaining aparticular solu- tion of(a)isknown astheprinciple ofsuperposition. Example 24.26. Usetheprinciple ofsuperposition toﬁndaparticular solution oftheequation (=1) (D2+1)y=iv’+we”+3- Solution. ByComment 24.25, asolution y,of(a)isthesumofthe particular solutions ofeach ofthefollowing equations. (b)(D2+1)z/=w’, (D2+1)y=we”, (D2+1)y=3- Particular solutions ofeach ofthese equations arerespectively—write (D2—|—1)yasy”+yandusethemethod ofLesson 21- (9) 1'/11>=$2"‘2» 3/21»=‘H2921 "‘ﬁzz)» l/39=3' Hence aparticular solution of(a)is (<1)y.=w’—2+see“—re”)+3=x”+1+toe"—an- LESSON 24-B. Algebraic Properties ofPolynomial Operators. Whenever P1(D) andP2(D) appear inthislesson, weassume that (24-3) P1(D) =IMD” +an-1D”_1 +'''+01D+(lo, =bmDm +bm-ID"-1 +'''+b1D +b0; aretwopolynomial operators oforders nandmrespectively, ngm. Whenever yappears weassume itisannthorder differentiable function, deﬁned onaninterval I. Deﬁnition 24.31. Thesumoftwopolynomial operators P1(D) and P2(D) isdeﬁned bytherelation (24-32) (P1+P2)?! =Ply-l"P21!- 256 Oransrons ANDLumen Tnmsroans Chapter 5 Theorem 24.33. LetP1(D) andP2(D) betwopolynomial operators of theforms (24.3). Then P1andP2canbeadded justasiftheywere ordinary polynomials, i.e.,wecanaddthecoeﬁicients oflikeorders ofD. Proof. By(24.32), (24.3) andtherules ofdifferentiation, (9.) (P1+P2)?! =(anD" —|—'-'—|—<12D2 -l"111D +ac)?! +(bmDm -if'''+b2D2 +b1D -|'be)?! =(anD"-l" "'+bmDm+"'+a2D2+b2D2 —|—a1D +b1D +110+be)?! =[¢lnD"+"'+bmD'"+'---l'(¢l2+b2)D2 +((11+b1)D +((10—|—bo)l1/- Example 24.331. If (a)P1(D) =31)“+D2+31>-1,P2(D) =51)’-7D+3, ﬁndP1(D) +P2(D)- Solution. ByTheorem 24.33, (b) P1(D) +P2(D) =3D“ +6D2 —4D+2. Asanexercise, prove that thefollowing identities follow from Deﬁnition 24.31. (24-34) P1(D) +P2(D) =P2(D) -l-P1(D) (commutative lawofaddition). P1(D) +[P2(D) +Pa(D)l =[P1(D) +P2(D)] +Pa(D) =P1(D) +P2(D) +Pa(D) (associative lawofaddition). Deﬁnition 24.35. The product ofafunction h(:z:) byapolynomial operator P(D) isdeﬁned bytherelation (24-36) lh(w)P(D)]1/ =h(w)[P(D)y]- Example 24.37. Evaluate (a) [2¢2(3D’ +1)]e=‘=. Solution. Comparing (a)with (24.36), weseethat h(:c) =22:2, P(D) =3D2 +1,y(z) =ea’.Therefore, by(24.36), (24.14) and(24.11), (b)[2a:2(3D2 +l)]e3"=2:z:2[(3D2 +1)e3‘] =2:z:2[3(9e3“) +e3‘]=2:z:2(28e3‘) =56:z:2e3". Lesson 24B ALGEBRAIC Pnornrrrnas orPotvnomuu. Ornmxrons 257 Example 24.38. Evaluate (a) ox“+2>[(1>’+so+2)+<21)’—1)]e“- Solution. ByTheorem 24.33 and(24.36), wecanwrite (a)as (b) (3103+2)l(3D2 +3D+1)¢2’]- Carrying outtheindicated differentiations in(b),weobtain (c) (3:03 +2)(12 +6+1)e2’ =l9e2‘(3a:3 +2). Comment 24.381. By(24.32), wemay alsowrite (24-39) h(@=)[P1(D) +P2(D)l?/ =h(w)lP1(D)y +Pz(D)y] =h(w)[P1(D)yl +h(w)lP2(D)y]- Example 24.4. Evaluate (a)ofExample 24.38, byuseof(24.39). Solution. (a)(3:03+2)[(D=' +31>+2)+(202-1)]e2" =(3a:3+2)[(D2 +3D+2)e2’]+(3:c3+2)[(2D2 -1)e’=] =(313+2)(-1+6+2)e2’+(3:03+2)(s-1).?” =19e2‘(3:c3 +2). Deﬁnition 24.41. The product oftwo polynomial operators P1(D) andP2(D) is’deﬁned bytherelation (24-42) [P1(D)P2(D)ly =P1(D)lP2(D)!/l- Example 24.43. Evaluate (a) [(302-5D+3)(D2 -2)](x3+2x). Solution. By(24.42), (24.14) and(24.11), (b)(3D’—5D+3)[(D’ —2)(w°+2%)] =(3D2 -5D+3)(6a: -—2:03—42:) =3(-12$) —5(6-611:2-4)+18¢—613-12¢ =-615’+30¢’-30¢-10. Prove asexercises that thefollowing identity follows from Deﬁnition 24.41, (24-44) P1(D)lP2(D)Pa(D)l =lP1(D)P2(D)lPa(D) =P1(D)P2(D)Ps(D) (associative lawofmultiplication), 258 Ormwrons ANDLumen Tmnsronms Chapter 5 andthefollowing identity from Deﬁnitions 24.31 and24.41, (24-45) P1(D)lP2(D) +Ps(D)] =P1(D)P2(D) +P1(D)Pa(D) (distributive lawofmultiplication). Theorem 24.46. If (24.47) P(D) =a,,D" —|—a,,_1D"'1 +----1-a1D +ao, an¢0, where ao,a1,---,anareconstants, then (24-43) P(D) =a..(D —"1)(D —T2)---(D—1'»), where r1,r2,---,r,,aretherealorimaginary roots ofthecharacteristic equa- tion (20.14) ofP(D)y =0,i.e.,apolynomial operator with constant c0e_ﬂ’i- cients canbefactored justasifitwere anordinary polynomial. Proof. Weshall prove thetheorem only forn=2,i.e., weshall prove (9-) D2*‘(T1+1‘2)D +T172 =(D'"T1)(D "'T2)- Intheequalities which follow, wehave indicated thereason after each step. Besuretorefer tothese numbers. Westart with theright sideof (a)andshow that ityields theleftside. (b)[(D—r1)(D -—r2)]y (D"-"1)l(D —T2)?/l (D"-T1)(D!/ *T21/) (D—*"1)Dy *‘(D—1‘1)("2Z/) Dzy—nDy—r2Dy+T1T29 D2?! -“(T1—|—?‘2)Dy —|—1'11'21/ lD2_(T1-l"T2)D -l"T172]?!By (24.42), (24.14), (24.21), (24.14), Theorem 24.33, (24.14). Corollary 24.481. IfP(D) isthepolynomial operator (24.47), then (24.482) P(D) =P1(D)P2(D). where P1(D) andP2(D) maybecomposite factors ofP(D), i.e.,P1andP2 may beproducts offactors of(24.48). Theproof follows from Theorem 24.46. Theorem 24.49. Thecommutative lawofmultiplication isvalid for polynomial operators, i.e., (24-5) (D-T1)(D —1'2)=(D—r2)(D —1'1)- Lesson 24B ALGEBRAIC Paornarrns orPonvnomnu. Oranxrons 259 Proof. Intheproof ofTheorem 24.46, interchange thesubscripts 1 and2ofr1andr-2.Since theﬁnalformula ontheright of(b)intheproof willremain thesame, theequality (24.5) follows. Example 24.51. Evaluate (D2—2D—3)(sin :1:+2:2). Solution. Method 1.Byapplication ofTheorem 24.2. (a) (D2—2D—3)(sinx +x2) =(D2-2D—3)sinx+ (D2-2D-3):c2 =—sinx -—2cos:c —3sin:c+2 —4x —31:2 =2—4:c— 3:02—4sina: —2cosx. Method 2.Byapplication ofTheorem 24.46 and(24.42). (b) (D2 —2D—3)(sin:z: +2:2) =[(D+1)(D—3)l(sinw +$2) =(D+1)l(D —3)(SiI1w +$2)l =(D+l)(cosx +2::—3sinx —312) =——sinx+ 2—3cos:r —62+ cosa: +21: ——3sin:c —32:2 =2—4x—3:2—4sin:c —2cos:r. Deﬁnition 24.52. If (24.521) P(D) =a,,D" +---+a1D +ao isapolynomial operator oforder n,then (24-522) P(D+<1)=<l»(D +<1)"+---+¢11(D +11)+(10, where aisaconstant, i.e., (24.522) isthepolynomial operator obtained byreplacing Din(24.521) byD—|—a. Summary 24.523.* Polynomial operators canbeadded, multiplied, factored, andmultiplied byaconstant, justasifthey were ordinary polynomials. Furthermore, the (24.53) Commutative law: PIP; =P,P1, Associative law: P1(P2P3) = (P1P2)P3 =P1P2P3, (24.55) Distributive law: P1(P2 +P3)=PIP2 —|—PIP; areallvalid. ‘Some oftheproperties here summarized donotapply topolynomial operators with nonconstant coeﬂicients. See,forexample, Exercise 24,17and18. 260 Ornnuons ANDLumen Tnausronus Chapter 5 LESSON 24C. Exponential Shift Theorem forPolynomial Opera- tors. Ifthefunction tobeoperated onhasthespecial form ue”, where uisannthorder differentiable function ofxdeﬁned onaninterval I, then thefollowing theorem, called theexponential shift theorem because theformula weshall obtain shifts theposition oftheexponential e",willaidmaterially inevaluating P(D)ue‘“’. Theorem 24.56. (Exponential Shift Theorem) If (24-57) P(D) =a..D”+a.._1D”" +---+111D+(10, an#60,isapolynomial operator withconstant coeﬂicients andu(:z:)isan nthorder diﬂerentiable function ofxdeﬁned onaninterval I,then (24.58) P(D)(ue‘“) =e‘"P(D +a)u, where aisaconstant. Proof. Weshall ﬁrstprove byinduction thatthetheorem istruefor thespecial polynomial operator P(D) =D2. ByComment 24.1 wemust show that: 1.(24.58) istrue when lc=1,i.e.,when P(D) =D. 2.If(24.58) isvalid when lo=n,then itistrue when lc=n'+ 1,i.e., if(24.58) isvalid when P(D) =D”,then itistruewhen P(D) =D"+1 Proof of1.When P(D) =D,theleftsideof(24.58) isD(ue“"). By (24.11) and(24.14), (a) D(ue“’) =aue“ +e“2u' =e“"(u' +au)=e“2(D —|—a)u. ByDeﬁnition 24.52, ifP(D) =D,then P(D -1-a)=D+a.Substitut- ingthese values ofDandD+aintheﬁrstandlastterms of(a),we obtain (24.58). Proof of2.Weassume that (24.58) istrue when P(D) =D". We must then prove that thetheorem istrue when P(D) =D"+1. By Deﬁnition (24.52), ifP(D) =D",P(D —|—a)=(D+a)". Substituting these values ofP(D) andP(D +a)in(24.58), weobtain (b) D"(ue“") =e“2(D+a)"u, which, byourassumption, isatrue equality. Operating on(b)with D, weobtain By (0)DlD"(1w“’)l =Dl¢‘“’(D +¢1)"ul (b), =e‘“’D(D -1-a)"u +ae“‘(D +a)"u (24.11), =e“[D(D +a)"—|—a(D —|—a)"]u (24.39), =e‘“‘[(D —-l—a)"(D +a)]u Corollary 24.481, =e"2(D +a)"+1u. Corollary 24.481. Lesson 24C Exrousnmn Snnrr THEOREM—POLYNOMIAL Orsmrrons 261 Theleftsideof(c)isD”+1(ue”). Hence wehave shown by(c)thatthe validity of(b)leads tothevalidity of Dn+1(,ueaz) =ea::(D +a)n+1u. Ittherefore follows byComment 24.1, that (24.59) D"(ue‘") =e"“(D +a)"u, forevery lc. LetP(D) betheoperator (24.57). Then By (6)P(D)(1w“2) =a..D"(W“2) +---+w1D(1/8"’) +110114" (24-14), =¢“’[a..(D +<1)”+---+a1(D+<1)+aolu (24-59),=e““P(D +a)u Deﬁnition 24.52. Corollary 24.6. (24.61) (D—-a)"(ue“‘) =e”D"u. Proof. LetP(D) =(D—a)”. Then byDeﬁnition 24.52, P(D +a) =(D+a-—a)"=D".Substituting these values ofP(D) andP(D+a) in(24.58), weobtain (24.61). Corollary 24.7. Ifcisaconstant, andP(D) isthepolynomial operator (24.57), then (24.71) P(D)(ce“") ==ce‘"P(a). Proof. By(24.58) andDeﬁnition 24.52, (a) P(D)ce“2 =e“2P(D +a)c =@'“”[a..(D +11)”+a.._1(D +a)""‘ +---+11016- ByTheorem 24.46, (24.42) and(24.14), (b) (D+a)"c=(D+a)’°"‘(D +a)c=(D+a)'°"1ac =(D+a)'°"2(D +a)(¢w) =(D+a)"'2(<12¢) =(D+a)(a'°_1c) =a'°c. Hence, by(b),wecanwrite thesecond equation in(a)as (c) P(D)ce‘“‘ =e'“[a,,a” -1-a,,__1a"_1 +---—|—ala+a]c. ByDeﬁnition 24.52, (c)isequivalent to P(D)ce“ =e“‘P(a)c =ce‘"P(a). 262 Ornmrons ANDLumen Tmnsronus Chapter 5 Example 24.8. Evaluate (a) (D2+2D+3)(e2‘sin1). Solution. Comparing (a)with (24.58) weseethat P(D) =D2+ 2D+3,a=2,u=sin1:.Hence byDeﬁnition 24.52, P(D +2)= (D+2)’+2(D+2)+3=D2+6D+11.Therefore by(24.58), (b) (D2+2D+3)(e2‘ sin1:)=e2’(D2 +6D+11)sin1: =e2‘(10 sin1:+6cos1:). Example 24.81. Evaluate (a) (D2 -—D+3)(1:2e‘22). Solution. Comparing (a)with (24.58), weseethat P(D) =D2— D—|—3,a=—2, u=1:3. Hence byDeﬁnition 24.52, P(D —2)= (D-2)2-(D—2)+3=D2—5D+9.Therefore, by(24.58), (b) (D2 —D+3)(1:3e'2‘) =e'2'(D2 —5D+9)1:3 =e'2’(61: -15¢’+913). Example 24.82. Evaluate (a) (D—2)2(e2‘ sin:0). Solution. Comparing (a)with (24.61), weseethat a=2,u=sin1:, n=3.Hence by(24.61), (b) (D—2)3(e2" sin1:)=e22D3 sin1:=—e22 cos1:. Example 24.83. Evaluate (9-) (D3-3D2+2)(5¢_4’)- Solution. Comparing (a)with (24.71), weseethat P(D) =D2— 3D2 +2,c=5,a=-4. Therefore P(—4) =—-64 —-48+2=-110. Hence by(24.71) (b) (D3-3D2+2)(5e"'22) =5e-*=(-110) =-5503'". LESSON 24-D. Solution ofaLinear Differential Equation with Constant Coeflicients byMeans ofPolynomial Operators. In Lessons 21and22weoutlined methods forﬁnding thecomplementary function y,andaparticular solution y,ofthenonhomogeneous linear equation, (24-9) an/‘") +a.._1y‘"'” +---+aw’+at=Q(w), at#0. Inthis lesson weshall solve (24.9) bymeans ofpolynomial operators. Weillustrate themethod bymeans ofexamples: Lesson 24D Sontrrron orLINEAR Equxrron BYPOLYNOMIAL OPERATORS 263 Example 24.91. Find thegeneral solution of (11) 1/”’+2y”—1/’—22/=6“- Solution. Inoperator notation, (a)canbewritten as (b) (D3+2D’ —D—2)y=e2”. ByTheorem 24.46, (b)isequivalent to (c) (D—1>(D+1)(D+2):;=e“- Let (d) H=(D+1)(D+2)?!- Then (c)becomes (e) (D-—1)u=ea’, u’—-u=e2‘, which isaﬁrst order linear differential equation inu.Itssolution, by Lesson 11B,is (f) u=e2’+cle‘. Substituting thisvalue ofuin(d)gives (5) (D+1)(D+2);!=6””+01¢‘- Let (11) v=(D+2);!- Then (e)canbewritten as (i) (D+1)v=e2‘+cle’; v’—|—v=e2‘—|—cle‘, anequation linear inv.ByLesson 11B, itssolution is (j) v=$2"+6%e’+c2e". Substituting thisvalue ofvin(h)gives (k) u’+21/=£8“+9;e‘+czf‘, anequation linear iny.Itssolution, byLesson 11B, is (1) 1/=-1*-re”+2-10‘+02¢"+caf”, which canbewritten as (111) 31=11292‘ +C19‘ +626-’ +039-2’- 264 Oranxroas ANDLumen Tmusronms Chapter 5 Comment 24.92. The solution (m)could have been obtained much more easily ifwehadfound y,,bymeans ofLesson 20,andused theabove method toﬁnd only theparticular solution e2”/12. The solution (f) would then have read u=62:; thesolution (j) v=ﬁe“, andthesolution (1), up=r‘ze"~ The roots ofthecharacteristic equation are, by(c),1,—l, -2. We could, therefore, easily have written thecomplementary function yc=cle’—|—c2e_’ +c3e"2". Example 24.93. Find thegeneral solution of (=1) 1/"+y=6‘- Solution. Inoperator notation, (a)canbewritten as (b) (D2+l)y=6‘- ByTheorem 24.46, (b)isequivalent to (C) (D+i)(D—ily=6‘- Let (d) u=(D——i)y. Then (c)becomes (e) (D+i)u==e‘, u’+iu=e‘, anequation linear inu.Itssolution, byLesson 11B, is 11,= 6:+01'6-£1. Hence (d)becomes . 1 _- (g) y’—1?!=75¢‘ +c1'e‘”, whose solution is y= -5-6: +%C1"l~6—"z +62,6“. This lastequation canbewritten with newparameters as (i) y==36‘+elf“ +626*’- Lesson 24—Exe1-cise 265 Again weremark that (i)could have been obtained more easily, ifwe hadfound yebymeans ofLesson 20andused theabove method toﬁnd only theparticular solution e”/2. Ingeneral, ifthenonhomogeneous linear differential equation (24.9) oforder nisexpressed as (24-94) (D-T1)(D —T2)'''(D-My=Q(x), where r1,r2,---,r,,aretheroots ofitscharacteristic equation, then a general solution (or,ifarbitrary constants ofintegration areignored, a particular solution) canbeobtained asfollows. Let (24.95) u=(D—r2)---(D——r,,)y. Then (24.94) canbewritten as (24-96) (D—T1)“=Q(1), anequation linear inu.Ifitssolution u(x) canbefound, substituting it in(24.95) willgive (24-97) (D—T2)(D —Ts)'''(D—my="(II)- Let (24.98) v=(D—r3)---(D--r,,)y. Then (24.97) becomes (24.99) (D—r2)v =u(:c), anequation linear inv.Ifasolution forv(x)canbefound, substituting thisvalue in(24.98) willgive (24.991) (D—r3)---(D—r,,)y =v(:c). The repetition oftheabove process anadditional (n—2)times will eventually lead toasolution fory. EXERCISE 24- 1.Prove byinduction that 12+22+32+...+n2= . 2.Find D°y, Dy,D21/, Dayforeach ofthefollowing: (a)y(z)=31¢’, (b)1/(w)=3sin2w. (0)y(w)=V5- 3.Find D91, Dr,D2rforeach ofthefollowing: (a)r(0)=cos0+tan0. (b)r(0)=02+sin0. 1 (0)N9)=55' 266 OPERATORS moLAPLACE Tmnsronus Chapter 5 4.Find D°:c, Dz,D22:foreach ofthefollowing: (a)a:(t)=t2—|—3t+1. (b)a:(t)=acosﬁt +bsin08. (c)a:(t)=acos (0t+ b). 5.UseDeﬁnition 24.13 toevaluate each ofthefollowing: (a)(D2—2D—3)cos22:. (b)(D2—-6D—|—5)2e3‘ (c)(D4—2D2)4a:3. (d)(D2-—4D+4)(x2 +x+1). 6.Use(24.22) toevaluate each ofthefollowing: (a)D3(cos aa:+sinbx). (b)(D2+D)(3e“ +22:3). (c)(D2—2D+4)(:ce‘ +52:2+2). 7.Usetheprinciple ofsuperposition toﬁndaparticular solution ofeach ofthe following equations. (a)y"+ 3y’+ 2y=8+6e“’+ 2sin:2:. (b)11::—2y’—8y=9:ce'+ l0e“’. (C)y+2y’+ 10y=¢‘+2- 8.Evaluate P1(D) —|—P2(D), where (a)P1(D) =D2+2D—-1,P¢(D) =3D3 —|—4D2 -—D+3. (b)P1(D) =D4-2D2, P2(D) =D2—6D—|—5. 9.Evaluate P1(D) +P2(D) —|—P3(D), where (a)P1(D) =D2—2D—3,Pg(D) =D2-D, P3(D) =D4+2D3——2D2+3D—|—4. (b)P1(D) =3D3+2D2+1, P2(D) =2D4—l—D3+2D2—3D+ 2, P3(D) =3D5 -4D2+7. 10.Evaluate bymeans of(24.36) (a)[r2(D2 +l)l(2¢’)- (b)Kw-l)(D3 +D2)l(¢2‘ +I2)- (¢)le"(3D +4)](mn w+1)- (d)[(12+31+4)(D3 +1)l(=v3 +2)- ll.Evaluate twoways, byTheorem 24.33 andby(24.39), (11)r2l(D2 +1)+Dl(2@'2’)- (b)(Z—1)[(D3 +D2)"l"(ZDZ —6D+5)l($3 -l‘2)- (c)sin:c[(2D2 ~—D+1)+(D2—|—D—-1)]cotav. 12.Evaluate twoways, by(24.42) andbyTheorem 24.46, (9-)(D-'1)(D +1)(3¢“)- (b) (D2 —D-l-1)(D +1)(3$3 —|—22:2 ——52:—6). (c)(D2+1)(D —3)(cos 2:+24:3’). 13.Evaluate, bymeans ofTheorem 24.56, (a)(D2-—2D—3)(e2‘ cos22:). (b)(D2—D—|—3)(3:v2e'2"). (c)(D—2)2(e"‘ tanx). (d)(D2+2D)(3e2‘ csc2:). (e)(D2——2D—|—6)(e‘3" log2:). 14.Evaluate, bymeans ofCorollary 24.6, (a)(D—1)e'sin:c. (b)(D-l—1)e"‘ cos2:. (c)(D-—2)2(e2’ log2:). (d)(D+2)2(e"2” tanac). (e)(D—3)2(e3‘ Arcsin2:). (f)(D—f-3)2(e'3‘ cot2:). Lesson 24—Answers 267 15.Evaluate, bymeans ofCorollary 24.7, (a)(D3+3D+1)(5e2‘). (b)(D3+3D—|—1)(2e‘2"). (c)(D—2)3(3e4”). (d)(D2—3D+7)5e8’. 16.Prove Theorem 24.46 forn=3.Hint. (D—r1)(D —r2)(D —r3)= D3-—(r1+12+r3)D2 +(T11'2 —|—r1r3+r2r3)D —r1r2r3. 17.Prove Theorem 24.46 isfalse ifthecoefficients ofP(D) in(24.47) arefunc- tions of2.Hint. Show, forexample, that(22D2 —-1)#(2D—1)(2D+ 1) byevaluating (22D2 -—1)e2‘ and(2D——l)[(2D —|—1)e2"]. 18.Prove thecommutative law(24.5) isfalse ifthecoefficients ofP(D) in (24.47) arefunctions of2.Hint. Show, forexample, that(D2—l— 2D)(3D)1/ 95 3D(D2 —|—2D)y, where yisasecond order differentiable function of2. Find thegeneral solution ofeach ofthefollowing differential equations. Follow themethod ofLesson 24D. —— —2y=e‘. 29.y”'—3y'+ 2y=e"‘. 2y=12e". 30.y”+4y=42sin22. =e“. 31.y"’-—3y"—|—3y’—y=e’. =sin2. 32.y”'—y’=e2’sin22. =cos2. 33.4y"—5y’=22e"’ 24. =8+6e"+2 sin2. 34.y"'—y"+ y’—y=4e‘3’+ 224. 25.y"— —81/=100" +92e‘. 35.y"'—5y”+8y’—4y=3e2‘. 26.y”— =2e2’sin2. 36.Prove theequalities in(24.34). 27.1/(4)—2y”+y=2—sin2. 37.Prove theequality 111(24.44). 28.y"+g’=22+22. 38.Prove theequality in(24.45).mannaS'°!°'."'.°!° §=§:§:Q:@:Q:+++++co .g@.;=~=.%’%°=%°%’°=.g~+2\\\N++++‘Qmum=e=e@ ANSWERS 24 2.(a)322,62,6,0. (b)3sin22,6cos22,~—12 sin22,-24 cos22.(0)\/Zix-1/2, ___&x-3/2’ £4;-5/2_ 3.(a)cos0+tan0,—sin 0—|—sec20,—cos 0+2sec20tan0. (b)02—|—sin0,29+cos0,2——sin0. (c)0'2, -——20_3, 60". 4.(a)t2+3t—|—1,2t+3,2.(b)acos01+ bsin0t,—-a0sin0t—|—b6cosﬁt, —a02 cos0t—b02sin0t. (c)acos(0!+b),-119sin(ﬁt+b),—-a02 cos(Ht—|—b). 5.(a)4sin 22—7cos22. (b)--8e3‘. (c)—-482. (d)422—-42+2. 6.(a)a3sinaa:-—b3cosb2. (b)6e’+622—l-122. (c)32e‘+18—-202+2022. 7.(a)y,=e‘+ 4+§(sin2 —3cos 2). (b)y,=—2e“ —-22". (c)1/,=23/3. (9)1'19=115914“ 8.(a) 3D3+5D2+D+2. (b) D4-D2—6D-l-5. 9.(a) D2+ZD3 —l—1. (b) BD5 —|—ZD4 +4D3 -—3D+10. 10.(a)4x2e=. (b)2(2—1)(6e2‘ +1). (c)e_‘(3 sec22+4tan2+4). (<1)($2+31¢+4)(13 +3)- ll.(9.)622e'2’. (b)(2—~1)(523 ——1822—l-182—|—16). (c)6cot2csc2. 12.(a)0. (b)323+222--52+12. (c)0. 13.(a)e2"(-4 sin22--7cos22). (b)3e"2’(2 -—102—|—922). (c)e"‘(2 sec22tan2—6sec22—|—9tan2). (d)302'csc2(csc2 2+cot22—-6cot2—|—8). (e)e"3’(21log2 —2'2—82"). 268 OPERATORS ANDLumen Tmnsronms Chapter 5 14.(a)e‘cos2. (b)—-0" sin2. (c)——e2‘2'2. (d)2e"2’ sec22tan2. (e)2e3‘(l ——22)"3'2. (f)2e_3‘ csc22cot2. 15.(a)75e2’. (b)—-26e-2”. (c)24¢“. (d)235e3”. 19.y=c1e"’ +0202‘ ——ﬁe‘. 20.y=2e’+c1e'2‘ —|—age". 21.y=~115(e“ -—3ie“) +c1e'2" +me". 22.y=-115(sin 2—-3cos2)-l-c1e‘2’ —|—age". 23.y=-115(3 sin2+cos2)—|—ole-2' +cge-*'. 24.y=4+e’—|—§(sin 2—-3cos2)—|—c1e‘2’ —l-020". 25.y=c1e4” +c2e‘2’ —2e’-2e’”. 2: 26.y=—2?(3sin2 +cos2)—|—c1+6263:. , ._,, sin2 27-1/=(v1 +cz2)e +(ca+c-me +2—T- 3 28.y=01+ c2e"'—|—£3-- 29.y=(c1+cz2)e= +c3e'2‘ —|—ﬁe". 30.y=c1cos22—|—02sin22—E(22cos22.—- sin22). 3 31.y=(01+62$+0322+ e'. , _,, 2— '231,,=,,+,,,, +,,, +(%+ ),=._ as.y=c1+nah" +(s1¢2+2341+266)$5; 32 34.y=c1e'—|—c2sin2+c3cos2 -—(eT+224+823+48)- 222 35-y=($1+c.w>e2'+ cse'+ LESSON 25. Inverse Operators. InLesson 21,wefound, bythemethod ofundetermined coefficients, a particular solution ofthenthorder linear differential equation (25-1) P(D)y =Q(x), where P(D) isthepolynomial operator (25.11) P(D) =a,,D" +---—|—a1D —|—ao, ansé0, and Q(2) isafunction which consists only ofsuch terms asb,2",e”, sina2,cosax,andaﬁnite number ofcombinations ofsuch terms. Here aandbareconstants andkisapositive integer. Inthislesson weshall show how inverse operators may furnish arelatively easy andquick method forobtaining thissame particular solution. Let (25-12) 1/.=cm+~-+en:/1. Lesson 25A MEANING orANInvnnss Orsnxron 269 bethecomplementary function of(25.1), i.e.,lety,bethegeneral solu- tionofP(D)y =0,andlety,,,-beaparticular solution of(25.1). Therefore thegeneral solution of(25.1), byTheorem 19.3, is (25-13) y=l/c+yr- Byfollowing themethod ofundetermined coefficients asoutlined in Lesson 21,weobtained aparticular solution y,,that contained noterm which was aconstant multiple ofaterm inyc.However, there arein- ﬁnitely many other particular solutions of(25.1). ByDeﬁnition 4.66, each solution which satisﬁes (25.1) and does notcontain arbitrary con- stants isaparticular solution of(25.1). Forexample, thecomplementary function ofthedifferential equation (2) (D2—1)y=12 is (b) y.=me’+26""- Aparticular solution of(a),found bythemethod ofundetermined coeffi- cients, is (c) y,,=-22 —2. Therefore thegeneral solution of(a)is (d) y=-22 —2+ole’—|—c2e_2. You canverify that thefollowing solutions, obtained byassigning arbi- trary values totheconstants c1andc2of(d)are,byDeﬁnition 4.66, also particular solutions of(a). (e) 1'/P = _z2 _21 y? = —x2 —2_36:; y,,=-22 —2+6e‘—e_‘, y,=-22 —2+3e"“, etc. Note, however, that every twofunctions differ from each other byterms which areconstant multiples ofterms iny¢.Intheproof ofTheorem 65.6, weshow that this observation holds forallparticular solutions of (25.1). Intheremainder ofthislesson andthenext, whenever werefer toapar- ticular solution y,,of(25.1), weshall mean that particular solution from which allconstant multiples ofterms inychave been eliminated. Forin- stance, intheabove example, ourparticular solution of(a)is-22 —2. LESSON 25A. Meaning ofanInverse Operator. Deﬁnition 25.2. LetP(D)y =Q(2), where P(D) isthepolynomial operator (25.11) andQ(2) isthespecial function consisting only ofsuch 270 Orsnxrons ANDLAPLACE Tmnsronms Chapter 5 terms asb,2",e“,sinax,cosax,andaﬁnite number ofcombinations of such terms,* where a,bareconstants andlcisapositive integer. Then the inverse operator ofP(D), written asP'1(D) or1/P(D), isdeﬁned as anoperator which, when operating onQ(x), willgive theparticular solu- tion y,,,of(25.1) that contains noconstant multiples ofaterm inthe complementary function ye,i.e., <25-21> P"‘(D)Q(w) =y.orﬁes) =y... where y,,istheparticular solution ofP(D)y =Q(2) that contains no constant multiple ofaterm inyo. Comment 25.22. Ifwearegiven P(D) andQ(2), wenow know how toﬁnd P_‘(D)Q(2). ByDeﬁnition 25.2, itistheparticular solution y,, ofP(D)y =Q(2) that contains noconstant multiples ofterms inyc. Example 25.23. Evaluate (a) (D2 —3D—|—2)"12. Solution. ByDeﬁnition 25.2, (D'2 -—3D+2)"12 =yp,where y,is aparticular solution of (b) (D2—3D+2)y=2, 1/"—3y’+21/=2. that contains noconstant multiples ofterms inyc. Bythemethod ofLesson 21or24D, weobtain theparticular solution 3(<>) up=3+1' Hence _ 3(<1) <1>’—s1>+2>*x=§+;- Comment 25.24. ByDeﬁnition 25.2, weconclude that (25.25) D*”Q(:e) Eintegrating Q(x) ntimes andignoring constants ofintegration. Proof. ByDeﬁnition 25.2, D“"Q(2) =y,,,where y,,istheparticular solution of (*1) D"?/=Q(I) thatcontains noconstant multiples ofterms inthecomplementary func- ‘This restriction onQ(2) isadrastic one. Ourdeﬁnition, however, would notbemean- ingful ifQ(z) were notthus restricted, since itmight then bedifficult toexhibit y,ex- plicitly orimplicitly interms ofelementary functions. Lesson 25A MEANING orANInvsnsr-1 Orumvron 271 tionyeof(a).Thecomplementary function of(a)is (b) ye=C1+62$+631112+---+c»_1w"“, andthese terms result from retaining theconstants ofintegration when integrating (a)ntimes. Example 25.251. Evaluate (a) D_2(2:v +3). Solution. By(25.25), _2 __ _1 __ _1 2 _It3 3Z2(b) D(2a:+3)-D (2x+3)d:z:-D (2:+3:z:)-§+—§-- You canverify that 1:3/3 —|—3122/2isaparticular solution of (c) Dzy =2:1:+3, y"=2:1:+3, andthatthecomplementary function of(c)isye=cl+C213. Comment 25.26. Wedraw another important conclusion from Deﬁ- nition 25.2, namely thatifP(D)y =0,then (25.27) y,=P*‘(D)(0) =0ory,=P—(155(o) =0. Proof. ByDeﬁnition 25.2, P_1(D) (0)=yp,where 31,,istheparticular solution of (Q) P(D)y =0 that contains noconstant multiple ofaterm inthecomplementary func- tion ycof(a). This particular solution isy,=O. Theorem 25.28. (25.29) P(D)[P"‘(D)Q] =QorP(D) Q]=Q- Proof. Lety,beaparticular solution of <2») Pom=Q- Therefore (b) P(D):/p =Q- By(a)andDeﬁnition 25.2, l (0) 1/1»=REQ- In(b)replace y,byitsvalue in(c).The result is(25.29). 272 Ormwrons ANDLumen TRANSFORMS Chapter 5 LESSON 25B. Solution of(25.1) byMeans ofInverse Operators. Asstipulated atthebeginning ofLesson 25,thefunction Q(x) of(25.1) may contain onlysuch terms asb,:c'°,e“,sinax,cosax,oraﬁnite com- bination ofsuch terms, where aandbareconstants andIoisapositive integer. Weshall ﬁrst consider each ofthese functions individually and then combinations ofthem. 1.IfQ(x)=bx“andP(D)=D-ao,then(25.1)becomes (a) (D—my=br'°, y’—any=W‘,"0¢0- The complementary function of(a)isyc=ce“°". Hence atrial solution yp,bythemethod ofLesson 21A, is (b)y,=A1:z:'°+A2:z:'°"1+ A3¢’*"=' +AM-3 +---+Aka:+A,,+,. Differentiation of(b)gives (c)y,,'=A,lcx'°_‘ +A2(k-1):v"‘2 +A3(k-2)x"_3 + +A,,. Therefore, by(b)and(c),ypwillbeasolution of(a)if (d)Up’—(lo!/p=-llofliwk +(A175 ""<loA2)$k—l +[A209 '-1)—¢loAsl$'°_2 +lAs(k —2)—aoA4lfBk_3 +-''+(Ah—a0Ak+1) =bxk. Equation (d)willbeanidentity in1:,if b(9) "@0111 ==b,A1=""E6" A=M=_l’l°_. A110-(1°A2=0, 2 ao (Z02 Ak-1 bkk——1 A2(k"-1)"-aoAa=0»As=$=-"%1§-'2' Ak—2 blck—1Ic-2 Aa(k'"2)—¢loA4=0, A4=i(5:i=— - A bk!Al:-'¢loAk+1 =0,Ak+1= Z5="-%' Substituting (e)in(b),wehave (f)yp=—%[x'°+%x""‘+%w""+---+£%,]» at#0- Lesson 25B So1.trrroN or(25.1) BYMEANS orINVERSE OPERATORS 273 Weshall now prove that thesame particular solution results ifwe formally expand 1/(D-—ao)inascending powers ofDandthen perform thenecessary differentiations. ByDeﬁnition 25.2, aparticular solution of(a)is (8) 1/p="171-E (Wk)=% (bwk) —a»(1——)"0 __1 DD’D3 0”] ,.——a[1-l-F0-l-I0;-l-E35-l-"'-l-m(b$), where thelast series was obtained byordinary division. Note that in making thisdivision, itisnotnecessary togobeyond theD'°/aok term since D'°+‘:c" =O.Performing theindicated differentiations, wehave (h) yr:_£;[xk+%xk—1+ xk—2+..--|- 7 (lo?50, a1. which isthesame as(f).Ingeneral, ithasbeen proved thatif (i) P(D)?! =(<1..D" +---+a1D+ac)?!=bx“, then <25-3)y.=$<1>@’“> =a(1+911>+1Zl0’+---+1“-D" (Mk)0 G0 (lo (lo ) =%(1+ b1D+b2D2+---+b;,D")x", a0¢0. where (1+b1D +b2D2 +---+b;,D'°)/ao istheseries expansion ofthe inverse operator 1/P(D) obtained byordinary division. Ifk=0,then (i)becomes P(D)y =b.Hence, by(25.3), 1 byp= ='a—o: (lo7'50. Example 25.32. Find aparticular solution of (a) y”-2y’-3y=5,(D2-2D-3)y=5. Solution. Comparing (a)with (i)above weseethatao=—3,b=5, k=0.Therefore, by(25.31), <1») y.=‘T5- 274 Ormmvrons ANDLumen Tmnsronus Chapter 5 Example 25.33. Find aparticular solution of (a) 4y”—3y’+9y=51:2, (4D2 —3D+9)y=52:2. Solution. HereP(D)=41>”-31)+9,<10=9,b=5.Hence, by (25.3), OOQF.(bl Up=ii (5$2) 9(1--+302 2 Note thatwedidnotneed togobeyond theD2term since D3(x2) =0. 2.IfQ(x)=bx“andP(D)=a,,D"+---+a,b,sothatao=0, then Disafactor ofP(D). Therefore, byTheorem 24.46, wecanwrite P(D) =D(a,,D"'1 +---+a2D —|—a1),where a1sf0.Ifboth ao=0 anda1=0,then D2isafactor ofP(D), sothatwecanwrite P(D) = D2(a.,.D"'2 +---+a3D+a2).Ingeneral, letD’beafactor ofP(D). Then P(D)y =bx"canbewritten as (a) P(D)y =D’(a,,D""' +---+a,-+1D +a,)y =b:c'°, a,-#0. Therefore, byDeﬁnition 25.2, (b) y,=Dr(anDn_r +MID+af)(bat), .1.¢0. Weshall now stipulate that theinverse operator in(b)means 1 1(25.34) y,=Dr[anDn_r + +‘MID +ar(bxr=)], a,¢0. Comment 25.35. Since polynomial operators commute, wecould also have written, inplace of(b)above, (°) y"=<a.1>»—' +--~i<1...»+a.)1>'("di) _ 1 1 1,_anDn_r+___+a’+1D+ar[D’_(b{I3 )]» G,-#0. Ineffect wewould now beintegrating ﬁrst, see(25.25), andthen differ- entiating. Although noharm results, following thisorder may introduce terms inthesolution y,,that areconstant multiples ofterms iny,,.In that event, wemerely eliminate such terms. Asexercises, follow theorder ofprocedure given in(c)toﬁndaparticular solution ofeach ofthetwo Lesson 25B SOLUTION or(25.1) BYMmns orInvnnsn Ornnxrons 275 examples below. Inboth cases youwillobtain aterm iny,,that isa constant multiple ofaterm inyo. Example 25.36. Find aparticular solution of (2) y”—2y’=5. (D2——2D)y=5- Solution. Here P(D) =D2—2D=D(D —2).Therefore byDeﬁ- nition 25.2 and(25.34) (b) 2.= <2]- By(25.31), with b=5,ao=-2, <0) 7%:<5)=—2- Substituting (c)in(b),andthen applying (25.25) totheresult, weobtain 1 5(<1) y.=5(—5)=-2»- Example 25.37. Find aparticular solution of (a) y(5) ___y(3) =21:2’ (D5 __D3)y =2x2_ Solution. Here P(D) =D5—D3=D3(D2 —1).Therefore, by Deﬁnition 25.2 and(25.34), 1 1 (b) 1/p=ﬁ[DT:—1(2$2)]- By(25.3), with ao=——1,b=2, 1 1 (<5) Wig (212) =_i(1_i1)2) (2132) =“"2(1 +D2)$2 =——2(:v2 +2). Substituting (c)in(b),andthenapplying (25.25) totheresult, weobtain 1 5 s (<1) y.=ﬁ[—2(w’+2)]=-2(%+$3)- 3.IfQ(x) =be“, then (25.1) becomes P(D)y =be“. Weshall now prove thataparticular solution ofthisequation is (25.4) y,,=RIF)be”=11%;. P(a)-50. Note that theainP(a) isthesame astheexponent aine“. 276 OPERATORS ANDLxrmcn TRANSFORMS Chapter 5 Proof. By(25.1) and(25.11), P(D)y =be“isequivalent to (a) (a,,D" +a,,_1D"'1 +---+a1D +a0)y =‘be“. Since P(a) 260,(D—a)cannot beafactor ofP(D). This means that a cannot bearootofthecharacteristic equation of(a).Thisinturnimplies that thecomplementary function of(a)cannot have aterm e“"init. Hence thetrialsolution y,,of(a),byCase 1ofLesson 21A, is (b) 1/»=A6”- Differentiating (b)ntimes, weobtain (0) Up’=“A6”. I/p"=<l2A¢°”,ypur =a3Aea::, ___’yp(n) =anAea:c. The substitution of(b)and(c)in(a)gives (d) a,,a"Ae“‘ +a,,_1a”'1Ae“’ +---—|—a1aAe“" —|—a0Ae““ =be“, Ae“”(a,,a" +a,,_1a"'1 +---—|—ala+ao)=be“. ByDeﬁnition 24.52, thequantity inparenthesis isP(a). Thelastequa- tion in(d)therefore simpliﬁes to (e) AP(a) .=b,A=F25- Substituting this value ofAin(b)gives theexpression ontheextreme right of(25.4). Hence be“/P(a) isaparticular solution ofP(D)y =be“. Example 25.41. Find aparticular solution of (a) ylll _yn +yr+y=3e——2z, (D3 __D2 +D+ =3e—2z_ Solution. Here P(D) =D3—D2—|—D+1.Therefore, by(25.4), withb=3,a= -2, 1 _3—_ 3-21¢ — 3--2:: (bl1”=P<1>>3° 2‘Pf-2)'<—2>=*—<i2>=—2+1=_I3§e 22:- 4.IfQ(x) =bsinaxorbcosax,nospecial difficulty arises. For, by (18.84) and (18.85), wecanchange these functions totheir exponential equivalents anduse(25.4). Easier, perhaps, istoapply themethod of Lesson 21B. Weshall usethislatter method tosolve thefollowing problem. Example 25.42. Find aparticular solution of (a) y”—3y’+2y=3sin22:, (D2 —3D+2)y=3sin22:. Lesson 25B SOLUTION or(25.1) BYMEANS orInvnnsn OPERATORS 277 Solution. Since em=cos2:t—|—isin22:,theimaginary part ofa particular solution of (b) (1)2-31>+2);,=36*“ will beasolution of(a). Here P(D) =D2——3D—|—2.Therefore, by (25.4), with b=3,a=2i, () — 382i! _. 3821:»? _. 362%: ‘ _ °3"’“P(2i)_(2.)?-61+2*-2(1+3.")(1-st) =§1{%’)@=*'= =-,=*,,(1 -3t')(cos2:t +1811. 22) =——§5[(cos 2x+3sin22:)—|—'i(sin2x—3cos2a:)]. Theimaginary partof(c)is (d) y,=ﬁ;(3cos2:0—sin2:0), which isaparticular solution of(a). 5.Exponential Shift Theorem forInverse Operators. Theorem 25.5. IfP(D)y =ue”, where P(D) isapolynomial opera- toroforder nanduisapolynomial in2:,then 1 ax__ ax 1yp —--}?(*l-)-)-1.66 ——6 +a)U. Proof. (a) M“ ==e““u. Byapplying (25.29) toeach sideof(a),wecanwrite (a)as (b) P(D)[F1DS(w“‘):| =e“"P(D +<1)[P-(5-1;‘; By(24.58), e‘“’P(D +a)u=P(D)(e“"u). Applying thisequality tothe right sideof(b),with S uplaying theroleofu,weobtain, by(b), (0) P(D) (1w°”)] =P(D) [e“”ID-(E-l_-,_-E5 12]. By(24.21), wecanwrite (c)as U8“ —GM:‘P-('1—j-fin; '11.]=O. 278 Ornnurons ANDLAPLACE Tnxnsronus Chapter 5 Letyrepresent thequantity inbrackets. Then, by(d)andComment 25.26, <2yr=lﬁ<"”“>" =ﬁ<°>=°’from which wededuce, 1 G, 1 <‘> 1/P=m<"‘"°=“Example 25.52. Find aparticular solution of (a) y"—2y’-—3y=x2e”, (D2—-2D-—3)y=xzef‘. Solution. Here P(D) =D2—2D—3.Therefore Bu (b) yp=$5 (232621) =35f- (232023) Def. 25.2, =enj,(T1_|-_-55 2:2 (25.51), =enH)jT) 1:2 Def. 24.52, I =oz’i——- 12 Theorems 24.46 D2""2D"3 and24.33,I=ea‘—————i-——T 2:2 2D D -3(1-T-T) =—<1+§D+aw (25.3). 2: =-93-(22+gt+1;) Def.24.13and(24.11) 6.Formula (25.4) canbeused only ifP(a) ;é0.What ifP(a) =0? IfP(a) =0,then (D—a)isafactor ofP(D). Assume (D—-a)’isa factor ofP(D). Therefore wecanwrite P(D) =(D—a)'F(D), F(a) 940. Weshall nowprove thatifP(D) =(D—a)'F(D), where F(a) 950,and P(D)y =be“,then 1 1 b:t'e°‘25.6 =Zbe“‘Ei—be°”=——» F 0. Proof. Westipulate thattheinverse operator in(25.6) shall mean Lesson 25B SOLUTION or(25.1) axMmns orInvsnsn Ornnxrons 279 (Remark. Ascommented in25.35, noharm results, other than more labor, iftheorder ofperforming theinverse operations isinterchanged. Asre- marked there, using thisinverted order may introduce terms iny,that areconstant multiples ofterms iny,.Inthatevent wemerely eliminate such terms.) By(25.4), 1.1_K(b) WIN —F(a)| 75 Substituting (b)in(a),weobtain ByDeﬁnition 24.52, ifP(D) =(D-—a)’,thenP(D+a)=(D+a—a)’ =D’.In(25.51), letP(D) =(D-—a)’andu=b/F(a). Then (c)is equivalent to raz-.1»By(25.25), D"'[b/F(a)] means integrate [b/F(a)] rtimes, ignoring con- stants ofintegration. The ﬁrst integration gives b:t/F(a), thesecond bx’/2!F(a), thethird b:c3/3!F(a), andﬁnally therthintegration gives bx’/r!F(a). Hence (d)becomes (e) Fe)as0. which isthesame asthelastexpression in(25.6). Example 25.61. Find aparticular solution of (a)y"'—5y"+8y’—4y=3e“, (D3—5D+8D—4)y=3e“. Solution. Here P(D) =D3-—5D’+8D—-4.Therefore, byDeﬁ- nition 25.2, 1 <'°> 2»=DT‘_-rte:-§1"§ <a.=.,_ Ifwenowattempted toapply (25.4) totheright sideof(b),wewould ﬁndthatP(a), which hereequals P(2), iszero. Hence wemust resort to (25.6). Since (D3—5D”+8D—4)=(D—2)2(D —1),wehave by (b), (c) y,=ili— 3e”.(D—2)’(D —1) Comparing (c)with (25.6), weseethat b=3,a=2,r=2,F(D) = 280 Ormwrons umLumen Tnmsronms Chapter 5 D—1sothatF(a) =F(2) =2—1=1.Hence by(25.6), (c)becomes 3222 (<1) y.=—’”§%—- Comment 25.62. Theabove solution (d)could alsohave been found bythemethod ofLesson 21,Case 3.Agood way todiscover how much easier theabove method is,istotrytoobtain (d)bymeans ofthatlesson. Remark. Asanexercise, show thatifinplace of(c),wehadinverted theorder oftheinverse operators andwritten (6) Up=% (3@2z)]» andsolved itbyﬁrstapplying (25.51) andthen (25.4), thesolution y,, would have contained additional terms that were constant multiples of terms iny,. Example 25.63. Find aparticular solution of (a) y"+4y’+4y=52-2“. Solution. Here P(D) =D2+4D+4=(D+2)”. Therefore, by Deﬁnition 25.2, (b) Ilp=5% (5522)- Ifwenow attempted toapply (25.4) totheright side of(b),wewould ﬁnd P(a) =P(—2) =O.Hence wemust resort to(25.6). Comparing (b)with (25.6), weseethata=——2,r=2,F(D) =1,b=5.Hence, by(25.6) and(b), 52—21 (a) y.= 7.Let Q(x) =Q1(x) +Q;(x) +---+Q,,(x). Therefore P(D)y = Q(x) =Q1(a:) +Q2(x) +---+Q,,(a:). Weshowed incomment 24.25, that asolution y,,ofP(D)y =Q(x) isthesum oftherespective particular solutions ofP(D)y =Q1,P(D)y =Q2,---,P(D)y =Q...Itfollows, therefore, that ifP(D)y =Q,then 1_1 1 1(25-7) l/P=Hl7)'Q=ﬁQ1+WQ2+"'+'1T(F)Qn- Because of(25.7) andtherules developed inthislesson, wearenow in aposition tosolve alinear differential equation (25-71) a»y‘"’ +---+aiy’+(loll=Q(rv). Lesson 25B Sommon or(25.1) BYMmns onInvnnsn Ornmvrons 281 where Q(x) may contain such terms asb,x",e‘“’,sinax,cosax,andcom- binations ofsuch terms. Here aandbareconstants andkisapositive integer. Example 25.72. Find thegeneral solution of (a) y”—y=a:3—|-3a:—4, (D2——1)y=:z:3+3:z:—4. Solution. Here P(D) =D2—1.Therefore byDeﬁnition 25.2 and (25.3) <1»)yp=Filrlo“ +ax-4)=—<1+1>”><x“+3»~4). By(25.7), wemay apply 1/(D2 —1)=—(1 +D2) toeach ofthe terms inQ(x), namely to2:3,3x,——5. Hence, by(b)and(25.7), (c) y,=—-(:03 —|—3:1:—-4+6:0)==—(:c3 +9:0-—4). The complementary function of(a)is (d) y¢=cle‘—|—c2e_‘. Thegeneral solution of(a)istherefore thesumof(c)and(d). Example 25.73. Find thegeneral solution of (a) ylll __y//+ yr___y=4e—3z +23:4, (D3-D2+D-1)y=4e_3‘—l-211:4. Solution. Here P(D) =D3—-D2+D-—1.Therefore by(25.7) 1 _, 1(b)up= (4” 3)+ (214)- By(25.4)1 _2‘ 4-33 (°)D3-1>2+1> -14”’ 3“(-3)8-(—§)=+(—3) -1 _ e—3a: _ 10 By(25.3) 21¢‘)=-(1+D+D‘)(2¢*) =-(2¢‘+81:3+4s).(d)i<1%1>1L11>=%1>=> < Therefore, by(b),(c),and(d), (e) y,,=-(%;+2x4+8:z:3+48>. 282 Orsnxrons mmLumen Tmnsromas Chapter 5 Theroots ofthecharacteristic equation of(a)are1,2',—i.Hence (f) y,,=ole’+Cgsin1:+03cos1:. Thegeneral solution of(a),istherefore thesumof(e)and(f). EXERCISE 25 l.Evaluate (seeComment 25.22). @<0_m*@L+u_wy (M(D—1Y%?(c)(D2-—3D+2)‘1 sin22:. (d)(D2-1)‘1(2z). (e)(4D2 —5D)'1(a:2e"‘). 2.Evaluate (seeComment 25.24). (a)D‘1(2:c+ 3). (b)D‘3a:. (c)D‘2(3e3"). (d)D“’(2sin2:|:). Find aparticular solution ofeach ofthefollowing differential equations bymeans oftheinverse operator; usetheappropriate method outlined in numbers 1to7ofLesson 25B. 3.y:'—|—3y’—|—2y=4. 14.y"—y=sin:0.Hint. See15. '+y’+y=:c2. 15.y"_--y=cos:c. "—yI=2x. For 14and 15,solve -—3y+2y=:|:. y——y=e". -1/=32’. 16.y"—3y’+2y=3sinx. y=2:2+22:. 17.y”+azy=sinax. Hint. See18. "'—— =22:3. 18.y"—|—azy =cosax. —"'+y”=6. Hint. For17and18,solve y"+ +2y=12e‘. a2y=e“",a940. —|—2y2= e". 19.4%"—-'5y' =:r2e‘;. =3e-‘. 20.;/+1/+1/=3a:e‘. —2y’— 8y=9a:e". 22.y”'+3y”+3y’+y=e""(2 ——2:2). 23.y”+4y=42:sin22:. Hint. See24. 24-.y”+4y=42:cos22:. Hint. For23and24,solve y"—|—4y=4:te2"‘. 25.y"—y=2e‘. 26.y"+y’-—2y=3e'2‘. 27.y”'-—5y"+8y’—4y=e’. 28.11(4)-—3y"' --6y"+281/’—-24y=e2‘. 29.y::'—11,11" +39y’ —45y=e3’. 30.y—2y +11 =72‘. 31.y”’+y’=sin2:.Hint. See32. 32.y”'+y’=cos1:. F0231 and32,solve y'”+y’=e“. 33.y —3y +3y —-y =2e". 34-.y”-I— 31/+ 2y=8+6e’+ 2sin:c. y:;)++3g/if I=2(;"Z+. 12); .y y y=:cslna: cosx. 37.y’-3y =x3+3:c—5. 38.Solve Examples 25.36 and25.37 byfollowing theorder suggested inCom- ment 25.35. Show that theresults differ from thetextanswers, ifthey do differ, byaterm which isaconstant multiple ofaterm in11,.?§§F5PPa@w+Q<ﬁ‘§‘§\‘§'§<Q\§_Q<ﬁ°§z::_Q\:‘@9900@\‘§\<q‘§:‘ Lesson 26A PARTIAL FRACTION Exrmsrou THEOREM 283 39.Solve Example 25.61, byinterchanging theorder oftheoperators inthe denominator of(c).Show thatyour answer differs from thetextanswer by aterm which isaconstant multiple ofaterm inya. ANSWERS 25 1.(a)—§17(9:2:3 +9:02+332:—34). (b)—(z2 —|—22:+2). 2(0)9600321 --215sin2:c. (<1)-21. (e)9%(812:+2341+ 266). 2.(8.)12+31.(b)$4/24. (.)ea‘/3. (<1)-4sinzt. 2..y,=2. 21. 4.J-21. 22. 5.—2z. x36' yp=§+Z' 24- 7.11,,=-—3(a:2 +2x+2). 8.y,=:3/3. 5 4 9.1,,=-2(%+”I+¢“+s¢”) 10.y,=32:2. ll.y,=2e’. 29. 1-3.‘ iz12.y,= e. 30.10 13.y,=§e'2‘. 14.y,=-—§sin2:. 15.y,=——}cos2:. 16.y,=330-(sina: —|—3cos2:). 17.y,=—-itcosax. 18.y,=Y2sinax.Z8.-re‘. :c3e_’ 2 Z(sin2:—2:cos22:). Z4 me’. -—:ce"2‘.(22:sin22:—|—cos22:). x3e2"/30. —a:2e3’/ 4. 7x2e'/ 2. —-$2: sin1. —§a: cos1. a:3e"/ 3. 4+e’+§(sin 2:-—3cos:0). 2:2—32:+-1--—2:ce_2" 2 12—|—Z?(cosa: —sin2:). 19.See1(e). 37.See1(a). 20.y,=§e’(3a:2 —62:+4). LESSON 26. Solution ofaLinear Differential Equation byMeans ofthe Partial Fraction Expansion ofInverse Operators. LESSON 26A. Partial Fraction Expansion Theorem. Inalgebra, anexpression ofthetype 2 3 (a) 1+5? 284 Orsnxrons moLxrmcn Tnmsronus Chapter 5 canbesimpliﬁed bytheusual method ofﬁnding theleast common de- nominator. There results 5:v—1 (bl -am‘ Conversely, ifwestart with (b)wecanexpand itintothepartial frac- tions (a)bymeans ofthepartial fraction expansion theorem of algebra. Wedothisbyfactoring thedenominator of(b),andthen deter- mining AandBsothat 522-1 A B <°> at-1-;'i+?-H‘ Putting theright sideof(c)under onecommon denominator, theequation becomes 51:-l__A:c+A+Bx—B (d) :c2—-1_ :z:2—-1 ' Since thedenominators onboth sides oftheequal sign arealike, theA's andB’smust bechosen sothat their respective numerators arealsoalike, i.e.,sothat (6) x(A+B)+(A—B)E5a:——1. Equation (e)willbeanidentity in2:ifthecoeﬁicients oflikepowers ofx onboth sides oftheidentity signareequal. Hence wemust have (f) A+B=5, A—B=—L Solving (f)simultaneously, weﬁndA=2andB=3.Substituting these values in(c)gives usthepartial fraction expansion of(b),namely 5x-—1_ 2 3_ (9 222-1 x--1+:c+1 Wethus areable togofrom (a)to(b)orfrom (b)to(a). Comment 26.1. The general rule fordetermining theform ofeach numerator inapartial fraction expansion ofaquotient P(a:)/Q(x), where P(:c) isofdegree lessthan Q(x), willbeevident from thefollowing example. Let (11) Q(=v)=(1+<1)($3 +b)(12 +¢)2(¢+(1)3. andletP(:t) beapolynomial ofdegree lessthan Q(z). Then thepartial Lesson 26A PARTIAL Fnxcrron EXPANSION Tnnonnm 285 fraction expansion ofP(:z:)/Q(x) willhave theform .P(a:)_ A Bx2+C:c+D E:c+F Ga:+H (1)Q(:t)_x+a+ x3+b +:c2+c+(x2+c)2 I J K +x—l-d+(x+d)2+(a:+d)3 Note that: 1.Each numerator isapolynomial ofdegree onelessthan thedegree of theterm inside theparenthesis ofitsdenominator. 2.Aterm such as(x2+c)2hastheexponent two outside theparen- thesis. Hence ($2+c)appears twice inthedenominator, once as ($2—|—c),thesecond time as($2+c)2. 3.Aterm such as(as—|—d)3hasexponent three outside theparenthesis. Hence (:0+d)appears three times inthedenominator, once as(at+d), thesecond time as(:1:+d)2,andthethird time as(:1:+d)3. Comment 26.11. Ifthenumerator ofafraction isaconstant and thedenominator hasdistinct zeros, then there isaneat method which willquickly givethenumerators ofapartial fraction expansion. Let (a) f(w)=(w-1‘1)(w —1'2)'''(W—Th)'''(1—T»). where thezeros 1'1,r2,---,r,,aredistinct. By(a) (b) f(Tr=)=(Th—n)(n. —T2)---0-''(rt-T»)=0- Bycomment 26.1, thepartial fraction expansion of1/f(z) willhave the form 1 ___.A2 ___.__A'= i.(C) f(:c)—:z:—r1+:v—r2+ +2;-r;,+ +:t—r,, Multiply (c)by(2:—17,). Since by(b),f(r;,) =0,there results - _A(-> _____A»<-> Let:1:—>rk.The leftside of(d)willapproach 1/f'(r;,) anditsright side willapproach A1,. Hence, (6) A,,=f,+. k=1,2,---,n. Substituting (e)in(c),weobtain 1_ 1 1 _______1___.‘zmlf(z)—r'<r.><x-r.>+r'<r.><w—r.> ++f'<r..><w-r.) 286 Ornnxrons ANDLAPLACE TRANSFORMS Chapter 5 Example 26.13. Find thepartial fraction expansion of 3(#1) 9-_-_—1' Solution. Comparing (a)with (26.12), weseethatf(z)=2:’—1= (:1:--l)(:z: —|—1). Therefore, r1=1,12=-1, f’(:z:) =21:,f’(r1) = f'(1) =2,f’(r2) =f’(—1) =-2. Hence, by(26.12), 3 1 1 (b) I2-1"‘3l2(x -1)2(x+1)l Example 26.131. Find thepartial fraction expansion of <a> 4-———(iv—2)(12 +iv+1) Solution. Wecanifwewish factor x2+x+l=[(x+ )(x+ )] andthenuse(26.12). Orwecanusetherules ofpartial fraction expansion given inComment 26.1. Using thislatter method, wehave 3 __ A Bx+C (b) (a:—2)(x2+x-l—1)_x—2+:c'*’+:c+1 _Ax2+Aa:+A+Bx2+Cx—2Bx—2C'_ _ (w-2)(w’+w+1) For(b)tobeanidentity in11:,thenumerator inthelastfraction must equal 3.Hence wemust choose A,B,C,sothat (c) (A+B)I2+(A—2B+C)$+(A—2C)E3. Equating coefficients oflikepowers of:0onboth sides oftheidentity sign, weobtain (d) A+B=0, A—2B+C=0, A—2C=3. Solving (d)simultaneously forA,B,Cgives (e) A=-?, B=-5}, C=-1}. Substituting these values in(b),wehave (f) 3 =3_3(x+3) _ (iv-2)($"’+w+1) 7($-2) 7(r”+a=+1) Lesson 26A PARTIAL FRACTION EXPANSION Tnnoanu 287 Example 26.132. Find thepartial fraction expansion of 1 (a) (.1+1)(:c-1)2 Solution. Here F(x) =(x+1)(a:—1)2hasarepeated root. Hence wecannot use(26.12), butmust fallback ontherules ofpartial fraction expansion given inComment 26.1. Therefore, 1 A B C (b) (x—|—1)(a:-—1)2_:z:—|—1+x—1+(a:—1)2 Ac-1)’+Be”—1)+c'(x+1)_(iv—1)2(w +1) For(b)tobeanidentity inan,thenumerator inthelastfraction must equal one. Hence wemust choose A,B,andCsothat (c) A(x —1)”+B(:v2 —1)+C(:t—|—1)E1. Instead ofequating coefficients oflikepowers of:2:andthen solving for A,B,Caswedidintheprevious example, analternate simpler method in thiscase, istolet:1:=1in(c).There results (d) 2C=1, C=Q. Ifweletz=-1in(c),weobtain (e) 4A=1, A=i. Ifweletas=0in(c),weobtain (f) A-B+C=1. With A=1,C=Q,weﬁndB=—;}. Substituting these values in (b),weobtain () 1 = 1_ 1+ 1_ g (1+1)(t-1)2 4(:c+1)4(@-1)2(:c-1)» Comment 26.14. Analogously itcanbeshown that: 1 1.Ifyp=ﬁQ(a:), then theinverse operator canalso beexpanded into partial fractions justasifitwere anordinary polynomial. 2.Applying each member ofapartial fraction expansion of1/P(D) to Q(x) andadding theresults willgive thesame answer aswillapplying 1/P(D) toQ(x). 3.IfP(D) hasdistinct factors, then itispermissible totake advantage of(26.12) toﬁnditspartial fraction expansion. 288 Ornnxrons ANDLAPLACE Tnmsronns Chapter 5 LESSON 26B. First Method ofSolving aLinear Equation by Means ofthe Partial Fraction Expansion ofInverse Operators. Weillustrate themethod bymeans ofexamples. Example 26.15. Find thegeneral solution of (a) y’”-—5y”+8y’——4y=2e“, (D3—5D2 +8D-—4)y=2e“. Solution. Here (b) P(D)=D3-50’+81>-4=(D-2)’(D -1). Therefore, byDeﬁnition 25.2, 1 . 1 .<°)yr= <2e‘ )=iii”- Following thepartial fraction expansion method outlined above, weﬁnd 1 1 1 1 (‘D (1>-2)2(1>-1)*1>-1*»-2+(1>-2)2' Hence by(d)andComment 26.14, (c)canbewritten as (6) tip=% (2641) —% (2643) —|—F55 (284,)- Applying (25.4) toeach term ontheright of(e),weobtain, with b=2, a=4,andP(D) equal totherespective denominators, 242 242 24: <f> y»=%—-%—+%=%@“»which isaparticular solution of(a). The roots ofthecharacteristic equa- tionof(a)are2,2,1.Hence (g) y.=(61+cmez‘ +cw‘- Thegeneral solution of(a)istherefore thesumof(f)and(g). Example 26.16. Find aparticular solution of (a) y”—-y=2e3", (D2 —1)y=2e3". Solution. Here P(D) =D2—1.Therefore byDeﬁnition 25.2, (b) tip=§l_—; (29%) =(j)'f1j1(D*"_',_*'ﬁ (2933)- ByComment 26.14 and(26.12), with f(z)EP(D) =D2-—1,f’(x) E Lesson 26B SOLUTION BYMEANS orAPARTIAL FnAc'rIoN EXPANSION 289 P'(D) =2D,1'1=1,r2=—1,P’(r1) =2,P’(r2) =-2,wecanwrite (b)as 1 81 1 3:: <°> 1/»=a1T1><2“ >-Wm >-Applying (25.4) toeach term intheright of(c),weobtain 32 3:: 3:: w n=%—%=%- Comment 26.2. Ifaninverse operator isofthesecond order andhas distinct factors, wecangeneralize itspartial fraction expansion. By(26.12), iff(z) =$2-—(r1—|—r2):z: +T1T2 =(:2:-—r1)(:z: —1'2),then 1_ 1 __ 1 + 1 (”rw*e—me-m_rwo—m Mmc—m’ where r1and r2aredistinct. Here f'(:c) =22:—(r1+r2). Hence f'(T1) =2T1—(T1-l"T2)=T1-"T2and f'(T2) =2T2—(T1+T2)= —r1 -l-1'2.Therefore (a)becomes (b) 1 _ 1 1_ 1 1_ (11-T1)($ —T2)_(T1'-T2)(Z-*T1) (T1'-T2)(Z-T2) Analogously, 1 l 1 1 1 (2621) (D——r1)(D -1'2) _r1— r2D-—r1_r1—r2 D—— r2 _1[__1___ 1]_r1—r2D-—r1 D——r2’ which may betaken asaformula forthepartial fraction expansion ofa second order inverse operator whose zeros r1and1'2aredistinct. Example 26.22. Find aparticular solution of (a) y"—|—2y’+2y=3xe‘, (D2—l-2D+2)y=3:re”. Solution. Here (b) P(D) =(D2+2D+2). Therefore byDeﬁnition 25.2 (0) 1'/n= § (31%?)- The roots ofD2—|-2D+2are r1=—1+i,r2 =-1-—i.Hence, (d) r1-—r2=—1—l—i+1+i=2i. 290 Orsnxrons ANDLAI>LAcn Tmnsronns Chapter 5 Therefore, by(26.21) andComment 26.14, wecanwrite (c)as 1 1 1 , <°) W= (W)"BET-Ft <3“ll" Applying (25.51) toeachterm ontheright of(e),weobtain, withu=32:, a=1, _1._1_ _=_1___ ](fl ”P—2tleD+2-H3”) “D+2+t(3“)' By(25.3), theseries expansions oftheinverse operators in(f)are[write 1/(D +2—i)=1/[(2 --i)(1+D/(2 —i))] and then useordinary division] 1 D 1 D 2-¢(1_ 2-i+"'> Mdﬁ7(1_2+t+"')' Therefore, by(25.3), (f)isequal to " 3 D (g) ”»=i;l2T.("ﬁ“-.+"')“ 3 D -21?-(1 -Tn+"')"l _3e’ :1: 1 _ x 1 _2."l2-1" (2-02 2+i+(2+i)’i _ _2)_2,,»_12,._2)‘5 25T5 25' which isaparticular solution of(a). LESSON 26C. ASecond Method ofSolving aLinear Equation by Means ofthePartial Fraction Expansion ofInverse Operators. Example 26.3. Find aparticular solution of (a) y"—3y’+2y=sinan, (D2——3D—|—2)y=sin2:. Solution. Here (b) P(D) =D2-—3D+2. Therefore, byDeﬁnition 25.2, (c) y=mu sin:1:p D2—3D+2 ’ where y,isaparticular solution of(a). The zeros ofD2—-3D+2are Lesson 26—Exercise 291 r1=2,1'2=1.Hence, r1—r,=2—1=1. By(26.21) andComment 26.14, (c)becomes 1 . 1 .’_l/p='lT:§81l'lI-'F:TS1IlSD. Let (e) y=——1—-sinx y=i1—sina:1"D—2 '2" 1)-1 ' Therefore (f) yr=3/In+1/21»- ByDeﬁnition 25.2, yl,andy2,,areparticular solutions respectively of (g) (D—2)y=sin2:, (D—l)y=—sin :c. Aparticular solution ofeach equation in(g)isrespectively 2. .(1,, gm:_%t2<.>:2, ,,,,__..s111g. Substituting these values in(f),weobtain (i) y,,=—-§sin:c—§cosx+~}sinx+§cosa: =-fgsina: +ﬁycosx, which isaparticular solution of(a). Comment 26.31. Ifwehadsolved (a)bythemethod ofpolynomial operators, asoutlined inLesson 24C, wewould have written (a)as (D—2)(D-1)y=Q(<v). andthen, ineffect, solved twolinear equations insuccession. Bytheabove method, wesolve twolinear equations independently. EXERCISE 26 1.Find thepartial fraction expansion ofeach ofthefollowing. 2 3 2:2:-l-1 (a)c?-1' (b)(z-2)(¢2-1)' (°).12-1' 2:c2+1 2: 2:-l-1 (d)ea-1' (°)(:v—1)2' (0(¢2+1)(¢-1)‘ 292 Ornnxrons ANDLArLAcn Tnmsronus Chapter 5 Find aparticular solution ofeach ofthefollowing equations. Use methods ofLessons 26BorC. 9°2*‘9‘9'P9°P§=§:Q=Q:Q='§:@-—-2y’—3y=3e“. . —2y’—3y=3cosa:. 4y=2e“‘. —3y’+2y=me". —-y’—61/=z+e2‘.—2y’—3y=3sin 1:. 11.y'” 12.g/(4)—1/”+y"=6. 13.gm—-311'” —6y"+ 281/—241/=02‘. 14-.y”’—111/’+ 39y’ —451/=ea’. 15.1/5)+2y"'+1;’=2::—|—sins: -1-cos2:.”+y =x'*’+e2‘. 9.11”’-311”-I-3y’ —y =e‘.10 ylll 11+ I e-11 11—u=4'3'+2x‘-—y" =22“. ANSWERS 26 1 1 1 3 1 1'(“)t—-_1_I' lblx-2_2<¢-1)+2(.-.+1)' 3 1 3 1 1 ‘">@+2<?+—1>' ‘°"”+2(¥I"1".»Ti)'1 — 1 (e),,,%f"l-(Y-;-IF‘ (f)ﬂi_,:f1+;:_-T‘ 2.y,=:c2——2+§e2‘. 3.—f°-(2 cos:0:+sin2:)—iﬁ;(2 sin:1:-—cos2:). 4.,5.See3. 6-Us=39-3- 7.y,=313-(6:te_‘+ 5e"). 8-Us=_3e2: _''318'(6$ ‘“1)- 9.x3e‘/6. -32: 3 4 310.1;,=-(T+2¢ +81;+48)11 =—2(£—|-£—l-:c3+3x2)-'1"’ 20412.y,=32:2. 13.y,=xaez’/30. 14.y,=—x2e3’/4. 2 37215.1/,=x—|—? (cosx —sinx). LESSON 27.The Laplace Transform. Gamma Function. LESSON 27A. Improper Integral. Deﬁnition ofaLaplace Trans- form. For aclearer understanding ofthematerial ofthis lesson, a knowledge ofthemeaning oftheimproper integral I:f(z)da:isessential. Weshall therefore brieﬂy review thissubject foryou. Letf(z) beacontinuous function ontheinterval I:0§:0§h, h>0.If,ash—>co,thedeﬁnite integral fof(z)dzapproaches aﬁnite limit K,wesaytheimproper integral J:f(z)dxexists and converges tothisvalue K.Inthatevent wewrite no ll (27.1) Io dz= die=K. Ifthelimit ontheright does notexist, wesaytheimproper integral on theleftdiverges anddoesnotexist. Lesson 27A Iurnornn INTEGRAL. Lumen TRANSFORM 293 Letf(z)beacontinuous function ontheintervalhl: 0<:0§h,h>0. Ifash—>ooande-—>0,thedeﬁnite integral Lf(z)da:approaches a. ﬁnite limit L,wesaytheimproper integral I:f(z)da:exists andcon- verges tothisvalue L.Inthat event wewrite an h (27.11) /Q dz:= dx=L. ¢—o0 Ifthelimit ontheright sidedoesnotexist, wesaytheimproper integral ontheleftdiverges anddoes notexist. Example 27.111. Determine whether thefollowing integral exists. ”1(8.) -/; 5? dﬁb. Solution. n (m £;i7n=m%m+n. Ash->oo,log(h+1)—>oo.Hence theimproper integral (a)diverges anddoes notexist. Example 27.112. Evaluate "1 Solution. h1 . . _Lr_(b) lino-/Z) 561? da:- (Arctanh —Arctan0)-2 Hence, by(27.1), theintegral (a)exists andconverges to1r/2. Wecan therefore write w /.1 1r w+1“_§' Thefollowing additional information willalsobeneeded. -—IZ (mnm (®ﬁm%T=0,Hs>Q25-000 —lZ @hm%;=QiM>Q2-000 -—lZ @umK-=ciu>c:4:-we 3 (d)lim:z:”e"” =0,ifs>0,nreal. 294- Ornnxrons ANDLumen Tnmsronus Chapter 5 Finally weshall need thefollowing theorem, which westate without proof. Theorem 27.12. Iftheimproper integral (27.121) /0e"“’f(a:) dx, 0§:1:<ac, converges foravalue ofs=so,thenitconverges forevery s>so. Theintegral (27.l21), ifitexists, isafunction ofs.Itiscalled the Laplace transform off(z) andiswritten asL[f(a:)]. Hence thefollowing deﬁnition: Deﬁnition 27.13. Letf(x) bedeﬁned ontheinterval I:0§at<co. Then theLaplace transform off(x) isdeﬁned by (27-14) L[f(¢)l =F(8)=I¢_"f(w) 11$,o where itisassumed thatf(z)isafunction forwhich theintegral onthe right exists forsome value ofs. Example 27.15. Find theLaplace transform ofthefunction f(z) =1, atgO. Solution. By(27.14), with f(z) =1, (a) L[l] =F(s) =/Q)e_"'d:c 1» -=lim/ie”"da:h—>w O =lim(—1e"")ll--ma 8 0 . —-e_'h 1) =£933. +E = ifs>0[by(27.113)(a)]. Example 27.151. Find theLaplace transform ofthefunction f(z) =:8, :0g0,where :2isthefunction deﬁned byy=ac,see(6.15). Solution. By(27.14), with f(z) =5:, (a) L[:2] =F(s) =/0:ve_" da: h =lim/:ee"“’ dxll—vw 0 Lesson 27B Pnormmns orTHELumen Tnmsrolm 295 Et»<»E-atr._... 8820 =(—-+Q= ifs>0[by(27.113)(@) and(b)]. LESSON 27B. Properties oftheLaplace Transform. Theorem 27.16. IftheLaplace transform off1(:c) converges fors>s1 andtheLaplace transform off2(x) converges fors>s2,thenforsgreater ‘than thelarger ofs1ands2, (27-17) Llclfl +¢2f2l =¢1Llf1l -1-02Llf2l» where c1andC2areconstants, i.e.,theLaplace transformation isalinear operator. Proof. By(27.14) andthehypothesis-of thetheorem, (3) 61Llf1l =61/ti) e_“f1($) din» 8>81» C2L[f2] =Cgifo 6_’zf2(Il3)d1li, 8>82. Hence byTheorem 27.12, both oftheabove integrals exist foralls>s1 ands2.Therefore fors>s1ands2, (bl ¢1Llf1l+ 62Llf2l =/0¢_"¢1J'1(Il>) div-1'L6_“¢2f2($) div =/Qe_”l¢1f1 +¢2f2l d$=Llcifi +6212]- Byrepeated application of(27.17), itcanbeproved that (27.1.71) L[C1f1 +Ogfg +'--+6,,f,,] =¢1Llf1l +62Llf2l —|—-''—|—6»Llfnl- Westate thefollowing theorem without proof. Theorem 27.172. Iff1(:e) andf2(:t) areeach continuous functions ofat andf1=f2,W"Llfil =Llf2l- (ﬁmvmeltl ifl/[f1] =Life], “"df1,f2¢"'¢ eachcontinuous functions ofre,thenf1=f2. 296 Ornmvrons ANDLxrmcn Tnxnsronms Chapter 5 Deﬁnition 27.18. IfFistheLaplace transform ofacontinuous func- tionf,i.e.,if Llfl=F, then theinverse Laplace transform ofF,written asL'1[F], isf,i.e., L‘1[F] =f. Theinverse Laplace transform, inother words, recovers thecontinuous function fwhen Fisgiven. InExample 27.151 weshowed thatL[:i:]=1/s2. Hence thecontinuous function which istheinverse transform of1/s2 is:2,i.e.,L'1[1/s2] =5:. Theorem 27.19. Theinverse Laplace transformation isalinear opera- tor,i.e., (27.191) L'1[c1F1 +c2F2] =c1L‘1[F1] +c2L‘1[F2]. Proof. Let (3) F1=Llfil, F2=Llfzli where flandjgarecontinuous functions. Therefore byDeﬁnition 27.18, (b) L_1lFll =fi, L_1[F2] =1'2- By(27.17) (C) Llclfl +62f2l =°1Llf1l -1-62Llf2l, which, by(a),canbewritten as ((1) Ll¢1f1 -1'Czfzl =CF1-1'62F2- ByDeﬁnition 27.18, weobtain from (d), (9) L_1l61F1 +62F2l =¢1f1 '1'6212- Hence by(b),(e)becomes (f) L-1l61F1 +¢2F2l =¢1L_llF1l -1-¢2L_1lF2l- LESSON 27C. Solution ofaLinear Equation with Constant Co- efficients byMeans ofaLaplace Transform. The method weare about todescribe forsolving thelinear equation (27-2) 11»?/(“(11) +an-11/("'”(¢) +''-+<11!/(Iv) +1101/=f($). where ao,a1,-~-,anareconstants andan#50,isknown bythename of theLaplace transform method. Asthename suggests, itattains its objective bytransforming onefunction intoanother. However, unlike the differential operator, theLaplace transform accomplishes thetransforma- Lesson 27C SOLUTION BYMmns or.1Lumen Tmnsronm 297 tion bymeans oftheintegral in(27.14). Weremark that theLaplace method hasoneadvantage over theother methods thus farstudied for solving (27.2): itwillimmediately give aparticular solution of(27.2) satisfying given initial conditions. Bymultiplying (27.2) bye‘“’ andintegrating theresult from zero to inﬁnity, weobtain (27-21) A¢""[a»..y‘"’ +a.._1v‘"'" +---+aiy’+aov]dw =/Q) e_'zf(x)dx1 3>30: which isequivalent to (27.22) an/0 6_“’y(") dx-1-a,,_1];) e_“’y("_1)d:c +--- +a1‘/Z) e_”y’ dz+ao’/Q) e_"’y dx=/(.)e_”f(:c) dx,s>so. By(27.14), wenotethateach integral in(27.22) isaLaplace transform. Hence theequation canbewritten as (27-23) a..L[v‘"’] +<1,--11/[v"""] +-~-+a1L[z/'1 +<loLl!/l =Llf(f'=)l. 8>80- Comment 27.24. Equation (27.23) iseasily obtained from (27.2). Insert anLafter each constant coefficient in(27.2) and place brackets around y(z).7/(2), '''.1/")(1=) andf(16)- Ournexttaskistoevaluate L[y‘"’]. By(27.14), (27241) L[y‘">] =[0e"‘y‘"’ dx. Ifn=0,(27.241) becomes (remuember, nisanorder, notanexponent) (27.25) L[y] =/0e'"y dz, s>so. Ifn=1,(27.241) becomes Q (27.26) L[y’] =/‘e_“’y’ dx, s>so.o Integrating (27.26) byparts, weobtain with_u =e""’, dv=y’dx, (27.27) L[y']=Ilim[6_”y((l;)]'6 +810e'“ydx =lim[e_"'y(h) —y(0)] -1-s’/Qe“”y dx,s>so.ll»-no 298 Ornnzrrons ANDLumen TRANSFORMS Chapter 5 Wenow make anadditional assumption that y(z), which isthesolution weseek, isafunction suchthat* (27.28) lime—"y<*>(x) =0,It=0,1,2,---,1»-1,8>so.1-QM Then (27.27) becomes, with thehelp of(27.25), (27-29) L[v’]=-2/(0) +sL[v]- Ifn=2,(27.241) becomes Q (27.3) L[y"] =/0e_'”y" dx, s>so. Theintegral in(27.3) canbeevaluated bytwosuccessive integrations by parts. However aneasier method istomake useof(27.29). Ifinitwe replace ybyy’,weobtain (27.301) L[y”] =—y’(0) +sL[y']. By(27.29), wecanwrite (27.301) as (27-31) L11/"1 =-1/(0) +8(—1/(0) +sL[yl) =821117] —ly’(0)+sy(0)]- Similarly, by(27.31), (27-311) Ll?/"'1 =82Lly’] —[v"(0) +sy'(0)l- By(27.29), (27.311) becomes <27-32> L[v”’]=s*(—v<0> +an/1)—[1/"(0)+ 81/(0)1=s3L[v] —[1/’(0) +82/(0) +8211(0)]- Andingeneral itcanbeshown that (27-33) Llyml =8”L[1Jl —[y("'"(0) +81/"”2)(0) +'" +s""y'(0) +s""v(0)l- Hence, by(27.33), wecannow write (27.23) as (27-4) <1»s"L[v] —a»lv‘"""(0) +sy"“”(0) +--- +s""’1/(0) +3"“:/(0)1 +a.._1s”“L[v] —<1»-ilz/‘”"”(0) +81/‘""“’(0) +'-- +s"‘3v'(0) +8"";/(0)1 +<12s"L[vl -—wzlv'(0) +sv(0)]+aisL[yl —1117(0) +aoL[yl =I/[f(t)] ‘See D.V.Widder, Advanced Calculus, Prentice-Hall (1961), forproof that thesolu- tion y(:c), obtained bytheLaplace transform method, isafunction which satisﬁes (27.28). Lesson 27C SoLU'r1oN BYMnxns orALumen Tmnsronn 299 Collecting coeﬁicients ofliketerms, (27.40) becomes (27-41) fans"+a.._is"“‘ +~--+@252+ais+coll-[2/1 —la..s""‘ +a.._1s"_’ +---+1125‘+a1ly(0) —[a..s"'2 +a.._18"_“ +---+was+a2]y'(0) —[ans+a.._1ly("_2)(0) —(1-y<"_"(0) =L[f(w)]- Examine (27.41) carefully. L[y]istheLaplace transform ofthesolution y(z) weseek, butwhich wedonotknow asyet; y(0), y’(0), ---,y("_”(0) areconstants given byinitial conditions; L[f(x)] istheLaplace transform ofthefunction f(z)which appears inthegiven linear equation (27.2). Just asthere areintegral tables, there aretables ofLaplace transforms which willgive L[f(:z:)]. ByDeﬁnition 27.13, L[f(x)] isafunction ofs. Now look again at(27.41). Bysolving itforL[y], theright side ofthe resulting equation willbewholly afunction ofs.Letuscallthisfunction G(s). The problem ofﬁnding theparticular solution y(z) isthus reduced tooneofhunting inthetables forthat continuous function ywhose Laplace transform isG(s), i.e., L[y] =G(s); y=L"1[G(s)]. Inshort, theLaplace transform method haschanged theoriginal differential equa- tion involving derivatives, toanalgebraic equation involving afunction ofs. Note. Iftheinitial conditions give thevalues ofy,y’,y",---,y("'1) at:1:=xo960,itisalways possible totranslate theaxes byletting 2:=It+zco,sothat E=0when re==xo.Thegiven differential equation canthen besolved interms ofIt,andItreplaced afterwards by1:——mo. Insolving linear equations bytheLaplace transform method, wemay useeither thetwoequations (27.23) and(27.33), ortheequivalent equa- tion (27.41). Comment 27.42. Iff(z) =0,then by(27.14) (27.43) L[0] =’/7e_”0 dx=0.0 Example 27.431. Usethemethod ofLaplace transforms tosolve (2) 1/’+2v=0. forwhich y(0) =2. Solution. Method 1.Byuseof(27.41). Comparing (a)with (27.2), weseethat n=1,a1=1,ao=2.By(27.43), L[0]=0.Hence (a) becomes, with theaidofthegiven initial condition y(0)=2and(27.41), (11) (8+2)Lly] —(1)(2) =0- 300 OPERATORS ANDLumen Tmnsronus Chapter 5 Therefore <c> Lo]= Referring toatable ofLaplace transforms (there isashort oneattheend ofLesson 27D) weﬁnd, see(27.82), (d) L[e_2”] =-81% ; therefore L[2e”2'] =51% - Hence by(c),(d)andTheorem 27.172, (6) y=26'“, which istherequired solution. Method 2.Byuseof(27.23) and(27.33). Asnoted incomment 27.24, we canwrite (a)as (1) L[y'l+2L(y) =L10]- By(27.33), ordirectly from (27.29), (f)becomes (S) SL1!/l -y(0)+2L1!/l =L10]- Solving (g)forL[y] andnoting from (27.43) that L[0] =0,andfrom the initial conditions thaty(0)=2,weobtain <11) Lo]= which isthesame as(c)above. Example 27.44. Usethemethod ofLaplace transforms tosolve (2) v”+21/’+v=1. forwhich y(0)=2,y'(0) =-2. Solution. Comparing (a)with (27.2), weseethat n=2,a2=1, a1=2,ao=1.Hence by(27.41), (a)canbewritten as (b) (82+2s+1)I-[y] —(8+2)y(0) -1/(0) =Lill- InExample 27.15, wefound thatL[1]=1/s.Substituting thisvalue and theinitial conditions in(b),andthen solving forL[y], weobtain __2s2+2s+1_l 1 1 ‘°> Llyl-—-<?-W-;’“-.-t—1*<T-r'1>—2' Lesson 27C SOLUTION BYMmns orALumen Tmnsronn 301 Referring toatable ofLaplace transforms weﬁnd,see(27.8), (27.82) and (27.83), 1 _ 1 _, 1L[1.]=-gr L[81]==-g-iii» L[1li8 Hence by(27.171), -1 _, 1 1 1 (9) Ll1+6 —a18l='s'-l'P,%1—(_?,'_"1)—2.' Therefore by(c),(e)andTheorem 27.172, (f) y=1+e‘”-—me”. Alternate Method ofSolution. Asnoted inComment 27.24, wecan write (a)as (3) L[7"] +2Ll7'] +L17]=L11]- By(27.33), ordirectly from (27.31) and(27.29), (g)becomes (h) 821717] —7’(0)—87(0)—27(0)+2-SL171 +L17]=L11]. which simpliﬁes to(b)above. Example 27.45. Solve (2) 7"+37’+27=12¢”. forwhich y(0) =1;y’(0) =-1. Solution. By(27.41), (a)canbewritten as (b) (82+3s+2)Ll7l -(8+3)7(0) —7'(0) =Ll12@2"l =12Ll¢2"l- From atable ofLaplace transforms, weﬁnd [see(27.82)], L[e2‘] = 1/(s—2).Substituting thisvalue andtheinitial conditions in(b),we obtain _ 8”+s _3_3 1_ (°)L171‘ (s+2)(s+1)(s—2) _8+2 8+1+8-2 Referring toatable ofLaplace transforms weﬁnd,see(27.82), -22: __ 1 —:c __ 1 22:__ 1 _(d) L[e ]-——--8+2» L[e ]--—-—-8+1: L[e]-i-s_2 Hence by(27.171), (‘*1 L13”—3””+“"1=£727.-.1-ii+%'Therefore by(c)and(e) (f) y=3e‘2‘ —3e" +ea‘. 302 Ornmvrons moLumen TRANSFORMS Chapter 5 Alternate Method ofSolution. Asnoted inComment 27.24, wecan write (a)as (s) Ll7”] +3Ll7’] +2L[7l =12Ll¢2'l- By(27.31) and(27.29), (g)becomes (11) 8211171 —7'(0) -87(0) —37(0) +3sL[7l +2L[7l =12L[¢2”l. which reduces to(b)above. LESSON 27D. Construction ofaTable ofLaplace Transforms. Inthislesson, weshallﬁndtheLaplace transforms ofafewsimple functions. Q ll (27.5) L[lt] =Le"'lc dz=Itlim‘/Le_”dx h—vuo O —:eh __—h =1¢1im[l”’-1] =k1im[i-+1)ll-no 3 0 ll-Mn 8 8 =ifs>0[by(27.11s)(-1)]. us ll (27.51) L[x"] =‘/Qe_“‘x" dx=,1im/0e“"a:" dz ' I3 -lime — 2—_(-—a:" nz:”‘1 n(n—1):c”_2 h—m1 8 8 _ _nlx_nl " 818 sﬂ-I“-I 0 !=55, s>0,n=1,2,--- [Ifs>0,then by(27.113), e""h" —>0ash——>oo,andwhen x=0, each term excepting thelastequals zero.] a_ an_ -_ on(F4): _. e(a—8)¢]h (27.52) L[e']-foe ”e“‘da:-Le dz-hh_12°[——a_8° =L1im(e‘“-'>" -1)=-1_. ifs>a.aW8h—>un 8_a By(18.84) and(27.52) (27.53) L[sinax]=L _1(_1_._m)"278-17. 8+7“ _1(i')__<=.__._2i s2+a9 _s2+a2 Lesson 27D Consrnuerrou orATAn1.n orLumen TnANsro1ms 303 Many theorems exist which willaidinthecomputations oftheLaplace transforms ofmore complicated functions. Unfortimately wecannot enter intoadetailed study ofthem. However, toshow youthepower ofthese theorems, weshall state below arelatively simple one,anduseittoﬁnd theLaplace transforms offunctions which otherwise would bemore diffi- culttoobtain. Theorem 27.6. If Q (27.61) F(s) =L[f(x)] ='/Qe"”f(:c) dz, s>so, then Q (27.62) F’(s)=-—L[:cf(:z:)] =-‘/Z) e_":cf(:t) dz, s>so, F”(8) =L[w2f(w)l =[0¢'“1v2f(w) dw.8>80» F‘"’(8) =(-1)"L[w"f(w)l =(-1)"[0 ¢'"¢"f(w) dw,8>80- Note. Each Laplace transform in(27.62) canbeobtained bydiffer- entiating theprevious function ofswith respect tos.Thetheorem in effect states thatifF(s) =L[f(x)], then onecanﬁndtheLaplace trans- form ofa:f(:::) bydifferentiating —F(s); ofa:2f(:c) bydifferentiating F(s) twice, etc. Example 27.63. Compute (a) 1.L[xsinax], 2.L[:c2sinax]. Solution. By(27.53) (b) Lon0.1]=§,_;ﬁ =17(8). Hence by(b)andTheorem 27.6, (c) Lnsinax]=-17(8) =2“(82+112)’' and (22_2 (<1) Lu”sinax]=F"(s)= Another theorem, called theFaltung theorem, which ishelpful in evaluating integrals andonewhich weshall prove, isthefollowing. Theorem 27.7. If (27-71) F(-*1)=I-[f(t)] andG(8)=I-17(7)], 304 O1>nnA'ro1ts ANDLumen TRANSFORMS Chapter 5 then 1 z (27.72) LU;fa—t)g(t)at]=LUOf(t)g(¢ -oat] =I-[f(t)] -L[7(w)] =F(8)-G(s)- Proof. Let u=z—t.Therefore with zconstant, du=—dt; u=0whent =z,andu=zwhent =0.Making these substitutions intheﬁrst integral in(27.72), itbecomes O 1 (27.73) Llt/Z —f(u)g(z —u)du]=L[/0 f(u)g(z —u)du]. Inthesecond integral of(27.73), replace thedummy variable ofintegra- tionubyt.Theresult isthesecond integral in(27.72). Hence wehave proved theﬁrstequality in(27.72). By(27.14), (27.74) Ln:/T f(z—t)g(t) dt]=/we e_"'[/;zof(z —-t)g(t) dt]dz.-0 z-. - =‘/Z0 [/i°e""‘f(z —t)g(t) dt]dz. InFig.27.741, theshaded part indicates theregion inthe(z,t) plane over which theintegrations onthe right of(27.74) take place. Theﬁrst integration from t=0tot=zwill ,9 givetheareaofthevertical rectangle it" ofwidth dz.Thesecond integration 11“ from z=0toz=oowill give thet (°°r°°) -Im- . areaoftheentire shaded region. . Letusnow invert theorder ofin- I tegration, i.e.,letusintegrate ﬁrst _ over dzandthen over dt.Aﬁrst in- "Mn" ’ tegration, seeFig.27.741, from z=t (0,0) xtoz=oo,willgive thearea ofthe horizontal rectangle ofwidth dt.A Figure 27.741 second integration from t=0to t=oowillgivetheareaoftheentire shaded region. Hence wecanwrite (27.74) as (27.75) LU;f(z-t)g(t)dt]=[':o[/:1 e“"f(z -t)g(t)dz]dt =/:0 g(t) e"'f(z —t)dz]dt. Lesson 27D CoNs'rnUc'r1oN orATunn orLumen Tmmsronms 305 Inthelastintegral of(27.75), letw =z-—t.Then dw=dz(remember tisaconstant inthisintegration) f(z—t)=f(w) ande"“‘=e_"“’+”. A1sow=0whenz= t,andw—-> oowhenz—> oo.Makingall these substitutions inthisintegral andthen changing w,thedummy variable of integration, back tore,weobtain (27.76) L[/io f(z—~t)g(t) dt]=/Q g(t)[Lia e_”e_”f(z) dz]dt =[0¢_“9(1) df/0 e_“f($) div =Le_'”g(z) dz/L)e"”f(z) dz =I-17(2)] -L[f(w)l- By(27.71), thelastexpression ontheright of(27.76) isG(s)F(s). Example 27.77. UseTheorem 27.7toshow that.. slO s 8 ,, __ alb! ,,(8.) 1.lo (23-'l)tbdt-— $ +6-+1, I a>—1,b>—1,z>0. _,,,,_ a!b! _2.fo(1 1)¢d1_~i-(a+b+,), Solution. Let (b) f(z)=2". 7(2)=2"- Then (¢) f(w—t)=(w-1)“. v(l)=t"- By(27.72)I (<1) Ll/0 f(Iv—09(7)<11]=L[f(fv)l '1217(1)]- Substituting (b)and(c)in(d),weobtain (e) LUO(1-¢)"1'*a¢] =Lo“)-Lo‘). From atable ofLaplace transforms, weﬁnd[see(27.81)] ., a! b! !b!<0 1.1-1-Ll-"1=@-;.T=;,‘1.,T,-.5 = “"+°+‘l- 306 Ornnxrons ANDLumen TnANsrom.1s Chapter 5 Replacing theright sideof(e)byitsvalue in(f),wehave byTheorem 27.172, 2 ., a!b!(Q) /ti) (IE—t)lbdt= $a+b+1. Toprove (a)2,setz=1in(g). Remark. Thefunctions in(a)areknown asbeta functions. Short Table ofLaplace Transforms Iff(z) = Then L[_f(z)] =F(s) = (27.8) lc £31s>0 1(27.81) 1” s>0,n=1,2,--- (27.82) e“ 3-3;, s>a na l Z6: l 8>(l,1'l,=1,2,-.. . G (27.84) S111 01$ E? 8 COS GI ﬁ , 2asZS111 0-I F 2 2S'-G 23COS (11 az- b(27.88) eS111bz us__(1)2+b2 8—G (8-<1)”+52 (27-9) Af(=v—07(1)dt I-Lf(@)l -1117(2)] =F(8)G(8)(27.89) e“cosbz LESSON 27E. The Gamma Function. Bymeans ofafunction called thegamma function, itispossible togive meaning tothefactorial function n!—-ordinarily deﬁned onlyforpositive integers—when nisany number except anegative integer. Formulas such as(27.81) and (27.83) willthen have meaning when n=0or1}or-—§or2%,etc.Wedigress momentarily, therefore, togiveyouthose essentials ofthegamma func- tionwhich weshall need forourpresent andfuture purposes. Lesson 27E Tm: GAMMA FUNCTION 307 Deﬁnition 27.91. The gamma function ofk,Written asI‘(k), is deﬁned bytheimproper integral (27.92) I‘(k)=[0a:'°‘1e"d:c, k>0. IfIc=1,(27.92) becomes Q (27.93) I‘(l) =/ie_'da: =lim[—e_"]'(§ =1. O —vco Integration of(27.92) byparts gives, with u=e", dv=x"_1 dx, .e"a:" " 1an1,(27.94) I‘(k) =hm -T 0—|—E01:e'”dz, k>0.h—»n By(27.113), theﬁrstterm intheright of(27.94) approaches zero as h—->oo,andiszerowhen :1:=0.Thesecond term by(27.92) equals I‘(Ic+1)/k.Substituting these values in(27.94), weobtain (27.95) rug)=%I‘(k+1),k>0. Hence by(27.95) (27.96) I‘(k +1)=kI‘(k), k>O. By(27.93), I‘(1) =1.Therefore by(27.96), when (27961) =1;r(2)=1r(1)= =11=2;r(s)=2r(2)= =21 =3;r(4)=sr(s)= =31=4;r(5)=411(4)= =41 ??'??'?§"??' »#C.OlOr-1CON)»-Il0)-1 1- Andingeneral, when lc=n,where nisapositive integer, (27.97) I‘(n—|—1)=nl By(27.95) (27971) I‘(k)= It¢0. In(27.971) replace Icbyk+1.There results __I‘(k+2) _(27.972) rot+1)_---k+1,Io¢1. Now substitute in(27.971) thevalue of1‘(k+1)asgiven (27.972). We thus obtain __I‘(k+2) _(27973) r(k)_WT), k¢0,1. 308 OPERATORS ANDLumen Tnmsronms Chapter 5 Ifin(27.972) wereplace Icbyk+1,there results _1‘(k+3) _(27974) I‘(k+ 2)_k+2,k9'52. Now substitute (27.974) in(27.973). There results (27975) rut)= k¢0,-1,-2. Ingeneral itcanbeshown that (27.99) I‘(k)= PU“+"l7<>(k+1)(lc+2)"'(k+n—1)' lc#0,—1,—2,---,—(n—1). By(27.971) and (27.98) wecanextend thedeﬁnition ofI‘(k), which, by(27.92), wasdeﬁned only forlc>0,toinclude negative values ofk, provided Ic750,—l,—2,---.Forexample, ifIa=—-§,thenby(27.971), wecandeﬁne (27.981) I‘(-—§) =—2I‘(§). If,therefore, weknow thevalue ofI‘(§), wethen alsoknow thevalue of I‘(—~}). Andifweknow thevalue ofI‘(—§), wethen alsoknow thevalue ofI‘(—§-). Forby(27.971), wecandeﬁne (27.982) I‘(——§-) =—§I‘(—§), etc. Tables ofvalues ofthegamma function exist justasthey doforsin1:, logac,ore’.From such tables weﬁnd, forexample, I‘(§) =\/Tr. Hence by(27.981) and(27.982), (27983) r(-5) =-2‘/7:. = =Mr! etc- Bymeans ofthegamma function anditsextended deﬁnition, weare thus abletogivemeaning tonlwhen nisanynumber excepting anega- tiveinteger. For,by(27.97), tables ofvalues ofthegamma function, and (27.96), (27.984) 0!=P(1) =1. (-2)! =F(2)=\/7% <—%>!=r<—1)= —2~/Ir.<291=rs)=%I‘(%)=M- Lesson 27E Tun Gmm FUNCTION 309 InFig.27.985, wehave drawn agraph ofthegamma function. I'(k) I ,_ U3- 2»- 1.. -4 -é -2 -i i 2 é 4 k ~-1 ~-2 ~-a (W.-. l Figure 27.995 Example 27.986. Compute (a) L[:c'1/2]. Solution. By(27.14) w (5) L[:v‘1/2] =/0e'“”x”1/zdz. Youwillﬁndinatable ofintegrals, thatthevalue oftheimproper in- tegralontherightof(b)isr(5)/\/5. By(27.984), F(2)=(-5)!=\/T’. Hence (b)becomes \/_ _ 1r (0) Lli1/2]=E‘ Example 27.987. Compute (a) L[a:"‘1/2], n=1,2,3,---. Solution. By(c)ofExample 27.986 (b) L[:z:_1/2] =\/7rs"1/2 =F(s). 310 OPERATORS ANDLsrmca Tnmsronms Chapter 5 Therefore by(b)and(27.62) ofTheorem 27.6, (9) And ingeneral (d)L1x""’”1 =I»lw"(w"”)] =<—1>"1*"’<s) = s—<»+1/2>, ,,=1 Example 27.988. Compute (a) L[x"” 20”]. Solution. By(27.14) (b) Ll Theintegral ontheright of(b)hasthevalue I‘(§)/\/s—a.By(27984)L1x"*1=L1w<r"*>1 =—F'<s)=#8-3", L[x“”1=L[w”(w""’)] =F"(s)=‘”Q#'s-5", Ll:/9”]=L[w”(w'”’)] =—F"'<s) = 8"”- W x—1/2eu:c] =I e-(a-a.):cx—lI2 0 I‘(:}) =(—=})! =\/Tr. Hence (b)becomes (c) L[z’1/2e"] =i, 5>a.V8-G72) 3! Additional Table ofLaplace Transforms Iff(z) = Then L[f(:c)] =F(s) = (27.99) (27.991) (27.992) (27.993) (27.994) (27.995)x-1/2 L; \/§ x,,_.1/2 1-3-5.. ~—1)\/; 8._(..+1/2), n=1'2_ Vs-a ! zneaz i_7L___ (8_ a)n+ll 5 ‘n. I $11!. >"-1 -1/2 az \/;I 8 ii r1t>—l<,,_,,,-<»+1/9, ,,_.27 Lesson 27—Exercise 311 Following exactly themethod outlined inExample 27.987, youwillﬁnd (27.989) L[x"‘1/20”] =TL'3'5'' _1)‘/7' (s—a)_‘"+1'2), 8>a,n=1,2,3,---. Remark. Wesee,therefore, thatwiththegamma function, theLaplace transforms ofac"anda:"e“‘ asgiven in(27.81) and(27.83) nowalsohave meaning when n>—1. EXERCISE 27 1.Evaluate those ofthefollowing improper integrals which converge. wrcda: w2:cda: “dz(8),/; ' (b)/0 (0)/Q?‘ °°a (‘D/.ﬁ§'Find theLaplace transform ofeachofthefunctions 2-5. 2.sinhax,:cg sha:c,:vg0. 4.aa:+b,a:;0. 5-f(z)='6, With theaidofTheorem 27.6, ﬁndtheLaplace transform ofeach ofthe following functions.no >-z-x/\i—l |vv\-°s»-8 6.itsinhax. 7.2:coshax. 8.2:cosax. 9.me". 10.x2e". 11.x3e“. Usethemethod ofLaplace transforms toﬁndasolution ofeach ofthe following differential equations satisfying thegiven initial conditions. §@12.y’— =0,9(0) =1. 13.y’—- =e‘,y(0) =1. 14-9'+9=9*’.9(0)=1-15-9”+49’+49=0.9(0)=1.1/(0) =16-9"-29'+59=0,9(0)=2.9'(0)=4- 17-39"'+ 59”+ 9’—9=0.9(0)=0.1/(0) =1.9”(0)=-1-18-9"-—59’—69=ea‘.9(0)=2.9’(0)=1~19.9”-—y’-—29=5sin 2:,y(0) =1,1/(0) =-1. 20.y'”—2y"+y’=2e"+ 22:, y(0) =0,1/(0) =0,1/'(0) =0. 21-9”+9'+ 9=9’.9(0)=1.9’(0)=1-22.Evaluate each ofthefollowing gamma functions. (*1)P(6)- (b)F(7)- (0)P(—5/2)- (d)I'(5/2)- (8)F(7/2)- 23.Evaluate each ofthefollowing factorial functions. (*1)(*2)! (b)(-9)! (0)(2)! (<1)(iv)! 24-.Verify thecorrectness of(27.989). Hint.Seethesolution ofExample 27.987.g-1 312 OPERATORS ANDLxrmcn Tmnsronms chaptgf 5 25.InExample 27.986, weused thefactthat @ /(.)x"1I2e"“’d:t =PG‘)/\/E, s>0. Prove it.Hint. Make thesubstitution u=sx.Then use(27.92). 26.Prove, ingeneral, that x"e"“da: = . n>——1,s >0.s /.Seehintin25. ANSWERS 27 1.(a)Diverges. (b)1. (c)Diverges. (d)<}. 2. 1S>G. 3. 18>G. 4¢ 18>0. 5Z=_<_1_—L'> 6L. , ' 8 . .(82 _ a2)2 '(82 .._ a2)2 s2—a2 1 2 3! ' ' l0. ' 1l. ' 12.y=e’. 13.y=(a:—|- 1)e’. 14.y=(a:+ 1)e"’. 15.1/=(1—|—3a:)e'2'. 16.y=(2cos2:1:+sin2:z)e". 17.y=%eds+ — e"’I 18. y___.Heﬂz +%_g_e—z __112,e3z_ 19.1/=§e2‘—|—-Q-e"" —--Qsin 2:+1}cos2:. 20.y=:c2+4:2:+4+ (xz—4)e". 21.y=0-1/2 (cos g2:+$ sin92:)—|—x2—22:. 22-<9)5!<1»)6!<9—1';~/? (3)iv? (el‘iv?-23-(*1)§\/F (b)—1’s\/7 (9)is/1 (d)1s"‘\/1r- Chapter 6 Problems Leading toLinear Differential Equations ofOrder Two Inthischapter weshall consider themotion ofaparticle whose equa- tion ofmotion satisﬁes adifferential equation oftheform 2 <9) %+21§+9.2.»=79>, where f(t)isacontinuous function oftdeﬁned onaninterval I,and r andwoarepositive constants. Iff(t)E0,then (a)simpliﬁes to dz d(b) -(-E;+WI: +wozx =0. Ifr=0,then (a)becomes d2 <9 75;‘+9.29=79>. Finally ifboth r=0andf(t)E0,then (a)becomes d2 32% +(00233 = Inthelessons which follow weshall name anddiscuss each ofthese four important equations (a),(b),(c),and(d). LESSON 28. Undamped Motion. LESSON 28A. Free Undamped Motion. (Simple Harmonic Mo- tion.) Many objects have anatural vibratory motion, oscillating back 313 314- Pnontnms LEADING T0Lmrxn EQUATIONS orOsman Two Chapter 6 andforth about aﬁxed point ofequilibrium. Aparticle oscillating inthis manner inamedium inwhich theresistance ordamping factor isneg- ligible issaid toexecute free umlamped motion, more commonly called simple harmonic motion. Two examples areadisplaced helical spring and apendulum. There arevarious ways ofdeﬁning thismotion. Ours will bethefollowing. Deﬁnition 28.1. Aparticle willbesaidtoexecute simple harmonic motion ifitsequation ofmotion satisﬁes adifferential equation ofthe formd2 (28.11) i+5.0’).=0, where weisapositive constant, andxgives theposition oftheparticle as afunction ofthetime t. Bythemethod ofLesson 20D, you canverify that thesolution of (28.11) is (28.12) 2:=clcoswot+c2sinwot. By(20.57) wecanalsowrite thesolution (28.12) intheform (28.13) 2:=(/15 sin(wot+3)=csin(wot+3), orby(20.58) with +8replacing -6,intheform (28.14) :1:=VEW cos(wot+8)=ccos(wot+6). Hence anequivalent deﬁnition ofsimple harmonic motion isthefollowing. Deﬁnition 28.141. Simple harmonic motion isthemotion ofa particle whose position atasafunction ofthetime tisgiven byanyofthe equations (28.12), (28.13), (28.14). Example 28.15. Aparticle moving onastraight lineisattracted to theorigin byaforce F.Iftheforce ofattraction isproportional tothe distance a:oftheparticle from theorigin, show that theparticle will execute simple harmonic motion. Describe themotion. Solution. Byhypothesis (a) F=—kx, where k>0isaproportionality constant. Thenegative signisnecessary because when theparticle isatP1,seeFig. 28.16, :1:ispositive and F acts inanegative direction; when itisatP2,2:isnegative andFacts in apositive direction. Fand 2:,therefore, always have opposite signs. Lesson 28A Fnsa Unnxnrnn MOTION. (SIMPLE HARMONIC MOTION) 315 Hence by(16.1) with sreplaced by1:,(a)becomes <1’ 71% 1.(b) F=mﬁ=—k:c, a;=——7;:z:. +--> l ' I l I -C P2 0 P1 c Figure 28.16 Since Itand marepositive constants, wemay in(b)replace It/m bya new constant wo2. Thedifferential equation ofmotion (b)isthen d2 (c) 5+<».=@=0. which isthesame as(28.11). ByDeﬁnition 28.1, therefore, theparticle executes simple harmonic motion. Description oftheMotion. Thesolution of(c),by(28.13), is (d) at=csin(wot+6). Differentiation of(d)gives (e) %=v=cwocos(wot—|—8), where visthevelocity oftheparticle. Since thevalue ofthesine ofan angle liesbetween —1and 1,weseefrom (d)that § Hence the particle cannever gobeyond thepoints cand——c,Fig.28.16. These points arethus themaximum displacements oftheparticle from the origin O.When =|c|,wehave, by(d), [sin(wot+6)|=1,which implies that cos(wot+6)=0.Therefore, when =|c|thevelocity v, by(e),iszero. Wehave thus shown that thevelocity oftheparticle at theendpoints ic,iszero. When as=0,i.e.,when theparticle isattheorigin, then, by(d), sin(wot+6)=0,from which itfollows that [cos(wot+6)]=1.Since thevalue ofthecosine ofanangle liesbetween —1and1,wesee,by(e), that theparticle reaches itsmaximum speed |v|=|cwo|, when itisat theorigin. Forvalues of:1:between 0and Ia],youcanverify, bymeans of(d)and (e),that thespeed oftheparticle willbebetween 0andits maximum value |cwo|; itsspeed increasing astheparticle goes from c, where itsspeed iszero, totheorigin, where itsspeed ismaximum. After crossing theorigin, itsspeed decreases until itsvelocity isagain zero at -c. The particle willnow move intheother direction——remember the force which isdirected toward theorigin never ceases toactonthepar- 316 Pnonnnms Lnxnmo T0LINEAR EQUATIONS orOsman Two Chapter 6 ticle-—its speed increasing until itreaches itsmaximum speed at0,and then decreasing until itiszero again atc.Theparticle thus oscillates back andforth, moving inanendless cycle from cto—ctoc. Example 28.17. Aparticle Pmoves onthecircumference ofacircle ofradius cwith angular velocity woradians persecond. Call Qthepoint ofprojection ofPonadiameter ofthecircle. Show that thepoint Q executes simple harmonic motion. Solution. SeeFig.28.18. Let Po(a:o,yo) betheposition oftheparticle attime t=0, P(z,y) betheposition oftheparticle atanylater time t, 8bethecentral angle formed byadiameter, taken tobetheasaxis, andtheradius OPo. P(x.y) C b Po(xo»yo) )3 L Q(x10) Q0(x0r0) xaxis Figure 28.18 Intime tthecentral angle through which theparticle hasrotated iswot. (For example ifwo=2radians/sec, then inQsecond, Phasswept outa central angle equal to1radian ;attheendof2seconds thecentral angle swept outis4radians; attheendoftseconds, thecentral angle swept outis2tradians.) The projections onthediameter ofthepositions ofthe particle att=0and t=t,arerespectively, Qo(:co,0) andQ(:z:,0). Itis evident from Fig.28.18 that (28.2) :0=ccos(wot+8), anequation which expresses theposition oftheparticle's projection Qon adiameter, asafunction ofthetime t.Acomparison of(28.2) with (28.14) shows that they arealike. Hence thepoint Qexecutes simple harmonic motion. The alternate form (28.13) may beobtained bymeasuring theinitial angle 8from theyaxis instead offrom the:1:axis, Fig. 28.22. From the ﬁgure weseethat (28.21) a:=ccos +8+wot)) =—csin(wot+8) '5Csin (wot+8). Lesson 28B Dsrmmons: SIMPLE Hnmomc Morron 317 yaxis Po(1o.J’o) P(=.y) \12Q(I.0) Qo(Io.0) 0(0.0) ""55 Figure 28.22 Comment 28.23. Description ofthe Motion. AsPmoves once around thecircle, itsprojection Q(Fig. 28.18) moves tooneextremity of thediameter, changes direction andgoes totheother extremity, changes direction again and returns toitsoriginal position headed inthesame starting direction. Note thesimilarity ofQ’smotion tothemotion ofthe particle inExample 28.15. LESSON 28B. Deﬁnitions inConnection with Simple Harmonic Motion. For convenience, werecopy thethree solutions of(28.11). Each, byDeﬁnition 28.141, istheequation ofmotion ofaparticle exe- cuting simple harmonic motion. (28.24) 2:=clcoswot+C2sinwot. (28.25) :1:=csin(wot+8). (28.26) at=ccos(wot+8). Ifyouwillrefer toFigs. 28.16 and28.18, andreread the“description of themotion” paragraphs ofExample 28.15 and Comment 28.23, the deﬁnitions which follow willbemore meaningful toyou. Deﬁnition 28.3. The center 0ofthelinesegment onwhich thepar- ticle moves back andforth iscalled theequilibrium position ofthe particle’s motion. Deﬁnition 28.31. Theabsolute value oftheconstant cin(28.25) and in(28.26) iscalled theamplitude ofthemotion. Itisthefarthest dis- placement oftheparticle from itsequilibrium position. Comment 28.32. By(28.13), |c|=\/cl’ -l-cf. Iftherefore thesolu- tion of(28.11) iswritten intheform (28.24), then theamplitude ofthe motion is\/cl? +C22. 318 Pnonmams LEADING roLINEAR EQUATIONS orOnnan Two Chapter 6 Deﬁnition 28.33. The constant 8in(28.25) andin(28.26) iscalled thephase orthephase angle ofx.[Note that when t=0in(28.26), :1:=ccos 8,andfrom Fig.28.18, weobserve that ccos 8istheinitial position (:1:o,0) oftheparticle’s projection.] If,in(28.26), t=0,21r/wo, 41r/wo, ---,2n1r/wo, naninteger, thenfor each t,:1;=ccos 8.This means that intime 21r/wo, theparticle has made onecomplete revolution around thecircle andisback atitsstarting position, headed inthesame starting direction. Orequivalently, thepar- ticlehasmade onecomplete oscillation along adiameter oralinesegment andisback atitsstarting position headed inthesame starting direction. Hence thefollowing deﬁnition. Deﬁnition 28.34. Theconstant (28.35) T=3’-’.W0 where woistheconstant in(28.24), (28.25), and (28.26), iscalled the period ofthemotion. Itisthetime ittakes theparticle tomake one complete oscillation about itsequilibrium position. Iftheconstant wois theangular velocity ofaparticle moving onthecircumference ofacircle, then theperiod isthetime required fortheparticle tomake onecom- plete revolution orequivalently thetime ittakes theparticle’s projection onthediameter tomake onecomplete oscillation about thecenter ofthe circle. Forexample, ifaparticle makes twocomplete revolutions, orequiva- lently twocomplete oscillations, inonesecond, i.e.,ifwo=41rrad/sec, then itsperiod by(28.35) is§second; ifitmakes 1ofarevolution orofa complete oscillation inonesecond, i.e.,ifwo=1r/2rad/sec, then itsperiod is4seconds. Note that thereciprocal ofTgives thenumber ofcomplete revolutions oroscillations made bytheparticle inonesecond. Forexample when theperiod T,which isthetime required tomake one complete revolution oroscillation, is=},theparticle makes two complete oscilla- tions inonesecond; when theperiod T=4,theparticle makes 1ofa complete oscillation inonesecond. Hence thefollowing deﬁnition. Deﬁnition 28.36. Theconstant _1_m (28.37) v-T-2,” iscalled thenatural (undamped) frequency ofthemotion. Itgives thenumber ofcomplete revolutions orcycles made bytheparticle ina unit oftime, orequivalently itgives thenumber ofcomplete oscillations made bytheparticle onalinesegment inaunit oftime. Lesson 28B DErmrr1oNs: SIMPLE Hxnmomc MOTION 319 Analternative deﬁnition ofnatural undamped frequency that isoften used isthefollowing. Deﬁnition 28.38. The constant woin(28.24) to(28.26), (which can bethought ofastheangular velocity ofaparticle moving onthecircum- ference ofacircle) isalsocalled thenatural (undamped) frequency ofthemotion. Comment 28.39. Weshall use9when wewish toexpress thefre- quency ofthemotion incycles perunitoftime andusewowhen weWish toexpress itinradians perunit oftime. Intheexample after Deﬁnition 28.34 where wo=41rrad/sec, thefrequency 11,by(28.37), isthus 2 cycles/sec; thefrequency wois41rradians/sec. Comment 28.4. ASummary. Thedifferential equation ofmotion of aparticle executing simple harmonic motion hastheform d2:tW +(00222 = Itssolutions may bewritten inanyofthefollowing forms. as=c1coswot+C2sinwot, x=ccos (wot+8), :1:=csin (wot+8). They aretheequations ofmotion ofaparticle executing simple harmonic motion. Themotion canbelooked atasthemotion oftheprojection ona diameter, ofaparticle moving with angular velocity woonthecircum- ference ofacircle ofradius c.Oritmay belooked atasthemotion ofa particle attracted toanorigin byaforce which isproportional tothedis- tance oftheparticle from theorigin. Theparticle moves back andforth forever across itsequilibrium position. The farthest position reached by theparticle from itsequilibrium position isgiven byc.Itsspeed atc and —ciszero; attheorigin itsspeed isgreatest. The time required to make acomplete oscillation isT=21r/wo; thenumber ofcomplete oscilla- tions inaunitoftime is1/T. Example 28.5. Aparticle executes simple harmonic motion. The natural (undamped) frequency ofthemotion is4rad/sec. Iftheobject starts from theequilibrium position with avelocity of4ft/sec, ﬁnd: 1.The equation ofmotion oftheobject. 2.Theamplitude ofthemotion. 3.Thephase angle. 4.The period ofthemotion. 5.Thefrequency ofthemotion incycles persecond. 320 PROBLEMS LEADING roLmsxn Eqwmons orORDER Two Chapter 6 Solution. Since theparticle executes simple harmonic motion, its equation ofmotion, by(28.26), is d .(a) 2:=ccos(wot+8), v=75=——woc sm(wot+8). The frequency of4rad/sec implies, byDeﬁnition 28.38, wo=4.Hence (a)becomes (b) :1:=ccos (4t+8), v=——4c sin(4t+8). Theinitial conditions aret=0,x=0,v=4.Substituting these values inthetwoequations in(b),weobtain (c) 0=ccos8, 4=—4c sin8. Since theamplitude c;-60,weﬁnd from theﬁrst equation in(c),8= :b1r/2. With thisvalue of8,thesecond equation in(c)gives c==F1. Hence theequation ofmotion (b)becomes (d) :1:=cos(4t—— or2:=—-cos (41—|— 1 which istheanswer tol.Theanswers totheremaining questions are: 2.Amplitude ofthemotion, byDeﬁnition 28.31, =1foot. 3.Thephase angle, byDeﬁnition 28.33, =-—1r/2 radians. 4.Theperiod ofthemotion, byDeﬁnition 28.34, =1r/2seconds. 5.Thefrequency ofthemotion incycles persecond, byDeﬁnition 28.36, equals 2/1rcps. Example 28.51. Aparticle executes simple harmonic motion. The amplitude and period ofthemotion are6feet and 1r/4seconds respec- tively. Find thevelocity oftheparticle asitcrosses thepoint :1:=-3feet. Solution. Since theparticle executes simple harmonic motion, its equation ofmotion, by(28.26), is (a) 2:=ccos (wot+8). The amplitude of6feetimplies, byDeﬁnition 28.31, that c=6(or——6). The period of1r/4seconds implies, byDeﬁnition 28.34, that 21r 1r6-8 ——Z1(.00 —— Hence theequation ofmotion (a)becomes, using c=6, d .(c) :1:=6cos(8t+8), i=——48s1n(8t+ 8). Lessons 28A and B—Exercise 321 When :2:=-3,weﬁndfrom theﬁrstequation in(c), (d) cos(8t+8)=—%, 8t+8=120° or240°, andfrom thesecond equation in(c) (e) v=%=-48 sin(120° or240°) =-48(4)»/3) ==F24\/3 ft/sec. The plus signistobeused ifthebody ismoving inthepositive direction, theminus signifthebody ismoving inthenegative direction. EXERCISE 28A AND B l.Verify theaccuracy ofeach ofthesolutions of(28.11) asgiven in(28.12), (28.13), and(28.14). 2.Ifaballwere dropped inaholebored through thecenter oftheearth, it would beattracted toward thecenter with aforce directly proportional to thedistance ofthebody from thecenter. Find: (a)theequation ofmotion, (b)theamplitude ofthemotion, (c)theperiod, and(d)thefrequency ofthe motion. Nora. This problem wasoriginally included inExercise 16A, 40. 3.Aparticle executes simple harmonic motion. Itsperiod is2-1rsec. Ifthe particle starts from theposition z=-4ftwith avelocity of4ft/sec, ﬁnd: (a)theequation ofmotion, (b)theamplitude ofthemotion, (c)thefrequency ofthemotion, (d)thephase angle, and(e)thetime when theparticle ﬁrst crosses theequilibrium position. 4.Aparticle executes simple harmonic motion. Att=0,itsvelocity iszero anditis5ftfrom theequilibrium position. Att=1,itsvelocity isagain 0 anditsposition isagain 5ft. (a)Find itsposition andvelocity asfunctions oftime. (b)Find itsfrequency andamplitude. (c)When andwithwhat velocity doesitﬁrstcross theequilibrium position? 5.Aparticle executes simple harmonic motion. Attheendofeach ifsec,it passes through theequilibrium position with avelocity of:1=8ft/sec. (a)Find itsequation ofmotion. (b)Find theperiod, frequency andamplitude ofthemotion. 6.Aparticle executes simple harmonic motion. Itsamplitude is10ftandits frequency 2cps. (a)Find itsequation ofmotion. (b)With what velocity does itpassthe5-ftmark? 7.Aparticle executes simple harmonic motion. Itsperiod isa1rsecandits velocity att=0,when itcrosses thepoint 2:=2:1,is=l=v1. Find itsequa- tionofmotion. 8.Aparticle executes simple harmonic motion. Itsfrequency is3cps. At t=1sec,itis3ftfrom theequilibrium position andmoving withavelocity of6ft/sec. Finds itsequation ofmotion. 9.Aparticle executes simple harmonic motion. When itis2ftfrom itsequilib- rium position, itsvelocity is6ft/sec; when itis3ft,itsvelocity is4ft/sec. Find theperiod ofitsmotion, alsoitsfrequency. 322 PROBLEMS Lsxnmc T0LINEAR Eouxrxons orORDER Two Chapter 6 10.Aparticle moves onthecircumference ofacircle ofradius 6ftwith angular velocity 11-/3rad/sec. Att=0,itmakes acentral angle of150°with aﬁxed diameter, taken asthe2:axis. (a)Find theequation ofmotion ofitsprojection onthediameter. (b)With what velocity does itsprojection cross thecenter ofthecircle? (c)What areitsperiod, amplitude, frequency, phase angle? (d)Where onthediameter istheparticle’s projection att=0? ll.Aparticle moves onthecircumference ofacircle ofradius 4ft.Itsprojection onadiameter, taken asthe2:axis,hasaperiod of2sec.At:2:=4,itsvelocity iszero. Find theequation ofmotion oftheprojection. 12.Aparticle weighing 8lbmoves onastraight line. Itisattracted totheorigin byaforce Fthatisproportional totheparticle’s distance from theorigin. If thisforce is6lbatadistance of-2ft,ﬁndthenatural frequency ofthe system. 13.Aparticle ofmass mmoving inastraight lineisrepelled from theorigin 0 byaforce F.Iftheforce isproportional tothedistance oftheparticle from 0,ﬁndtheposition oftheparticle asafunction oftime. NOTE. Theparticle nolonger executes simple harmonic motion. 14.Ifatt=0thevelocity oftheparticle ofproblem 13iszeroanditis10ft from theorigin, ﬁndtheposition andvelocity oftheparticle asfunctions of time. 15.Att=0thevelocity oftheparticle ofproblem 13is—a\/h ft/sec andit isafeetfrom theorigin. Iftherepelling force hasthevalue kz,ﬁndthe position oftheparticle asafunction oftime. Show that ifm<1,the particle willnever reach theorigin; thatifm=1,theparticle willapproach theorigin butnever reach it. 16.Aparticle executes simple harmonic motion. Att=0,itis-5ftfrom its equilibrium position, itsvelocity is6ft/sec anditsacceleration is10ft/sec2. Find itsequation ofmotion anditsamplitude. 17.Let|c|betheamplitude ofaparticle executing simple harmonic motion. We proved inthetext, seedescription ofthemotion, Example 28.15, that the particle haszerospeed atthepoints =l=candmaximum speed attheequilib- rium position 1=0. (a)Prove thattheparticle hasmaximum acceleration at2:=icandhas zeroacceleration at:2:=0.Hint. Take thesecond derivative ofthe equation ofmotion (28.25). (b)What isthemagnitude ofthemaximum velocity andofthemaximum acceleration? (c)If9,,isthemaximum velocity anda,,,isthemaximum acceleration ofa particle, show that itsperiod isT=21rv,,,/a.,,. anditsamplitude is A=v...’/aw 18.Two points AandBonthesurface oftheearth areconnected byafriction- less,straight tube. Aparticle placed atAisattracted toward thecenter of theearth with aforce directly proportional tothedistance oftheparticle from thecenter. (a)Show thattheparticle executes simple harmonic motion inside thetube. Hint. Take theorigin atthecenter ofthetube andlet2:bethedistance oftheparticle from thecenter atanytime t.Callbthedistance ofthe center ofthetube from thecenter oftheearth andRtheradius ofthe Lesson 28C Exammssz SIMPLE Hnmonrc Morrow 323 earth. Take a.diameter oftheearth parallel tothetube asapolar axis andlet(r,0)bethepolar coordinates oftheparticle atanytime t.Then show that thecomponent offorce moving theparticle isFcos0 and that cos0 =1:/r. Initial conditions aret=0,2:=\/R5 --5?, dz/dt =0. (b)Show thattheequation ofmotion is :2:=\/R2 —b2cos\/lc/mt, where kisaproportionality constant andmisthemass oftheparticle. (c)Show thatforagiven mass, thetimetogetfromAtoBisthesame for anytwopoints AandBonthesurface oftheearth. Hint.Show that theperiod ofthemotion isaconstant. ANSWERS 28A AND B 2.(a)1=(4000)(5280) cos(t/812) inft. (b)4000 mi. (c)85min. (d)0.7oscillation/hr. 3.(a)2:=4\/5 sin(t— . (b)4\/5 ft. (c)1rad/sec or ops. (d)—1r/4. (e)1r/4sec. 4.(a):2:=5cos(8-rt); v=-401’ sin(8-rt). (b)81rrad/sec, 5ft. (c)-11;sec,—40'l' ft/sec. 5.(a)x=2sin t)- (b)-3-sec,41'/3 rad/sec, or§cps,6/1rft. 6.(a)a:=10sin(41rt—|— 8). (b)=|;201r\/3' ft/sec. 7.x=\/1:12 +a2v12/4 sin(2t/a +8),where sin6=:01/\/1:12 +a2v12/4. B.2:=V9+1/12 sin(61rt+8),where 8=Arcsin(31/V 912+1). 9.1rsec,2rad/sec or1/1rcps. 10.(9.)as=6cos(rt/3 +51'/6). (b);l:21r f_t/sec. (c)6sec,6ft,1}cps,51/6 rad. (d)—3\/ 3ftfrom center. ll.2:=4cos (rt). 12.x/E rad/sec or\/§/1r cps. 13.:0:=c1e\/'75‘ +c2e"w/"/—'"'), where kisaproportionality constant. 14-.zc=5(ev'm‘+ e"1/"/—""), v=5\/U1? (ew/"/—’"‘ —e"~/W‘). 15.x=g[<1~mum‘ +<1+\/"fin-""’_'"‘1. 16.1=3\/§sin\/5t ——5cos\/§t;\/Eft. 17.(b)v=cwo,a=0:.-:02, where cistheamplitude andweisthefrequency of themotion inradians perunitoftime. LESSON 28C. Examples ofParticles Executing Simple Harmonic Motion. Harmonic Oscillators. Adynamical system which vibrates with simple harmonic motion iscalled aharmonic oscillator. Below we give twoexamples ofharmonic oscillators. 324- Pnonnnms Lmnmo 'roLmsan Eqoxrrons orOnnrm Two Chapter6 Example A. The Motion ofaParticle Attached toanElastic Helical Spring. Hooke’s Law. Theunstretched natural length ofan elastic helical spring islofeet,Fig.28.6(a). Aweight wpounds isattached toitandbrought torest, Fig.28(b). Because ofthestretch duetothe lo lo lo l I _y=0 yaQ Equilibrium position ofthe L4.y wring __Y_ (11) (b) (c) Figure 28.6 attached weight, aforce ortension iscreated inthespring which triesto restore thespring toitsoriginal unstretched, natural length. ByHooke’s law, thisupward force ofthespring isproportional tothedistance l, bywhich thespring hasbeen stretched. Hence, (28.61) Theupward force ofthespring =kl, where Ic>Oisaproportionality constant, called thespring constant orthestiﬁness coefﬁcient ofthespring. Thedownward force acting on thespring istheweight wpounds that isattached toit.Ifthespring isonthesurface oftheearth, then w=mg,where misthemass inslugs oftheattached weight andgistheacceleration duetothegravitational force oftheearth infeetpersecond persecond. Since thespring isinequilib- rium, theupward force must equal thedownward force. Hence by(28.61) (28.62) kl=mg. Lety=0betheequilibrium position ofthespring with theweight w pounds attached toit.Ifthespring with thisweight attached isnow stretched anadditional distance y,Fig.28.6(c), then thefollowing forces willbeacting onthespring. Lesson 28C Examrmcs: SIMPLE HABMONIC MOTION 325 1.Anupward force duetothetension ofthespring which, byHooke’s law, isnow k(l+y). 2.Adownward force duetotheweight wpounds attached tothespring which isequal tomg. ByNewton’s second lawofmotion, thenetforce acting onasystem is equal tothemass ofthesystem times itsacceleration. Hence, with the positive direction taken asdownward, 2 (28.621) m% =mg—k(l—|—1/)=my—-kl-Icy. By(28.62), (28.621) simpliﬁes to dz dz lc(28.63) m$=—ky, -‘Ti’+51/=0, which isthedifferential equation ofmotion ofanelastic helical spring. Since ithasthesame form as(28.11), wenow know that adisplaced helical spring with weight attached and with noresistance will execute simple harmonic motion about the equilibrium position y=0.The solution of(28.63) is (28.64) y=ccos (Vic/mt+ 6)ory=csin (\/lc/mt +8). Example 28.65. A5-pound body attached toanelastic helical spring stretches it4inches. After itcomes torest, itisstretched anadditional 6inches andreleased. Find itsequation ofmotion, period, frequency, and amplitude. Take thesecond foraunitoftime. Solution. Since 5pounds stretches thespring 4inches =§foot, we have by(28.62), with mg=5,Z=§, (a) lg=5, k=15, which gives thestiffness coeﬂicient ofthespring. With g=32,themass moftheattached body is355.Therefore thedifferential equation of motion, by(28.63), is 5dzy day Itssolution, by(28.64), orbysolving itindependently, is (c) y= ccos (\/96t+6), %== -—\/96csin (\/96t+6). Theinitial conditions aret=0,y=Q,v=dy/dt =0.Inserting these 326 Pnonu-zus LEADING T0LINEAR EQUATIONS orOnnnn Two Chapter 6 values inboth equations of(c),weobtain (d) Q=ccos8, 0=——\/96 csin8. Since c,theamplitude, cannot equal 0,thesecond equation in(d)implies 8=0.With thisvalue of6,theﬁrst equation in(d)gives c=Q.Sub- stituting these values intheﬁrstequation of(c),weﬁndfortheequation ofmotion (e) y=%cos\/961. The period ofthemotion, byDeﬁnition 28.34, istherefore 21r/\/ 96 seconds; thefrequency, byDeﬁnition 28.36, is\/96/21r cps,orbyDeﬁni- tion28.38, \/96rad/sec; theamplitude, byDeﬁnition 28.31, isQfoot. Example 28.66. Abody attached toanelastic helical spring executes simple harmonic motion. The natural (undamped) frequency ofthemo- tion is2cpsanditsamplitude is1foot. Find thevelocity ofthebody as itpasses thepoint y=%foot. Solution. Since thebody attached tothehelical spring isexecuting simple harmonic motion, itsequation ofmotion by(28.64) is (a) y=ccos (\/Ic/mt —|—6). The amplitude of1foot implies, byDeﬁnition 28.31, that c=1.The frequency of2cpsimplies, byDeﬁnition 28.36, that 2=\/lc/m/21r, or \/lc/m =41r. Substituting these values in(a)itbecomes d .(b) y=cos(41rt —|—8), v=7?:=—41rsm (41rt +6). When y=§,weobtain from theﬁrstequation in(b) (c) =1‘=cos(41rt +6), 41rl+6=60°,300°. Hence, by(c),when y=Q, (d) sin(41rl +8)=:b Substituting (d)inthesecond equation of(b),weobtain (e) v=5%=:b2\/8 1r, which isthevelocity ofthebody asitpasses thepoint y=1}.The plus sign istobeused ifthebody ismoving downward; theminus sign ifitis moving upward. Lesson 28C EXAMPLES! SIMPLE Hanuomc MOTION 327 Example B.The Motion ofaSimple Pendulum. Adisplaced simple pendulum oflength l,with weight w=mgattached (Fig. 28.7) willexecute anoscillatory motion. Iftheangle ofswing isvery small, then asweshall show, themotion willclosely approximate asimple har- monic motion. Theeffective force Fwhich moves thependulum isthecomponent ofthe weight acting inadirection tangent tothearcofswing. From Fig. 28.7, 0ispositive iftheweight wis K totheright ofO;negative iftotheleftofO. 0 Q-d0/dtispositive iftheweightwmoves counterclockwise; I negative ifitmoves clockwise. Fispositive ifitactsinadirection tomove thependulum ,, counterclockwise; negative 8=1 “’mg ifthedirection isclockwise. 0 0 Note thatwhen 6ispositive, __ . Fisnegative; when 0is F’mgSm0 mg negative, Fispositive. mgsin9 Figure R-7 weseethat thevalue ofthiscomponent ismgsin0,where 0istheangle ofswing and mand ghave their usual meanings. (The tension inthe string andthecomponent oftheweight inthedirection ofthestring do notaffect themotion.) Therefore byNewton's second law (28.71) F=mg =-mg sin0, %=-gsin0. The distance sover which theweight moves along thearcis (28.72) s=I0,thereforeg =l%- Since v=ds/dt, weobtain from thesecond equation in(28.72), 2do dv d6 Substituting in(28.71) thevalue ofdv/dt asgiven in(28.73), wehave 2 (28.74) 1%:+gsino=0. Equation (28.74) does notasyethave theform (28.11) ofsimple harmonic motion. However, weknow from thecalculus that sin0=0—03/3! + 328 Pnonnnus LEADING T0LINEAR EQUATIONS orOnnen Two Chapter 6 0°/5! —---.Wemaytherefore write (28.74) as 1120 go’ g05(28.75) tum-‘l-Q9--5!--i-'3!---"'=(). Hence, forapendulum swinging through asmall angle, itisnotunreason- abletoassume thattheerror made inlinearizing theequation bydropping terms in03andhigher powers of0willalsobesmall. Hence (28.75) becomes d’a d20 g(28.76) 17,-t;+go_0, E-+7a_0. Butthislastequation isnowthedifferential equation ofaparticle exe- cuting simple harmonic motion. Thesolution of(28.76) is (28.77) 0=ccos(\/g/ll +6), where cistheamplitude of0inradians. Example 28.771. Asimple pendulum whose length is5ftswings with anamplitude of316radian. Find theperiod ofthependulum andits velocity asitcrosses theequilibrium position, i.e.,theposition when 0=0. Solution. The given amplitude of116radian implies that c=-116in (28.77). Therefore, withl =5andg=32,(28.77) becomes (a) 0=-115cos(\/32/5t +8), ﬁg’=-,1,“/32/'5' sin(\/32/5 7+a). Theperiod ofthependulum, byDeﬁnition 28.34, is (b) T=21“/5/32 ={V10 sec. When 0=0,wehave by(a), (c) 0=cos(V32/5 t+8), which implies that ((1) sin(V32/5 t—|—5)=:|:1. Substituting (d)inthesecond equation of(a),there results <e> ff}:=d=.*.-/W =iam- Bytheﬁrstequation in(28.73) and(e),weobtain with l=5, (f) v=5(a={;\/FJ) =:|=§\/1T)ft/sec, Lesson 28C—Exercise 329 which isthevelocity ofthependulum asitcrosses thecentral position. Theplussignistobeused iftheweight ismoving counterclockwise; the minus signifitismoving clockwise. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10EXERCISE 28C Verify theaccuracy ofthesolution of(28.63) asgiven in(28.64). Verify theaccuracy ofthesolution of(28.76) asgiven in(28.77). Solve (28.63) foryasafunction oftifatt=0,1/=yo,v-vo. 2 Show that thesum lgyz+g ,where yisthesolution of(28.63) as given in(28.64), isaconstant. Nora. Theﬁrstterm gives thepotential energy ofamass attached toahelical spring displaced adistance yfrom itsequilibrium position. Thesecond term gives itskinetic energy. Since thesum ofthetwoenergies isaconstant, amass oscillating onahelical spring with simple harmonic motion satisﬁes thelawoftheconservation ofenergy. Prove thelawoftheconservation ofenergy fortheundamped helical spring (seeproblem 4)bymultiplying (28.63) bydy/dt andthen integrating the resulting equation with respect tol.Forhint, seeanswer. A12-lb body attached toahelical spring stretches it6in.After itcome torest,itisstretched anadditional 4in.andreleased. Find theequation J motion ofthebody; alsotheperiod, frequency, andamplitude ofthemotion. With what velocity willitcross theequilibrium position? A10-lb body stretches aspring 3in.After itcomes torest,itisstretched an additional 6in.andreleased. (a)Find theposition andvelocity ofthebody attheendof1sec. (b)When andwith what velocity willtheobject ﬁrstpasstheequilibrium position? (c)What isthevelocity ofthebody when itis-3in.from equilibrium? (d)When willitsvelocity be2ft/sec andwhere willitbeatthatinstant? (e)What aretheperiod, frequency, andamplitude ofthemotion? Aspring isstretched 8in.bya16-lb weight. Theweight isthen removed anda24-lb weight attached. After thesystem isbrought torest,thespring isstretched anadditional 10in.andreleased with adownward velocity of 6ft/sec. (a)Find theequation ofmotion, period, frequency, amplitude andphase angle. (b)Answer thesame questions iftheweight were given anupward velocity of6ft/sec instead ofadownward one. Aheavy rubber band ofnatural length loisstretched 2ftwhen aweight W isattached toit.Find theequation ofmotion andthemaximum stretch of theband, ifatthepoint lotheweight isgiven adownward velocity of6ft/sec. Assume therubber band obeys Hooke’s law. Thespring constant ofahelical spring is27.When aweight of96lbisat- tached, itvibrates with anamplitude of3/2ft. (a)What areitsperiod andfrequency? (b)With what velocity does itpassthepoint y=—lft? 330 Pnosnsus L1-zxmnc T0LINEAR EQUATIONS orORDER Two Chapter 6 ll.Apiece ofsteel wireisstretched 8in.when a128-lb weight isattached toit. After itisbrought torest,itisdisplaced from itsequilibrium position. Find thefrequency ofitsmotion. Assume thewireobeys Hooke’s law. 12.Ahelical spring isstretched 2in.when a5-lbweight isattached toit.Ifthe 5-lbweight isremoved andaweight ofWlbattached, thespring oscillates with afrequency of6cps. Find theweight W. 13.When twosprings, suspended parallel toeach other, arecormected byabar attheir bottom, they actasifthey were onespring. Thespring constant of thesystem isthen equal tothesumofthespring constant ofeach spring. Assume aweight of8lbwillstretch onespring 2in.andtheother spring 3in. (a)What isthespring constant ofeach spring? (b)Ifthetwosprings aresuspended parallel toeach other andconnected byabarattheir bottom, what isthespring constant ofthesystem? (c)Ifaweight of8lbisattached tothesystem, brought torest, andthen released after being displaced 6in.,ﬁnd theamplitude, period, and frequency ofthemotion. 14.When twosprings areconnected inseries, i.e.,when onespring isattached totheendoftheother, thespring constant kofthesystem isgiven bythe formula 1 1 l i'E+E’ where k1andkgaretherespective spring constants ofeach spring. Assume thesprings ofproblem 13areattached inseries. (a)What isthespring constant ofthesystem? (b)Ifaweight of8lbisattached tothebottom spring, brought torest, displaced 6in,andthen released, ﬁndtheamplitude, period, andfre- quency ofthemotion. The solutions toproblems 15-17 willbefacilitated ifreference ismade tothesolution given inproblem 3. 15.Aspring isstretched lftbyaweight Wandbrought torest. Itisthen given adownward velocity ofvoft/sec. (a)Find thelowest point reached bytheweight andtheelapsed time. (b)Find theamplitude, frequency, andperiod ofthemotion. 16.Aspring isstretched lftbyaweight Wandbrought torest. Itisthen stretched anadditional aftandgiven adownward velocity vo. (a)Find theamplitude, frequency, andperiod ofthemotion. (b)Answer thesame questions iftheinitial velocity isvoupward instead of downward. 17.Ahelical spring hasaperiod of4secwhen a16-lb weight isattached. If the16-lb weight isremoved andaweight Wattached, thespring oscillates with aperiod of3sec. Find theweight W. Inproblems 18-28, itisassumed that 0,theangle apendulum makes with thevertical, i.e.,theangle thependulum makes with itsequilibrium position, issufﬁciently small sothat equations (28.76) and(28.77) are applicable. Remember that inthese problems angular velocity w=d0/dt, Lesson 28C—Exercise 331 andlinear velocity v=ld0/dt =lw,where listhelength ofthependulum from point ofsupport tothecenter ofmass ofthebob. Forpositive and negative directions, seeFig.28.7. 18.Asimple pendulum oflength lftisgiven anangular velocity ofwerad/sec from theposition 0=0e. (a)Find theposition ofthebobofthependulum asafunction oftime. (b)Find amplitude, period, frequency, andphase angle. (c)With what velocity, angular andlinear, does thependulum cross the equilibrium position? 19.Asimple pendulum oflength lftisreleased from theposition 0=0e. (a)Find theposition ofthependulum asafunction oftime. (b)Find amplitude, period, andfrequency. Compare results withproblem 18. (c)With what velocity, angular andlinear, does thependulum cross the equilibrium position? 20.Asimple pendulum oflength 2ftisreleased from theposition 0=112-rad attime t=0. (a)Find thevalue of0when l=1r/16. (b)When andwith what velocity, angular andlinear, does thependulum cross theequilibrium position‘? 21.Asimple pendulum oflength 6in.isgiven anangular velocity ofmagnitude Qrad/sec toward thevertical from theposition 0=-115rad. Find theequa- tionofmotion, amplitude, period, frequency, andphase angle. Hint. At t=0,w=-Qrad/sec. 22.Asimple pendulum oflength 2ftisgiven anangular velocity ofQrad/sec toward thevertical from theposition 0=——-1%; rad. Find theequation of motion andtheamplitude. 23.Solve problem 22,ifthependulum isgiven anangular velocity ofQrad/sec away from thevertical from theposition 0=315rad. 24.Asimple pendulum oflength lftisstarted bygiving itanangular velocity ofwerad/sec from itsequilibrium position. (a)Find theequation ofmotion. (b)Find thevalue of0andthetimewhen thependulum reaches itsmaximum displacement. 25.Aclock pendulum isregulated sothat 1secelapses each time itpasses its equilibrium position. (Itsperiod istherefore 2sec.) What isthelength of thependulum? 26.Itisdesired that aclock pendulum cross itsequilibrium position each second andhave anamplitude of315rad. What should itsangular velocity beasitcrosses theequilibrium position? Hint. Thelength ofthependulum hasalready been found in25,theequation ofmotion inproblem 24. 27.Theamplitude ofapendulum is0.05rad. What fraction ofitsperiod has elapsed when 0=0.025 rad? Forhint, seeanswer. 28.Aclock pendulum hasalength of6in.Each time itcrosses theequilibrium position, itmakes atick. How many ticks willitmake in1hr? 29.Inthependulum example inthetextandintheproblems above, theweight ofthewiresupporting themass wasconsidered negligible andtherefore ig- nored. If,however, thisweight isnotnegligible, then, instead of(28.71), 332 PROBLEMS LEADING T0LINEAR Eomrrons orORDER Two Chapter 6 thedifferential equation ofmotion ofthependulum andbobis (28.772) I-Z-2=—(mg sin6)l, where Iisthemoment ofinertia* ofthependulum andbobabout anaxis which passes through thepoint ofsuspension andisperpendicular tothe plane ofrotation ofthependulum; misthemass ofpendulum andbob; listhelength ofthependulum from thepoint ofsupport tothecenter of gravity ofpendulum andbob. [For thecase ofapendulum ofnegligible mass, themoment ofinertia Iofaparticle ofmass mattached atadistance Ifrom theaxisofrotation ismlz. Substitution ofthisvalue ofIin(28.772) willyield (28.74).] With theaidof(28.772), ﬁndtheequation ofmotion andtheperiod ofa pendulum consisting ofauniform rodoflength landswinging through a small angle. Hint. Themoment ofinertia Iofauniform rodoflength l andmass mabout anaxisthrough oneendisgiven by n I 2 I=lim2at.-“At. =/0Mat=%witha=m/l.1HW-_;‘_ Since therodisuniform, itsmass may beconsidered asconcentrated atits center, i.e.,itscenter ofgravity isatitsgeometric center. Inproblems 30-34, wenolonger assume that theangle ofswing is small. Hence wecannot, in(28.74), replace sin0by0. 30.Solve (28.74) ford0/dt with theinitial conditions l=0,0=0e,w=0. Hint. Multiply theequation by2110/dt andthen make useoftheidentity 2ﬂdz_182’.anatIdzdzAnswer is (28.78) w=d0/dt ==!=\/2g/IV cos0—cosBe. 31.Byintegrating (28.78), remember theinitial condition isl=0,0=0e, show that 0 (28.19) t(o)==i=\/z/2, x--4"l--;-0‘VCOB u—008 0 Since att=0,0=Be,w=0,one-half ofaperiod elapses when 0=—0e. Usethisfact, (28.79), andthefactthat cosuisaneven function, i.e., cos(-—u) =cosu,toshow that theperiod Tofthependulum isgiven by ,'0T z <1(28191) 5=\/ggL. “00\/COS ‘ll.—'COS () '0duT=2\/2-\/l/g‘/i ————-———-- °\/cosu-—cos0e ‘For deﬁnition ofmoment ofinertia, seeLesson 30M-B. Lesson 28C—Exercise 333 In(28.791), replace cosubyitsequal 1—2sin25,cos0ebyitsequal 1—2sin2%).Show thattheresulting equation is 90 (28.7911) T=2\/W/Q ‘\/S111 e —-SID u Finally in(28.791l), make thesubstitutions . .0. .0(28.79l2) S111lg=S111E0sin¢, Qcos 5du=singcos¢d¢. Show thatwhen u=0,dz=0;when u=0e,¢=1r/2andthat (28.7911) therefore becomes 1/2 (287913) T=4\/l/g/S. k=sin93-°V1—-ls?sin?¢ 2 Theintegral in(28.79l3) isknown asthecomplete elliptic integral ofthe ﬁrst kind. Itcannot beexpressed interms ofelementary functions, but itcanbeevaluated* bywriting theintegrand asapower series QBS a IQ»-I >502OQUI 3Qt-5 [O,“°;ta-(287914) (1-13S1112¢)“”2 =1+it’SiI12¢+—-Itsin4¢+---. Substituting theabove series in(28.7913) andcarrying outtheintegration, weobtain, since, when nisanevenpositive integer, 8,,4,d_,,=m<’*:_1l:, k2 1-3’.(281915) T=21r\/lE1+§+ 5-; lt+--~. which gives theactual period ofapendulum. By(28.77), theperiod ofapendulum, approximated byreplacing sin0 by0,is (28.7916) T=21$ ~ Assume theinitial displacement ofapendulum att=0,is0e=4°.Then by(28.7913), k=sin2°=0.0349 andI02=0.00122. Hence by(28.7915), themore exact period ofthependulum is (281917) T(exact) =2“/E (1+E +ww +-- =21r\/ﬁ(1.000305)set, andby(28.7916) theapproximate period is (28.7918) T(approx.) =21r\/I7; sec. ‘Atable ofvalues ofthisintegral canbefound, justasonecanﬁndatable ofvalues ofsin:1:;see,forexample, Pierce’s Tables. 334- Pnonnnms Lnnnmo 'roLINEAR EQUATIONS orORDER Two Chapter 6 By(28.79l7) and(28.7918), wesee,therefore, that (28.7919) T(exact) =1.000305T (approx.). 32.Suppose now, wedesign aclock with pendulum ofsuﬂicient length Iso thatitsperiod, using theapproximation formula (28.7918) is2sec,andits amplitude is4°.Therefore, by(28.7919), itsmore exact period is2.00061 sec. This means thatineach 2sectheclock isincorrect by0.00061 sec,itssecond hand hasmoved 2secwhereas itshould have moved 2.00061 sec;theclock isthus tooslow bythismuch. Calculate theerror intheclock attheend ofeach 24hr. Am. 26secslow. Solve (28.74) ford0/dt with theinitial conditions l=0,0=0,w=we. Seehintin30.Answer is d02 22g 2 29=(00 '—T(1—008 0)=(00 1—W (1—O05 0) ' 33.In(28.792), make thesubstitution (28.793) k2=4i and sin22=§(1—cos0).lwez Show thatattime t,remember att=0,0=0, o 0/2 (287931) 1(0)=410 '1“ E42'/i '14’ 34. (a) (b)we\/1 —I02sin?(u/2) "'°°V1—k2sin?upl Theintegral in(28.7931) iscalled anincomplete elliptic integral ofthe ﬁrst kind. Inthetext, wederived thedifferential equation forapendulum ofnegligible weight swinging along thearcofacircle. Theresulting equation (28.74) was notsimple harmonic. Assume now that theendofthependulum moves along acurve which isgiven bytheparametric equations a(20) —|—asin20, a—acos20,$= y= where aisapositive constant and0istheinclination ofthetangent tothe curve atapoint P,Fig.28.794. Hence, thedifferential equation ofmotion ofthependulum is miv=-mg sin0.dt Replacing vbyitsequal dc/dt, weobtain dzs .E=—gsin0, where sisthedistance along thecurve measured from thelowest position 0. Itcanbeshown that thedistance salong anarcofthecurve (a)from its lowest point iss=4asin0.Solve forsin0andsubstitute thisvalue in(b). Show thattheresulting equation isnowsimple harmonic. Solve forsasa function oft,ﬁndtheperiod ofthemotion andestablish thefactthat the period isaconstant. Lesson 28C—Exercise 335 The parametric equations (a)above aretheequations ofaninverted cycloid traced byapoint onacircle ofradius a.(When acircle rollsalong astraight line,thecurve traced byapoint Ponitscircumference, iscalled acycloid.) In(a)above, 20isthecentral angle made bylines drawn from thecenter ofthecircle, onevertical, theother toP.Itcanthen beproved that theinclination ofthetangent tothecycloid atPis0.If,therefore, youcandevise apendulum whose bobwillmove along thepath ofanin- verted cycloid, thependulum willswing insimple harmonic motion regard- lessoftheangle ofswing, anditsperiod willbeindependent ofitsamplitude. x=a(20)+a sin20 y=a -acos20 20 s=4a sin0X P 9 0 mgsin0=F 0 mg Figure 28.794 Archimedes’ principle states that abody partially ortotally sub- merged inaliquid isbuoyed upbyaforce equal totheweight ofthe liquid displaced. Forexample, ifa5-pound body ﬂoats onwater, then theweight ofthewater displaced is5pounds, i.e., V(inftsofwater displaced) X62.5 (weight ofaftaofwater) =5. The fact that thebody isresting orﬂoating onwater implies that the downward force due totheweight W=mgofthebody must bethe same astheupward buoyant force ofthewater, namely theweight of water displaced. Call y=0theequilibrium position ofthelower endofabody when it isﬂoating inaliquid. Ifthebody isnow depressed adistance yfrom its equilibrium position, anadditional upward force willactonthebody due totheadditional liquid displaced. Ifthebody’s cross-sectional area isA, then thevolume ofadditional liquid displaced isAyand theweight of thisadditional liquid displaced is(Ay)p, where pistheweight perunit volume oftheliquid. Since thisadditional upward force isthenetforce acting onthedepressed body, wehave, byNewton’s second lawofmo- tion, with positive direction downward, d2 (28.795) mwg =—-pAy. 336 Pnonmms Lmnmo T0Lmmn EQUATIONS orOnnnn Two Chapter 6 Aﬂoating body, therefore, when depressed from itsequilibrium position executes simple harmonic motion. With theaidof(28.795), solve thefollowing problems: 35.Abody ofmass mhascross-sectional areaAsqft.Itisdepressed yoftfrom itsequilibrium position inaliquid whose weight perftaispandreleased. (a)Show thatitsequation ofmotion is (28.796) y=yocosVpA/m t, where yisthedistance ofthebody from equilibrium. Hint. Att=0, y=yo,v=0.(b)Ifthebody isaright circular cylinder with vertical axis andradius r,show thatitsperiod is (28.79?) T=3,Tn 36.Acylindrical buoy ofradius 2ftandweighing 64lbﬂoats inwater with its axisvertical. Itisdepressed 1}ftandreleased. (a)Find itsequation ofmotion, amplitude, andperiod. (b)How farbelow thewater lineistheequilibrium position. 37.Acubic block ofwood weighing 96lbhascross-sectional area.4ft. Itisde- pressed slightly inaliquid which weighs 50lb/cu ftandreleased. What is itsperiod ofoscillation? 38.Acylindrical buoy ofradius 1ftﬂoats inwater with itsaxisvertical. Its period, when depressed andreleased, is4sec. What isitsweight? 39.Acubical block ofwood is2ftonaside. When depressed andreleased in water, itoscillates withaperiod of1sec;What isthespeciﬁc gravity ofthe wood? Hint. Use(28.796) toﬁndthemass mofthecubical block. Then ﬁnditsweight percubic foot. Then compare thisﬁgure with theweight ofa. cubic footofwater. 40.When acylinder with vertical axisisdepressed andthen released inwater, itvibrates with aperiod Tosec. When water isreplaced byanother liquid, itvibrates with aperiod of2T0sec. Find theweight oftheliquid percubic foot. Hint. Use(28.797) toﬁndp. 41.Abody whose cross-sectional areaisA,displaces, when inequilibrium, lft ofaliquid. Thebody isthendepressed andreleased. Show thatitsdiﬁeren- tialequation ofmotion is 1121/_0(28/798) W --i1/, where yisthedistance ofthebody from equilibrium attime t.Hint. In equilibrium, theweight ofthebody equals theweight ofthewater displaced. Theweight ofdisplaced water is(Al)p. Therefore Alp=mg.Solve forp andsubstitute in(28.795). 4-2.Solve (28.798). What istheperiod ofvibration‘? 43.Acylinder with vertical axesvibrates inwater with aperiod of2sec. How farfrom thesurface isitsequilibrium position? Hint. Make useofthe solution of(28.798) asfound inproblem 42. 4-4.Aspherical body ofradius risinequilibrium when halfofitissubmerged in aliquid. Itisgiven adisplacement andreleased. Find itsdifferential equa- tionofmotion and, ifthedisplacement issmall incomparison with the radius r.alsoﬁnditsapproximate period. Forhints seeanswer. Lesson 28C—Answers 337 3. 5. 6 7 9 10 11 12 13 14 15 16 17 18. 19ANSWERS 28C y=v0Vm/ksin(Vic/m t)+yocos(Vlc/m t) EV1/02+(m/k)vo2 $iI1(V 76/"Il+5), where 6=Arcsin—--—!mi—— - Vyo’+(M/k)vo2 Tointegrate theﬁrst term make thesubstitution u=dy/dt, du= (dzy/dig) dt. y=11;cos8t,1r/4sec,8rad/sec, or4/1rcps,§ft,=l:§-ft/sec. (a)0.16ft,-5.37 ft/sec. (b)\/2-ir/32 sec,-5.66 ft/sec, (c)=b2\/6 ft/sec, (d)0.31sec,-0.47 ft, (e)\/21:-/8 sec,8x/2 rad/sec, Qft. (1.)y=¥sin(4\/51)+gcos(4\/50Ela‘sin(4\/5¢+0.67),6\/5 -$5 sec,4\/2 rad/sec, V131/6r\/2 ft,0.67rad. (b)y=-%§ sin(4\/2 t)+gcos (4\/2 t);remaining answers arethe same asin(a). y==3-sin 4t—2cos4t.Amplitude is5/2ft.Maximum stretch is9/2ft. (a)21r/3 sec,3rad/sec or3/21‘ cps. (b)=i;§-\/5 ft/sec. 4x/§ rad/sec. 20§'T—5 lba _ _ (a)48and32. (b)80. (c)Qft,\/51r/20 sec,4\/5/1r cps. (a)19.2. (b)<}ft,21r/V 76.8sec,V76.8/21' cps. (a)y=v0Vm/k =vox/mft,t =gVm/k =gx/msec. (b)M/W rt,\/lc/m =\/g/lrad/sec, 2n/W sec. (a)A=Va2+Ive?/g, period andfrequency arethesame asin15. (b)Same answers asin(a). 91b. (11)9=90¢0$\/9/l¢+wo\/1/9$iI1\/0/llE\/902+ (l/9)wo2$i11(\/9/l¢+5), wherea=Arcsin(00/\/002 +(l/g)w02). 0»)\/002+(l/0)wo’rt,an/Wsec,<1/21»/W cps-6isgiven in(a). (c)w==l=V(g/D002 —|—mo?rad/sec, v==l=Vlg0o2 -I-lzwoz ft/sec. (a)0=00cosVg/lt. 1 (b)A=0°rad,T=21-\/W sec,v=5V9/I cps. (c)w=:1:00Vg/I rad/sec, v==l=00\/3 ft/sec. 338 Paonums LEADING T0Lmmn Eotmrrons orORDER Two Chapter 6 20. 21. 22. 23. 24. 25. 27. 28. 29. 34-. 36. 37. 42. 43. 44.(a)0=\/2/24 rad. (b)t=§(2n +1)1r,n=0,1,---,5:1}rad/sec, =!=§ft/sec. 0=-ﬁ5cos8t --173-sinstE\/E/so sin(st_4.15),\/so/so rt,1/4sec,4/1rcps,4.15rad. 0=-116cos4:+ﬁsin4:,A=\/iﬁ/40 ft. 0=-ﬁgcos 4:+isin4:,A=\/ET/40 rt. (a)0=om/T/Z sin(\/571z).(b)o=wqx/iﬁra.d,t =(1/2)\/Wm.32/1r2ft. 26.0.21rad/sec. T/6. Hint. Inthesolution given inproblem 18,00=0.05andtakewo=0. Find twhen 0=0.025. 9167. 0=clcosV3g/2l t+62sinV3g/2l t,T=21rV2l/3g sec. s=01sin(t/2)V g/a+62cos(t/2)Vg/a, T=41rVa/g. (a)y={rcos<5./5; 1),A=5rt,r=§\/1r/5sec. (b)o.osrt.0.77 sec. 38.2546 lb. 39.0.405. 4-0.One-fourth theweight of water. y=ccosVg/lt, T=21I'\/T/Igisec. g/1r2 ft=3}ftapprox. Letybethedisplaced distance ofthesphere from equilibrium. Thebuoyant force oftheliquid, therefore, isequal totheweight ofliquid displaced, i.e., itistheweight ofavolume ofliquid equal toone-half thevolume ofthe sphere minus thevolume ofasegment ofasphere ofheight r—1/.The volume ofasegment ofasphere is(1rh2/3)(3r —h),where ristheradius ofthesphere andhistheheight ofthesegment. And since thebody isin equilibrium when only one-half ofitissubmerged, theweight oftheliquid perunitvolume must be2p,where pistheweight perunitvolume ofthe sphere. Thedifferential equation motion is L51__2[32_(2)3].dtz_ 2 r 1' Ifthedisplacement gissmall incomparison with r,then itisreasonable to assume that theerror made inlinearizing theequation ‘bydropping (y/r)3 willalsobesmall. Theapproximate period istherefore T=21rV2r/3g. LESSON 28D. Forced Undamped Motion. Themotion ofaparticle ofmass mthatsatisﬁes adifferential equation oftheform 2 2 (28.8) 71%?+mm,=fa).-‘§_,,—§’+My=$0)where f(t)isaforcing function attached tothesystem andmoisdeﬁned in28.38, iscalled forced undamped motion, incontrast tothefreeun- damped motion (i.e., simple harmonic motion) when f(t) E0.Let us assume thattheforcing function f(t)=mFsin(wt+B)where Fisacon- stant. Then (28.8) becomes2 (28.81) ‘Z7?+wfy=Fsin(wt+B). Lesson 28D Foncan Unoxnrao Morrou 339 IfWesettheleftside of(28.81) equal tozero, and solve theresulting homogenous equation, weobtain thecomplementary function (28.82) yc=csin(wot+8). Aparticular solution y,,of(28.81) will then depend ontherelative values ofthenatural (undamped) frequency cooofthesystem and the impressed frequency woftheforcing function mFsin(wt+13). We shall treat each ofthetwopossibilities inthetwocases below. Case 1.w96wo.Ifw;£wo,then aparticular solution of(28.81) is F .y,,=0?‘? S111(wt + Hence, by(28.82) and(28.83), thegeneral solution of(28.81) is (28.84) y=csin(wt+5)+L811. (wt+B).° (.002—(.02 The motion ofthesystem isnow thesum oftwo separate and distinct motions, each ofwhich issimple harmonic. The displacement ordepar- tureoftheparticle from itsequilibrium position istherefore thesumof twoseparate (harmonic) displacements with respective amplitudes ofc andF/(w02 -wz). Themaximum value ofthedisplacement ordepar- ture, however, cannot exceed [cl+IF/(@002 —w2)|. Since allthese letters denote constants, thedisplacement ordeparture hasﬁnite magnitude. A motion inwhich thedisplacement ordeparture ofaparticle from its equilibrium position remains ﬁnite with time iscalled astable motion. If,however, woand warenearly alike, then ((002 -0:2)will besmall, andsince thisterm appears inthedenominator of(28.84), thedeparture ordisplacement oftheparticle from equilibrium will belarge, i.e.,the vibrations ofthesystem willbebig,andifsufficiently large, abreakdown ofthesystem may result. Thebehavior oftheparticle’s motion isagain inﬂuenced bytwomo- tions with different frequencies, thenatural (undamped) frequency wo and theforcing frequency w.Ifmo/to isarational number, say3,then themotion duetoy,willmake three revolutions while themotion dueto ypismaking only one. Therefore, during thetime interval 21r/w (the period ofonerevolution duetotheypmotion), themotion ofthesystem willbeerratic. However, attheendofthistime interval, theposition of theparticle duetothey,andy,motions willagain beitsstarting one, andthesystem willagain repeat itserratic behavior. The motion ofthe system willthus have anappearance somewhat likethat shown inFig. 340 PROBLEMS Lsxomo 'roLINEAR EQUATIONS orOaonn Two Chapter 6 28.85. If,however, wo/w isanirrational number, then themotion will nothave arepetitive pattern. t ¢=21 ,=21O (0 Figure 28.85 Case 2.w=wo. Ifw=wo,the differential equation ofmotion (28.81) becomes 2 (28.9) ?,§+wozy=Fsin(wot+8). Now, however, aterm intheyepart ofthesolution asgiven in(28.82), agrees, except forphase andconstant coeﬂicient, with thefunction onthe right of(28.9). Hence thetrial function y,,,byLesson 21A, Case 2,must beoftheform (28.91) y,,=Atsin(wot +)3)+Btcos(wot +5). Following themethod described inLesson 21A, Case 2,weﬁnd (28.92) y,,=—Zlumtcos (wot+B). Hence thegeneral solution of(28.9), by(28.82) and(28.92), is . F(28.93) y=csin(wot+6)-$8tcos(wot-l-I3). . .. FThe maximum departure ordisplacement ofthemotion 1S|c|+5-t . we The presence ofthevariable tinthesecond term implies that thedepar- tureordisplacement duetothispartofthemotion increases with time, seeFig.28.94. Amotion inwhich thedeparture ordisplacement increases beyond allbounds astime passes iscalled anunstable motion. Insuch cases, amechanical breakdown ofthesystem isbound tooccur. This condition, where w,thefrequency oftheforcing function, equals wo,the Lesson 28D Foncno Unomrno MOTION 341 natural (undamped) frequency ofthesystem, isknown asundamped resonance, andwoiscalled theundamped resonant frequency. J’ ri\ v"\%/°’° Fy=2—tcos (wot+B)“'0 0 t J’§\ F /2?,1/ Figure 28.94 i Comment 28.941. Inengineering circles, thefunction f(t)of(28.8) is referred toastheinput ofthesystem, thesolution y(t)of(28.8) asthe output ofthesystem. Example 28.95. The differential equation ofmotion ofasystem is (a) y”+4y=cos2t. Find itsequation ofmotion. Isthemotion stable orunstable? What is theundamped resonant frequency? Solution. Setting theleftside of(a)equal tozero, andsolving, we obtain thecomplementary function (b) yo=ccos (2t+6). Since theright sideof(a)agrees, except forphase, withthecomplementary function (b),thetrialfunction y,,,byLesson 21A, Case 2,must beofthe form (c) y,=Atsin2t+Btcos2t. Following themethod outlined inthislesson, weﬁnd (d) y,=itsin2t. Hence thegeneral solution of(a)is (e) y=ccos(2t+6)+itsin2t. Thepresence ofthefactor t/4intheamplitude ofthesecond term implies that thedisplacement increases with time. The motion istherefore un- 34-2 Pnostans LEADING roLINEAR EQUATIONS orOnoan Two Chapter 6 stable. Acomparison of(a)with (b)also shows that thecondition of undamped resonance ispresent since thefrequency oftheforcing function andthenatural (undamped) frequency ofthesystem areboth 2radians perunit oftime. Hence theundamped resonant frequency is2rad/unit oftime. Example 28.951. The differential equation ofmotion ofasystem is (a) y"+4y=6sint. Find itsequation ofmotion. Isthemotion stable orunstable? Solution. The general solution of(a),byany method you wish to use,is (b) y=ccos (2t+ 6)+2sint. Themaximum displacement is|c|+2,aﬁnite quantity. Hence themotion isstable. Example 28.96. A16-lb weight stretches aspring 6in.Aforcing function f(t)=10sin2tisattached tothesystem. Find theequation ofmotion. What isthemaximum displacement oftheweight? Isthe motion stable orunstable? What frequency oftheforcing function would produce resonance? Solution Because ofthepresence oftheforcing function, (28.63) must bemodiﬁed toread 2 (a) m%% -1-Icy=10sin2t. Here mg=16,m=§-§-==}.Since 16pounds stretches thespring 6 inches, wehave, by(28.62), (b) Ha=16, Ic=32. Hence (a)becomes 2 2 (8) é9%’+32y=10sin2:, %+64y=20sin2:. The general solution of(c)is (d) y=ccos(8t+6)+1}sin2t. Itsmaximum displacement is|c|+§,aﬁnite quantity. Therefore the motion isstable. Toproduce resonance, thefrequency oftheforcing function would have tobe8rad/unit oftime. Lesson 28D—Exercise 343 EXERCISE 28D 1.Verify theaccuracy ofthesolution of(28.81) asgiven in(28.84). 2.Verify theaccuracy oftheparticular solution of(28.9) asgiven in(28.92). 3-(E)S01v8 (28.81) with waswoand initial conditions t=0,y=yo, v=vo.Useforthey,solution theform y,=c1sinwot+62coswot. (b)Write thesolution ifB=0. 4.In(28.81), replace sin(wt+)9)bycoswt.Solve theequation, with w94wo, andinitial conditionst =0,y=yo,v=vo. 5.Solve (28.9) with13=0andinitial conditionst =0,y=yo,v=vo. When aforcing function f(t)isattached toahelical spring, thediffer- ential equation ofmotion, asgiven in(28.63), must bemodiﬁed toread 2 (28961) m‘:72+it=f(t). Usethisequation tosolve thehelical spring problems below. Take positive direction downward. 6.A4-lb body stretches ahelical spring 1in. Aforcing function f(t)= isin8\/6 tisattached tothesystem. Find theequation ofmotion. Isthe motion stable orunstable? What istheundamped resonant frequency? 7.A16-ll_1_ body stretches ahelical spring 4in. Aforcing function f(t)= sin4\/6tisattached tothesystem. After itisbrought torest,itisdisplaced 6in.andgiven adownward velocity of4ft/sec. Find theequation ofmotion ofthebody. Isthemotion stable orunstable? What istheundamped resonant frequency? Hint. Att =0,y=1},v=4. 8.An8-lbbody stretches ahelical spring 2ft.After itisbrought torest, a forcing function f(t)=sin6tisattached tothesystem causing ittovibrate. Find: (a)Distance andvelocity asfunctions oftime (Hint. Att=0,y=0, v=0). 1 (b)Thenatural frequency ofthesystem. (c)The forcing frequency. (d)Themaximum possible displacement ordeparture ofthebody from its equilibrium position. (e)Whether themotion isstable orunstable. 9.Answer thesame questions asinproblem 8,ifatt=0,thebody isheldat rest4ftbelow theequilibrium position andthengiven avelocity of-3ft/sec. Inaddition, write thecomplementary function intheform ya=ccos(kt+ 8). 10.Solve (28.8) iff(t)=Ftwhere I"isaconstant. Isthemotion stable or unstable? ll.Aparticle weighing 16lbandmoving onahorizontal line,isattracted toan origin 0byaforce which isproportional toitsdistance from 0.When the particle isatx=-2,thisforce is91b. Inaddition, aforcing function f(t)= sin3tisimpressed onthesystem. Ifatt=0,z=2,v=0,ﬁnd(a)the equation ofmotion oftheparticle and(b)theresonant frequency ofthe system. 12.Inproblem 11,change sin3ttocos2t.Find: (a)theequation ofmotion, (b)thenatural frequency ofthesystem, (c)theforcing frequency, and (d)themaximum possible displacement oftheparticle from 0. 34-4 PROBLEMS LEADING T0LINEAR EQUATIONS orORDER Two Chapter 6 13. 14. 15. 16.Abody attached toahelical spring oscillates with aperiod of1r/8sec. A forcing function attached tothesystem produces resonance. What isthe frequency oftheforcing function? Amass misattached toahelical spring whose spring constant isk.At t=0,itisbrought torestandaconstant forcing function f(t)=1/blb isimpressed onthesystem. After bsec,theforce isremoved. (a)Find theposition yofthemass asafunction oftime. Hint. First solve with f(t) =1/b. Find 1/(b) and y'(b). Now solve theequation with f(t)=0andinitial conditionst =b,y=y(b), dy/dt =y’(b). Remark. After theinput orforcing function 1/bisremoved, note thatit stillispossible tohave anoutput 1/(t). Note, too,thatifbissmall, say1/50, thenf(t)=501b, butthatitactsforonly 1/50sec.Itisasifthemass were given asudden blow byaforce thatwasimmediately removed. Finally note thattheoutput orresponse function y(t)iscontinuous fortg0,eventhough theinput orforcing function f(t)isdiscontinuous. Thelatter canbewritten as I—.0StSb, f(t)=b __ 0,t>b. Ifb=0,f(t)does notexist. Inengineering circles, however, theﬁctitious forcing function f(t)which results when b=0,iscalled aunit impulse; inphysics itiscalled aDirac 6-function. (b)Show that asb—>0,thesolution 1/(t)—let uscallityo(t)-approaches 1/o(t) =(1/k)Vk/m sinVlc/m t. Hint. Usethefactthatlim(sin0/0) =1.Now prove that yo(t) satis-'—>0 ﬁestheequation m(d2y/dtz) +ky=0with initial conditions t=0, 1;=0,dy/dt =1/m. Thefunction yo(t) iscalled theimpulsive re- sponse ortheresponse ofthe system toaunit impulse. A16-lb weight stretches aspring 8ft.Att=0,itisbrought torestanda forcing function f(t),deﬁned by f(¢)=:“' °§‘§1'0,z>1 isimpressed onthesystem. Find theequation ofmotion. (Seehintin14.) Aforcing function which hasadifferent formula foradifferent time interval iscalled anintermittent force. (a)Show thatthesolution of .1” .ﬁ—|—wo2y =Fsin (wt), w94wo, with initial conditionst =0,y=0,dy/dt =0,is F . w.(28.97) y=--2’; (SIDwt-Esinwot) -woz__ Lesson 28D—Exercise 345 (b)Show that ifw=wo—|—e,where e>0isassumed tobesmall, then thesolution (28.97) becomes y= ; [Sill wot —Sill (L00 +€)l] +z§;T SID wot. (c)Using theidentity sinA—sinB=2cos%—€ sinIL;-E ,show that ifweignore thelastterm ontheright of(28.971)-we shall refer toitagain later—(28.971) canbewritten as F e.et(28.972) y=-—ﬁ(w0+ 6[2cos(wo —|—§)tsin - (d)Aswepointed outinthislesson ofthetext, thephenomenon ofun- damped resonance occurs when w=wo.Theamplitude ofthemotion then increases with time sothat anunstable motion results. Inthis problem, wehave taken w=wo—|—e,eas0,sothatw9'5wo.However, ase->0,wapproaches theresonant frequency wo.Wenowmake the assumption that e,although notzero, isvery small incomparison with wo.Hence wecommit arelatively small error ifin(28.972) wereplace wo+e/2bywo.Show thatthen (28.972) becomes F.t(28.973) y=(—$0sm coswot. (e)Wecangetanideaoftheappearance ofthegraph ofthemotion given by(28.973), ifwelook atthefunction deﬁned byit,asaharmonic motion coswotwith atime varying amplitude (28974) A=_Lsini‘-ewo 2 Equation (28.974) itself deﬁnes asimple harmonic motion whose period is41/e.Since eisassumed small, theperiod ofAislarge. This means thattheamplitude of(28.973) isvarying slowly. Itis,therefore, called appropriately aslowly varying amplitude, andthefunction coswotis saidtobeamplitude modulated. Thegraph ofAisgiven bythe broken lines inFig.28.975. J’ WW ‘X Z/ F /// \\ L gin it ‘r // \{ ewo 2 0 \ / 2”/wo \\ // / 0 1I 1 I I it3'ﬁr1» /ax 'F 2wo‘ \2wo 2wo 2w0 s \\_ i \ / \ ‘mo v\ // F -at . \\\ I/, y(z)-— 5;sin€¢os¢,.;o¢ \\ Figure 28.975 346 PROBLEMS LEADING roLINEAR EQUATIONS orORDER Two Chapter 6 By(28.973), show that foreach value oftsuch that coswot==F1, y(t)==l=(F/ewo) sin(et/2). Hence show thatthegraph ofthesolution y(t)willtouch theupper partofthedotted curve inFig.28.975, foreach value oftsuch that coswot=-1andtouch thelower part ofthe dotted curve foreach value oftsuch thatcoswot=1. (f)Show thaty(t)=0when t_ 1r 31r 5-n" (2'n+1)1l', 2wo’2n>o’2wo’ ’2wo andthat theperiod ofcoswotistherefore 2-ir/wo. Since woismuch greater than e,thisperiod 2-ir/wo ismuch smaller than theperiod 411'/e oftheslowly varying amplitude Aof(28.97 4),whose graph isrepresented bythedotted lines inFig.28.975. Show, therefore, that thegraph of y(t)willthusresemble thesolid lines shown inFig.28.975. Thevariations intheamplitude ofy(t)areknown asbeats. When theamplitude islargest, thesound isloudest. Thisphenomenon ofbeats canbeheard when twotuning forks with almost butnotidentical fre- quencies aresetintovibration simultaneously. Note thatthelastterm in(28.971), which wasomitted inarriving at(28.973), represents asimple harmonic motion. Itdoes notaffect thephenomenon ofbeats. Remark I.Each musical note inaninstrument hasadeﬁnite frequency associated with it.When astandard note anditscorresponding musical notearesounded atthesame time, beats willresult iftheir frequencies differ slightly, i.e.,iftheyarenotintune. When themusical noteoftheinstrument isadjusted sothatbeats disappear, themusical noteisthenintune withthe standard note. Canyouseehowthisresult canbeused totune aninstru- ment? Remark 2.We assumed inthis problem w=wo—|—e.Therefore, as e->0,w->wo,thenatural frequency ofthesystem. Hence theundamped resonant casediscussed inthetextisthelimit oftheundamped modulated vibrations discussed inthisproblem. ANSWERS 28D 1 F . F‘(8)1/=go("O— Smwot+<1/0— coswot + sin (wt -l-'ﬂ). 1 Fw _ F _ (b)y=-56(1)0 - Smwot+yoCOSwot—|—$3 Sinwt. . F Fy=PiS111 wot +(yo — COS wot COS ml. F—|—2wv . Ftcoswty= Slnw0l+y0COSU0l— ' 6.y=csin(8\/61+8)-LL65:cos(8\/6:). Unstable. wo=8\/6rad/sec. Lesson 29A FREE Dmrno Morion. (D.uv1i>En HARMONIC Morion) 34-7 7.y= sin4¢6t+§cos4M6t—%/I-6tcos4\/6t. Unstable. wo=4\/6 rad/sec. 8.(a)y=-116(3 sin4t-2sin 6t),v=§(cos 4t—cos6t). (b)wo=4rad/sec. (c)w=6rad/sec. (d)<}ft. (e)Stable. 9.(a)y=-§95sin4t+ 4cos4t-§sin6t E4.03sin(4t+5)-§sin6t,where 8=-ir—Arcsin80/80.5. v=-16.1 cos(4t+ 5)—§cos 6t. (b),(c),(e)same asin8. (d)4.03+Q=4.23ft. 10.y=c1coswot+62sinwot+Ft/w2. Unstable. 11.(a)a:=ﬁsin 3t+2cos3t-§tcos3t. (b)wo=3rad/sec. 12.(a)2:=goes 2t+§cos3t. (b)3rad/sec. (c)2rad/sec. (<1)§+ =2. 13.16rad/sec. 14. ii(1—cosVk/mt), 0§t§b, 1/(t)=<ﬁ[cosVk/m (t-b)—cosVk/mt] 2. ‘b . b=Es1n[Vk/m (t—§)]sinVk/m-ii t>b. r4 15. 2%—§sin2t-§cos2t, O§t§ 1, ya)=l§(2esin2 —|—ecos 2-1)sin2t ‘—|—%(2ecos2 —esin2 —2)cos2t, t>1. LESSON 29. Damped Motion. Intheprevious lesson, weignored theimportant factor ofresistance or damping. Inthis lesson weshall discuss themore realistic motion ofa particle that issubject toaresistance ordamping force. Weshall assume, forillustrative purposes, that theresisting force isproportional tothe ﬁrstpower ofthevelocity. Frequently itwillnotbe.Insuch cases more complicated methods, beyond thescope ofthistext, willbeneeded to solve theresulting differential equation. LESSON 29A. Free Damped Motion. (Damped Harmonic Motion). Deﬁnition 29.1. Aparticle willbesaid toexecute free damped mo- tion, more commonly called damped harmonic motion, ifitsequation ofmotion satisﬁes adifferential equation oftheform .12 .1 .12 d(29.11) 41.71%+241.1% +mwozy =0,7%+21%+wozy=0, 348 PROBLEMS LEADING 'roLINEAR EQUATIONS orORDER Two Chapter 6 where thecoeﬂicient 2mr >0iscalled thecoefficient ofresistance of thesystem. Asbefore woisthenatural (undamped) frequency ofthe system andmisthemass oftheparticle. The characteristic equation of(29.11) ism2-1-2rm +wo2=0,whose roots are (29.12) m=—r=|=Vr2 —wo2. The solution of(29.11) willthus depend onthecharacter oftheroots of (29.12), i.e.,whether they arereal, imaginary, ormultiple. Weshall con- sider each case separately. Case 1.r2>woz. Ifr2>wo2, theroots in(29.12) arereal and unequal. Hence thesolution of(29.11) is y=c1e(—f+Vr2—uo2)l +c2e(—r—Vr3—oo3)t- Since both exponents in(29.13) arenegative quantities (verify it)wecan write (29.13) as (29.14) y=c1e‘“+ 62¢“, A<0,B<0. Ifc1#0,62960,andc1,c2 have thesame sign, then because e‘>0 forall2,there isnovalue oftforwhich y=0.Hence, inthiscase, the graph of(29.14) cannot cross thetaxis. If,however, cl#60,C2960,and cl,cohave opposite signs, then setting y=0in(29.14) andsolving itfor twilldetermine thetintercepts ofitsgraph. Therefore setting y=0in (29.14), weobtain (29.15) 6“-B" =-‘-2,61 (A-B)t=log , ‘=Tl"'§‘°g(:a§2)' From (29.15), wededuce thatthere canbeonly onevalue oftforwhich y=0.Wehave thus shown that thecurve representing themotion given by(29.14) cancross thetaxisonce atmost. Further, by(27.113), y->0ast->co.[Remember AandBin(29.14) arenegative.] Differentiation of(29.14) gives (29.16) ‘git’=c1Ae‘“ +6,348‘, A<0,B<0. Since this equation hasthesame form as(29.14), it,too, canhave, at most, only onevalue oftforwhich dy/dt =0.Hence thecurve deter- mined by(29.14) canhave atmost only onemaximum orminimum point. Lesson 29A FREE DAMPED MOTION. (Dxmrno HARMONIC MOTION) 349 Themotion istherefore nzmoscillatory anddiesoutwith time. InFig. 29.17 wehave drawn graphs ofafewpossible motions. y y y y (0,c1+c,) (oml+02) (0,C1+C2) 2 0 t 2 e (0.¢‘1'1'°2) Figure 29.17 Comment 29.18. Inthiscase, where r2>woa,theresisting ordamp- ingforce represented byroverpowers therestoring force represented by woandhence prevents oscillations. Thesystem iscalled overdamped. Example 29.19. Ahelical spring isstretched 32inches byanobject weighing 2pounds, andbrought torest. Itisthen given anadditional pullof1ftandreleased. Ifthespring isimmersed inamedium whose coeﬂicient ofresistance is1;§,ﬁndtheequation ofmotion oftheobject. Assume theresisting force isproportional totheﬁrstpower oftheve- locity. Also draw arough graph ofthemotion. Solution. Because ofthepresence ofaresisting factor, whose coeﬁi- cient ofresistance isQ,thedifferential equation ofmotion (28.63) forthe helical spring must bechanged toread .1’ 1d Inthisexample, since 2pounds stretches thespring 32inches =-§feet, wehave, by(28.62), (b) §k=2,Ic=2. Themass moftheobject is33;=11;.Hence (a)becomes Id” 1d 8 d’ d(c) ﬁ5‘,4+§%+;9=0. ,,—,Z+87i,’+12y=0. whose general solution is (d) y=010-2‘ +c2e_“‘, y’=-2c1e'2‘ -6c2c’°‘. Theinitial conditions aret=0,y=1,dy/dt =0.Substituting these values in(c),weobtain (9) 1=ci+92, 0=—2c1 -602. 350 Pnosnams Lmnmo T0LINEAR EQUATIONS orOannn Two Chapter 6 Thesolution of(e)is01==§,Q2=—§. Hence (d)becomes y=_;_e—2t ___%e—6t’ yl,_____ __3e—2t +36-6!- Setting y=0in(f),weﬁndt=—il0g 3=-0.27. Setting y’=0, weﬁndt=0andbytheﬁrstequation in(f),y=1when t=0.Hence thecurve hasamaximum att=0,y=1.Setting y"=0,weﬁnd thecurve hasaninﬂection point att=}1og3 =0.27, y=0.77. A rough graph ofthemotion isgiven inFig.29.191. Themotion isnon- oscillatory. The maximum displacement occurs att=0,i.e.,atthe beginning ofitsmotion; thedisplacement then gradually dieout. J’ 0IIIII0I I I -0.27 0.27 I Figure 29.191 Case 2.rz=woz. If1'2=wo’,theroots of(29.12) are—rtwice. Hence thesolution of(29.11), byLesson 20C, is (29.2) y=c1e_" +c2te_", y’=——rc1e"‘ +c2e_" —rczte-". Since r>0,by(27.113), bothe"‘andte"" —->0ast—->oo.Andasinthe previous case, there isonlyonevalue oftatmost forwhich yandy’=0. Therefore asintheprevious case, themotion isnonoscillatory, anddies outwith time. Thegraphs ofsome ofitspossible motions aresimilar to those shown inFig.29.17. Comment 29.21. Inthiscase, where r=we,theresisting ordamp- ingforce represented byrisjustasstrong astherestoring force repre- sented byweandhence prevents oscillations. Forthisreason thesystem issaidtobecritically damped. Case 3.rz<woz.Ifr2<(.002,theroots in(29.12) areimaginary and canbewritten as (29.3) m=-r:|=i\/oi;-51$. The solution of(29.11), byLesson 20D, istherefore (29.31) y=ce_"sin (\/W’-7 ¢+a). Lesson 29A FREE DAMPED Morrow. (Dunno HARMONIC MOTION) 351 Because ofthesine term inthesolution, themotion isoscillatory. Thedamped amplitude ofthemotion isce_" andsince r>0,this factor decreases astincreases and approaches zero astapproaches co. Hence with time, theparticle vibrates with smaller and smaller oscilla- tions about itsequilibrium position. Each function deﬁned in(29.31) isnotperiodic since itsvalues donot repeat. However, because themotion isoscillatory, wesaythefunction isdamped periodic and deﬁne itsdamped period tobethetime it takes theparticle, starting attheequilibrium position, tomake onecom- plete oscillation. Hence itsdamped period issaidtobe 21r(29.32) T X/Zd;5___iﬂ The damped frequency ofthemotion is\/“,0? _7-2radians perunit oftime, or\/(.002 —r2/21r cycles perunit oftime. Theexponential term e_"iscalled appropriately thedamping factor. Since thisfactor decreases with time, themotion eventually dies down. When t=1/r,thedamping factor is1/e.Thetimeittakes thedamping factor toreach thisvalue 1/eiscalled thetime constant. Hence the time constant 1'=1/r. Agraph ofthefunction deﬁned by(29.31) isgiven inFig.29.33. Itis anoscillatory motion whose amplitude decreases with time. y y=|¢e‘"| 6 y=ce_nsin(1/wo2:Ft+§) (0,csin) Damped period :1 9t.+T- z y=—l¢e'"l Figure 29.33 Comment 29.34. Inthis case, where r2<woz, thedamping force represented byrisWeaker than therestoring force represented bywe andthus cannot prevent oscillations. Forthisreason thesystem iscalled underdamped. 352 Pnostsms Lmnmo 'roLINEAR EQUATIONS orOnnsn Two Chapter 6 Example 29.35. Ifthecoeﬂicient ofresistance inexample 29.19 is -3-instead ofQ,ﬁnd: 1.Theequation ofmotion ofthesystem. 2.The damping factor. 3.The damped amplitude ofthemotion. 4.The damped period ofthemotion. 5.The damped frequency ofthemotion. 6.Thetime constant. Solution. The differential equation ofmotion (c)inexample 29.19 now becomes 1112 3d 3 d2 d <5‘) nﬁ+§rl+zy=°' ﬁ+“ri+‘2=°~ Itssolution is (b) y=ce“3‘ sin(\/3t +6). Differentiation of(b)gives (c) y’=-3ce_3‘ sin(\/3t +6)+c\/3 e'3‘ cos(\/3t +6). Theinitial conditions aret =0,y=1,y’=0.Substituting these values in(b)and(c),weobtain (d) 1-=csin6, 0=——3csin5+\/3ccos6. Substituting inthesecond equation of(d),thevalue ofcasgiven inthe ﬁrst equation, weobtain 3(e) 0=—3+\/3cot6, cot6=——=\/3,\/§ 6=%-or 7%.‘, sin6=:l=%- Hence, bytheﬁrstequation in(d),choosing sin6=Q,wehave, (f) c=2. The equation ofmotion (b)therefore becomes (g) y=2e_3' sin(\/3t + , which istheanswer to1.The answers totheremaining questions follow. 2.Thedamping factor ise'3‘. 3.Thedamped amplitude ofthemotion is2e“3‘ feet. Lesson 29A—Exercise 353 4.The damped period ofthemotion is21r/\/3 seconds. 5.The damped frequency ofthemotion is\/3rad/sec Egcps. 6.The time constant 'r=§sec. EXERCISE 29A 1.Verify theaccuracy ofthesolution of(29.11) with 1'2>4002, asgiven in (29.13). 2.Verify thateach oftheexponents in(29.13) isanegative quantity. 3.Verify theaccuracy ofthesolution (d)ofExample 29.19. 4.Verify theaccuracy ofthesolution of(29.11) with r2=0:02, asgiven in (29.2) 5.Verify theaccuracy ofthesolution of(29.11) with 1'2<woz,asgiven in (29.31). 6.Verify theaccuracy ofthesolution (b)ofExample 29.35. 7.Show thatthesystem whose differential equation is (1221 dz!W-+2a;i?—|-bzy =0,a>0, is:(a)overdamped andthemotion notoscillatory ifa2>b2,(b)critically damped andthemotion notoscillatory ifa2=b2,(c)underdamped and themotion oscillatory ifa2<b2. 8.(a)Solve thedifferential equation 1121/ dy Note thatherebisnotsquared asin7. (b)Show thatthemotion ofthesystem isstable onlyifa>0andb>0. (For deﬁnitions ofstable andunstable, seeLesson 28D). Hint. Show thatifa>0,b>0,each independent solution of(29.36) approaches zeroast—>w.Hence thedistance yfrom equilibrium approaches zero. Consider each other possibility a>0,b<0;a<0,b>0;a<0, b<0,andshow thatineach easey—><=<>ast—+w. (c)Show thatifa<0,b>0anda2<b,themotion, although unstable, is0scillatory;ifa >0,b<0,orifa <0,b<0,orifa <0,b>0, andineachcasea2>b,themotion, although unstable, isnotoscillatory. 9.With thehelpoftheanswers toproblems 7and8,determine, without solving, whether themotion ofthesystem, whose differential equation is: dzy dy dzy dy(8.)'zﬁ'-71?--2]/-0. (6)W+4E—‘4y—0. dzy dy 42;, dy dzy dy dzy dy(<5)'Et§'l'2E'l'5il-0- (8)W+6E+6ll—0- <d>"'—2”+4@+4=oas any' 354 Pnonuzms L1-zxnme roLmsxa EQUATIONS orORDER Two Chapter 6 10. ll. 12.isstable orunstable; oscillatory ornotoscillatory. Also determine whether thesystem isunderdamped, critically damped, oroverdamped. Check your answer bysolving each equation. Draw arough graph ofeach motion. Aparticle moves onastraight lineaccording tothelaw .121 atEl-5‘ +27'E+33-0, where risaconstant andacisthedisplacement oftheparticle from itsequilib- rium position. (a)Forwhat values ofrwillthemotion bestable; unstable; oscillatory; not oscillatory. Forwhat values ofrwillthesystem beunderdamped; criti- cally damped; overdamped. (b)Check your answers bysolving theequation with r=Q,r=1,r=2, r=—=},r=——1. (c)Forwhat value ofrwillthemotion beoscillatory andhave adamped period equal to311-? (d)Isthere avalue ofrthatwillmake thedamped period lessthan 21r? Aparticle moves onastraight lineinaccordance with thelaw é+4Q+13.-0dt2 dt _' Att =0,:z:=0,1)=12ft/sec. Solve theequation for2:asafunction oft. What isthedamping factor, thedamped amplitude, thedamped period, thedamped frequency, thetime constant? Find thetime required forthedamped amplitude—-—and hence alsofor thedamping factor—to decrease by50percent. Hint. Thedamped amglitude is4e'2‘. When t=0,4e"2‘ =4.You want tsothat 4e’‘=2. (d)What percentage ofitsoriginal value hasthedamping factor, andthere- forethedamped amplitude, after onehalf period haselapsed? Hint. Thedamped period is21'/3. Evaluate e-2‘ when t=1r/3. What is thedamped amplitude atthatinstant? Where istheparticle andwith what velocity isitmoving when t= 1r/6sec? Draw arough graph ofthecurve.(=1) (b) (<1) (e) (f) Aparticle moves inastraight lineinaccordance withthelaw dza: da:W-F 105‘-+ 1623 —0. Att =0,x=1ft,v =4ft/sec. (a)Find theequation ofmotion. (b)Isthemotion oscillatory? (c)What isthemaximum value of2:?When does 1attain thismaximum value? (d)Draw arough graph ofthecurve. Does thecurve cross thetaxisfor t>0‘? Lesson 29A—Exercise 355 13.Aparticle isexecuting damped harmonic motion. In10sec,thedamping factor hasdecreased by80percent. Itsdamped period is2sec. Find the differential equation ofmotion. , 14-.Aparticle ofmass mmoves inastraight line. Itisattracted toward the origin byaforce equal toktimes itsdistance from theorigin. Theresistance is2Rtimes thevelocity. Find themaximum value ofmsothatthemotion willnotbeoscillatory. 15.Aparticle moves inastraight lineinaccordance withthelaw dz: da:W-+6E—l6$ —0. Att=0,theparticle isat2:=2ftandmoving totheleftwith avelocity of10ft/sec. (a)When willtheparticle change direction andgototheright? (b)Williteverchange direction again? When thedamping orresisting factor ofasystem isnotnegligible, the differential equation (28.63) forthehelical spring must bemodiﬁed toread, with downward direction positive, (2937) mfg-|—r§g—|—ky=0' dtz dt ’ where wehave assumed that theforce ofresistance isproportional tothe ﬁrstpower ofthevelocity andr>0isthecoefficient ofresistance ofthe system. Note thathererreplaces 2mrof(29.11). Use(29.37) tosolve thefollowing problems, 16-25. 16.Aweight of16lbstretches ahelical spring 1%ft.Thecoefficient ofresistance ofthespring is2.After itisbrought torest,itisgiven avelocity of12ft/sec. (a)Find theequation ofmotion. Draw arough graph ofthemotion. (b)Find damping factor, damped amplitude, damped period, damped frequency, time constant. (c)When willtheweight stopfortheﬁrsttime andchange direction? How farfrom equilibrium willitthen be? (d)When willitstop forthesecond time? How farfrom equilibrium will itbe? (e)Write aformula which willgivethetimes when theweight crosses the equilibrium position andforthetimes ofitssuccessive stops. 17.A16-lb weight stretches aspring 6in.~Thecoeﬁicient ofresistance is8.After thespring isbrought torest, itisstretched anadditional 3in.andreleased. Find theequation ofmotion. Draw arough graph ofthemotion. 18.Inproblem 17,change thecoefficient ofresistance to10.Find theequation ofmotion. Draw arough graph ofthemotion. 19.(a)Solve (29.37) if12<4kmandtheinitial conditions aret=0,y=yo, v=0. (b)What isthedamped period ofthemotion? (c)When willthedamping factor, andtherefore thedamped amplitude, be ppercent ofitsinitial value? Hint. Thedamping factor ise""/2"‘. Att =0,thedamping factor ise""/2"‘ =1=100percent. Therefore want tsuch thate-"/2'" =p/100. Hence —-rt/2m =log(p/100), t= -—(2m/T) 10$(N109)- 356 PROBLEMS LEADING roLINEAR EQUATIONS orORDER Two Chapter 6 (2938)(d)Callthetime obtained in(c)tosec.Therefore thedamping factor atthe endoftosecise"‘°’2"' andthisdamping factor, andtherefore alsothe damped amplitude, isppercent ofitsvalue att=0.Show thatatthe endofevery period oftosec,thenewdamping factor isppercent 20fits value atthebeginning oftheperiod. Hint. Show that e"(‘°+‘v)/ "'= p2/104, i.e.,show that itisp2/104 oftheoriginal damped period and hence isppercent ofthedamped period attheendoftosec. Oryoucan lete"‘°/2'" =100percent. Then want t1such that e"‘l/2"‘ =p/100. Find t1=—(2m/r) logp/100 asin(c). (e)When t=T,thedamped period ofthemotion, thedamping factor is e"'T/2"‘. Ithasadeﬁnite value, sayqpercent ofthevalue ofthedamping factor att=0.Show thatattheendofeach period ofTsec,thedamp- ingfactor, andtherefore thedamped amplitude, isqpercent ofthedamp- ingfactor atthebeginning oftheperiod. Hint. See(d)above. This constant percentage, therefore, gives thepercentage decrease inthe displacement ofaparticle from equilibrium attheendofaperiod as compared with itsdisplacement atthebeginning ofaperiod. Hence, the damped amplitude attheendofaperiod ofTsec=qpercent ofthe damped amplitude atthebeginning ofthatperiod. Therefore, 10 Damped amplitude atthebeginning ofaperiod ofTsec) g Damped amplitude attheendofthatperiod =log(100/q) =aconstant D. Theconstant Discalled thelogarithmic decrement. Itis,asequa- tion (29.38) shows, theconstant positive difference between the logarithm ofthedamped amplitude atthebeginning ofaperiod ofT secandthelogarithm ofthedamped amplitude attheendofthatperiod. (f)Find thelogarithmic decrement ofthisproblem. Hint. In(29.38) sub- stitute thedamped amplitude when t=0andwhen t=Tasfound in(b). (g)When willthebody ﬁrstreach theequilibrium position? A20-lb weight stretches aspring 3in.After itcomes torest,itisgiven an additional stretch of2in.andreleased. Theinternal resistance ofthespring isnegligible buttheresistance duetotheairis1/50 ofitsvelocity. (a)Find theequation ofmotion anddraw arough graph ofitsmotion. (b)Find thedamped amplitude, damping factor, damped period, damped frequency, time constant. (c)When willthedamping factor have decreased by50percent? (d)Over what time intervals willthedamping factor attheendofaninter- valbe50percent ofitsvalue atthebeginning oftheinterval? (e)What percentage ofitsoriginal value doesthedamping factor have, and therefore alsotheamplitude, attheendofaperiod? Note by(e)of problem 19thatthedamping factor attheendofanyperiod isthissame percentage ofthedamping factor atthebeginning ofthatperiod. (f)Find thelogarithmic decrement. (g)When does theparticle ﬁrstcross theequilibrium position? Theoscillatory motion ofaspring isgiven by (1211 dz; 2?t§+2aE+by—0, G <17. Lesson 29A—-Exercise 357 Itisobserved thatthedamping factor hasdecreased by80percent in10sec andthatitsdamped period is2sec. Find thevalues ofaandb. 22.Thenatural frequency ofaspring is1cps. After thespring isimmersed ina resisting medium, itsfrequency isreduced to§cps. (a)What isthedamping factor? (b)What isthedifferential equation ofmotion? 23.Inproblem 21,ﬁndaandbiftheperiod ofthemotion is2secandtheloga- rithmic decrement is 24.Thedifferential equation ofmotion ofabody attached toahelical spring is given by(29.37). (a)Solve theequation ifitsmass m=1'2/4k andatt=0,y=yo,v=vo. (b)Isthemotion oscillatory ornotoscillatory? (c)When willitreach itsmaximum displacement from equilibrium? (d)Show that from itsmaximum displacement itwillmove toward the equilibrium position butnever reach it.Hint. Show that y—>0as t-—+0°[see(27.113)]. (e)Show thatifrvo=——2ky0, thebody willnever change itsdirection but willmove continually toward theequilibrium position. 25.Thedifferential equation ofmotion ofabody attached toahelical spring is given by(29.37). (a)Solve theequation if1'2>4kmandatt=0,y=0,v=vo. (b)Isthemotion oscillatory ornotoscillatory? (c)Show thatthesolution canalsobewritten intheform 2mvo —rl/21» .\/1-2 —4kmy=—-——————-— e sinh i— t. \/1'2 -—4km 2'" Hint. See(18.9). (d)When willthebody reach itsmaximum displacement from equilibrium? (e)Show that from itsmaximum displacement, itwillmove toward the equilibrium position butnever reach it. Wehave included below onlyafewpendulum problems because ofthesimilar- ityinform ofthependulum equation andthehelical spring equation—compare (29.37) with (29.381) below. Thesame questions asked forthespring could be asked forthependulum. Alloneneed dotoobtain asolution forthependulum istoreplace lcintheprevious answers bymg/l andyby0.Remember, linear velocity v=ld0/dt =lw,where wisangular velocity. 26.Asimple pendulum oflength l,withweight mgattached, swings inamedium which offers aresisting force proportional totheﬁrstpower ofthelinear velocity. Show thatthedifferential equation ofmotion is .120 d0(29.38l) W+-'";E+§la =0, where risthecoefficient ofresistance ofthesystem. Hint. Adjust (28.71) totake into account theresisting force, andremember linear velocity d0U—lEt'. 358 PROBLEMS LEADING roLINEAR EQUATIONS orORDER Two Chapter 6 27.(a)Show thatthependulum inproblem 26isoverdamped andthemotion notoscillatory if12/4m2 >g/l;critically damped andthemotion not oscillatory if1'2/4m2 =g/l;underdamped andthemotion oscillatory ifr2/4m2 <g/l. (b)Solve (29.381) ifatt=0,0=0,w=weand1'2/4m2 <g/l. 28.Aweight of2lbisattached toapendulum 16ftlong. Find thesmallest positive value ofthecoefficient ofresistance rforwhich thependulum willnot oscillate. 29.Aweight of4lbisattached toapendulum swinging inamedium which offers aresistance ofone-eighth ofthelinear velocity. Itisdesired thatthe period ofthependulum be21r.How longmust thependulum be? 30.(a)Solve (29.381) ifthependulum isreleased from theposition 0=00. Assume oscillatory motion. (b)When willthependulum ﬁrst reach theequilibrium position? ANSWERS 29A =c]e(—a+\/a3—b2)t+ c2e(—a—\/a7—b2)l, be<G2, = (C1 + 0209-“: b2 =a2: =e'°‘(c1 cos\/bf-ii-t +C2sin\/Wit), a2<b2. =c1e<-..+\/SE): _|_c2e(—-a—ﬁ)t, b<as, =(c1+c2t)e‘“‘, b=a2, y=e‘“‘(ci cosvb —a§t+ cgsinvb —agt), a2<b. 9.(a)Unstable, notoscillatory, overdamped, y=c1e2‘—|—cge"‘. (b)Unstable, notoscillatory, overdamped, y=c1e2‘ +age‘. (c)Stable, oscillatory, underdamped, y=cc"sin(2t+6). (d)Stable, notoscillatory, critically damped, y=oie"2‘ +czte-2‘. (e)Unstable, notoscillatory, overdamped, y=c1e(‘2 +2‘/5)‘ +cge(‘2‘i/7)‘. (f)Unstable, oscillatory, underdamped, y=ce‘cos(2t+6). (g)Stable, notoscillatory, overdamped, y=ce(_3+\(§)‘ —|—cge('3'1/33‘.9°2" Q<==e==@ 10.(a)Stable only ifrg0;underdamped and oscillatory if0<r<1; critically damped andnotoscillatory ifr=1;overdamped andnot oscillatory ifr>1;unstable andoscillatory if—1<r<0;unstable andnotoscillatory ifr§-1. (b)y=ce-‘/2 sin(\/3t/2+6), =,,e(— +)t+c2e(— —)1, _ =cie‘/2sin(t/3t/2+s), \=(c1_+ c2t)e‘. (C)1"=\/5/3. (<1)No. ll.(a)2:=4e‘2‘ sin3t. (b)e"2‘, 4e'2‘ ft,2-ir/3 sec,3rad/sec or3/2-ir cps, 1)sec. (c)t=%log2sec=0.35 sec. (d)12.3 percent, 0.49 ft. (e)1.4ft,-2.8 ft/sec. 12.(a):2:=2e‘2‘ —e'3‘. (b)No. (c)2:=1.19ft,t=0.12sec. (d)No.<Q<Q‘§ dzy dy 213.555+ 0.322 5+ 9.896y —0. 14.R/lc. 15.(a)0.223 sec. (b)No. Lesson 29B FORCED MOTION WITH Dxmrme 359 16.(a)y=3e'2‘ sin4t. (b)e"2‘, 3e-2‘ ft,1r/2sec,4rad/sec, Qsec. (c)0.277 sec, 1.54 ft. (d)0.277 —|—1r/4sec,—0.32 ft. (e)mr/4 sec;0.277 +nir/4 sec,n=0,1,2,---. 17. =e_8‘(1+ 2t) 18 —§e'4‘ —112-e"6‘ y . .y- . 4k _, ,,, \/4k — 219.(a)y=y0‘l e "2cos(%Lt+8)» where 8=Arctan(—r/\/ 4km —-12). (b)T=4-rm/\/4km —12sec. (f)logdecrement =21rr/\/ 4km —r2. (g)t=m(1r —28)/V 4km ——r2sec. 20.(a)y=ﬁe-°-°16‘ cos(8\/2t+ 5),approximately, where tan6=—0.0014, 6=—-0.0014 radian, _ _ (b)-§e'°-M6‘ ft,e‘°-016‘, 1r\/2/8 sec,8\/2 rad/sec, or4x/2/1r cps,62$sec. (c)and(d)43.3sec. (e)99percent. (f)0.009. (g)0.14sec. 21. a=0.161, b=9.896. 22.(a)e"4-68‘. (b)y”+ 9.371/—|— 39.511 =0. 23. (1=0.1, b=9.88. 24-(oy=[yo+<1/0+ i]e"”""'. (b)Notoscillatory. (c)t=vorz/2k(rv() +2ky()) sec. 25. (a) y= "W0 e—rtI2m(et Vr2—4km/2m __e—tVr3—4I:m/21») \/1'2 —-4km (b)Notoscillatory (d)tanhx/1'2 —4kmt _\/1'2 -—4km 2m _ r ' t—i—— tanh_1 4°'2—4km-\/rz —4km r l l —-rll2m .(9 1'227- 6=2wom § 6 Slllwi '-w C. 9 128.r=Z lb-sec/ft. 29.l=25.6ft. 4\/2 30.Inanswers to19(a) and(g),replace lcbymg/l andyby0. LESSON 29B. Forced Motion with Damping. The motion ofa particle that satisﬁes thedifferential equation .1“ it(29.4) mﬁ+2mr-dl: +m...,’y =f(z), .121; d 1W+27?+My=gm). where, asbefore, 2mr isthecoefficient ofresistance ofthesystem, weis thenatural (undamped) frequency ofthesystem, misthemass ofthe 860 Pnonums Lmnmo T0Lmmn EQUATIONS orOnmm Two Chapter 6 particle andf(t)isaforcing function attached tothesystem, iscalled forced damped motion incontrast tothefree damped motion (i.e., damped harmonic motion) when f(t)E0.Inengineering circles, f(t)is called theinput ofthesystem andthesolution y(t)of(29.4) theoutput ofthesystem. Letusassume theforcing function f(t)=mFsin(wt+13) where Fisaconstant. Then (29.4) becomes 2 (29.41) %+2r‘;-f+my=Fsin(wt+/3). Thedifferent possible complementary functions y,obtained bysetting theleftsideof(29.41) equal tozeroandsolving itwillbethesame as_ those given inthethree cases ofLesson 29A. Thetrialsolution y,forall such solutions y,is (29.42) y,=Asin(wt+B)+Bcos(wt+ Following themethod outlined inLesson 21A, weﬁndthat (29.43) A= F("’°2'""’2)[(0)02 -'W)’+(2"<v)’] ' B__ —F(2rw) _ _l(wo’—wz)’+(2rw)’] Let(seeFig.29.44), \",l_'1.xcva3keg~“‘ 2rw N02_w2 Figure 29.44 2_2 (29.45) cos<1=—--‘-"°—°’?—-, \V(@102-—~12)’+(2"w)’ 2rw _ V(4-=0’-—01”)”+(2rw)2 Substituting these values in(29.43) andtheresulting expressions forA andBin(29.42), weobtainSina = F (29-46) yp=(“>02—wz)’+(2rw)’ X[cosasin(wt+5)»-—sinacos(wt+8)]. Lesson 29B Foacan MOTION wrrn DAMPING 361 Hence thegeneral solution of(29.41) is <29-41) y=9.+--—-i—-—sin (wt+5—<1).V(1-"02—91”)’+(2w)’ where yeisanyoneofthefunctions given inLesson 29A. Aswesawthere, themotion duetotheycpartofthesolution (29.47), inallcases, whether oscillatory ornonoscillatory, diesoutwith time. Forthisreason thispart ofthemotion hasbeen called appropriately thetransient motion. The equation ofmotion (29.47) isthusacomplicated oneonlyforthetime in which thetransient motion iseffective. Thereafter themotion will be dueentirely tothey,partofthesolution asgiven bythesecond term on theright of(29.47). This part ofthemotion hastherefore been appro- priately named thesteady state motion. Comment 29.48. Inmany physical problems, thetransient motion is theleast important partofthemotion. However, there arecases where it isofmajor importance. By(29.41) and(29.47), weseethat thesteady state motion hasthe same frequency astheforcing function f(t),namely wrad/sec, butisout ofphase with itandthattheamplitude ofthesteady state motion is (29.5) A= V(“>52-<92)’+(21902 Ifw=wo[thecondition for(undamped) resonance], theamplitude re- duces totheinteresting form F(29.51) A_ﬁz- Ifw¢wo,then bydifferentiating (29.5) with respect towandsetting theresulting expression fordA/dw equal tozero, weobtain (29.52) 2(0),,”-w2)(—-2w) +We=0, from which weﬁnd (29.53) <5’=<50’-2%, w=\/was -2r2,<50’>2r2. Hence ifaresisting force ispresent, andifw,thefrequency oftheforcing function, isnotequal towo,thenatural (undamped) frequency ofasys- tem, then, forﬁxed F,theamplitude Aofthesteady state motion will beamaximum ifwhasthevalue given in(29.53). Aforcing function f(t), having thisfrequency w,isthen saidtobeinresonance with thesystem. Substituting thisvalue ofwin(29.5), weﬁnd that themaximum ampli- tude is (29.5s1) .4,,,,,=_-iii.21V (.002 —'7'2 362 Pnonmns LEADING T0LINEAR EQUATIONS orORDER Two Chapter 6 Assume now that 2r,thecoefficient ofresistance ofasystem perunit mass, issmall. Hence wecommit asmall error ifweomit ther2term in(29.531). Wethus obtain F(29.532) Am“ ~5;,-3» thesame amplitude obtained in(29.51) when w=wo. Further, we showed inLesson 29A, Case 3,that thenatural (damped) frequency ofa system is\/woz —r2,which, forsmall r,isclose totheresonant fre- quency \/wo2 —2r2,i.e.,itisclose tothefrequency which willproduce themaximum amplitude. Weinfer from alltheabove remarks that ifaresisting force ispresent and w,thefrequency oftheforcing function f(t), equals wo,thenatural (undamped) frequency ofasystem, orisclose to\/w02 —1'2,thenatural (damped) frequency ofthesystem, thentheamplitude ofthesystem isinversely proportional tothedamping orresisting factor 2r.Hence if2rissmall, Awillbelarge, and tremendous vibrations may beproduced. That is why soldiers crossing abridge may beordered tobreak step (although thechances arethat thisprecaution isunnecessary), foritisfeared that ifthefrequency which they create with their footbeat isthesame asthe natural (undamped) frequency ofthebridge, ornear itsdamped frequency, andifinaddition theintemal resistance ofthebridge issmall, thevibra- tions may become solarge astocause abreakage. The walls ofJericho, sosome assert, came tumbling down because thesound thetrumpeteers made with their trumpets caused awave motion whose frequency equaled thenatural (undamped) frequency ofthewalls. Students atCornell University used toﬁnditamusing either tocreate awave motion inthe oldsuspension bridge over thegorge ortogetittoswing violently from sidetoside. They would march across itinastraight linewith arhythmic beat orwalk with asailor’s gait, ﬁrstemphasizing oneside, then theother. Totimid souls, however, itwasnever very amusing—terrifying would be amore descriptive word. Onsuch occasions, itwas impossible towalk across thebridge with aneven steporinastraiglFine, depending on whether thebridge waswaving orswinging. Wecitetwomore examples ofthisphenomenon and ones which you caneasily experience ormay have already experienced. 1.Aswing, with achild seated onit,when displaced from itsequi- librium position, willmove back andforth across theequilibrium position with anatural (damped) frequency. Ifyou now apply aforce tothe swing with afrequency close tothisnatural (damped) frequency, then for aﬁxed Fandsmall r,themaximum amplitude willequal, approximately, F/2rw0. Hence, ifrissmall, theamplitude ofswing willbelarge. Ifyou want astilllarger amplitude, youmust increase F. Lesson 29B Foacnn MOTION WITH DAMPING 363 2.VVhen youjump offadiving board, theendoftheboard willvibrate about itsequilibrium position with anatural (damped) frequency. If instead ofjumping off,younow jump upanddown above theendofthe board with afrequency near this natural (damped) frequency, you will beable tomake themagnitude oftheoscillation large. Ifrissmall, the maximum amplitude, foragiven F,willequal, approximately, F/2rw0. The ratio Amplitude ofy,29.54 M=4-———-, ‘l F/we where Fand wozaregiven in(29.41), iscalled themagniﬁcation ratio ofthesystem ortheampliﬁcation ratio ofthesystem. By(29.54) and(29.47), thismagniﬁcation ratio is 2 (29.55) M=———;9°i———— V(‘"02—~12)”+(2Tw)’ 1 O,22 T2Q,2 J11—<9.)1+4<9.)<59Since woisﬁxed, theampliﬁcation ratio ofasystem depends onthefre- quency woftheforcing function f(t)andthecoeﬂicient ofresistance per unit mass 2r.Inpractical applications where wisalsoﬁxed, theresistance 2rismade large ifonewishes themagnifying response tobesmall as,for example, invibrations ofmachinery andinshock absorbers; theresistance 2rismade small, ifonewishes theresponse tobelarge, as,forexample, inaradio receiver. Ifin(29.55), welet (29.55) )1=Q30 and 9= theequation becomes (29551) M=ill-V(1—#2)’+41'2#’ The quantity u,by(29.56), isthus theratio oftheimpressed orinput frequency wtothenatural (undamped) frequency wo. The quantity 11 may belooked atasmeasuring theamount ofdamping present fora ﬁxed wo.Foreach ﬁxed value ofv,Misafunction ofu.Hence itispos- sible todraw agraph ofthemagniﬁcation Mforeach such ﬁxed value ofv.Forexample, if11isQ“,then, by(29561), 1(29.-562) M(p)= - 364 Pnonnsms LEADING T0LINEAR EQUATIONS orORDER Two Chapter 6 Ifv=0,which implies by(29.56) that r=0,then, by(29.56l), (29553) M(;1)= By(29.563), Weseethat as/.¢-—>1,which implies by(29.56) that w—->wo,M—>co. Example 29.564. Aforcing function f(t)=9]-cos 2tisapplied tothe motion given inExample 29.19. Find thesteady state motion andthe ampliﬁcation ratio ofthesystem. Isresonance possible? Solution. With f(t) =§cos 2t,thedifferential equation ofmotion (c)inExample 29.19 becomes 1.1’ 1.1 35<9 nd—1§'+§%+z”=§°°s2‘» .1’), d(b) W+87%+12y=40cos2:. Aparticular solution of(b)is (c) y,=2sin2t+cos2t, which isthesteady state motion. ByComment 28.32, theamplitude of themotion deﬁned by(c)is\/22 +12 =\/5. Comparing (b)with (29.41), weseethat wz=12,2r=8,F=40. Therefore by(29.54), themagniﬁcation ratio ofthesystem is (/5 's\/5<9 M=T/n=To' And since w°2=12<2r2=32,resonance isnotpossible. See(29.53). Comment 29.6. Foreasy reference, wehave listed inthetable on page 365thedifferent differential equations discussed thus farinthis chapter, andthepertinent information related toeach. —\ EXERCISE 29B 1.Verify thevalues ofAandBasgiven in(29.43). 2.Verify thesolution (29.46). 3.Verify (29.53). 4.Verify theaccuracy ofthesolution (c)ofExample 29.564. 5.Forwhat value ofwwillthemagniﬁcation ratio asgiven in(29.55) bea maximum? Find thismaximum value. 6.Aparticle moves according tothelaw dzy dy .9y =5811125. 683ﬁgskgm%%___N\3hoQ23E050888”kggw“O gsw__:°§E°WO “A35+NAN3|NO3V\(‘J1! AHOSOEwoﬁgg_V8_s_wH N35+“Q3Iﬂog, AulQII3EmK +£“_M €+“3:z"\_H_uWWEN3+%&+@_68$Moﬁg:2$3?LP‘ Nk“O3>_ hogm1I“O3> 63%__:3_mE°wO Q_TQ_2___§_m_bo§E3° Igz 302 _gﬂ_Osﬁwaaﬂw$5S8308umﬂOG_:_§_~HVQQESQ Q3VkéII1IN_§\(VEmti:" QI_IO3|___T3g+_T2Dl S-¢nx 3Ak:“°s|N__\'|__|VQQ+ I__" __N°?N;+_iQQ 3 Q"N°3+ W%§&+£_2o_3wﬂ5Aﬁggowwmv O3“Tl“ RN£5_ro§__m°Mo N3|No3l‘i+Q R QEOEwogadgﬂ _V8_s_m 3"O3ea an+33WOOQQMI{M" §3mmO3An‘I3ﬂmw +Qggsaw8gN ‘M G+3EmkH may‘MNO3+i N68$ﬁg: km3no:01__HO3 __OvsO3 _0_o_£AHOSOE_g_$&QnH QQNg+N8\( 50308“VQQQEQQ: $5__O2__Oa__2 25$ Q+“O3mgU" Q‘I5EQH“O3EmNo+“O3M8G" iQQ 1N:_QI5%g2_Uo___m G232MO 3; %__£9__< 8302MO Qadz _HO$_:Ow __O$_gvmQ“15““NDGOWWSQ:qQWm=QWmA___migﬁﬂzvg_ﬂm“=QH9_umQ 366 PROBLEMS LEADING T0LINEAR EQUATIONS orORDER Two Chapter 6 (a)Find thesteady state motion; alsotheamplitude, period, andfrequency ofthesteady state motion. (b)What isthemagniﬁcation ratio ofthesystem? (c)What frequency oftheforcing function willproduce resonance? 7.Aparticle moves according tothelaw 4’ amKg+2mr3%+mwozy =f(t). (a)Find theequation ofmotion iff(t)=mFcoswt.Assume r2<5:02. (b)What isthetransient motion; thesteady state motion? (c)What istheampliﬁcation ratio ofthesystem? 8.Aparticle moves inaccordance with thelaw @+4@+16 =19) dz? dz y ' (a)What frequency ofthefunction f(t)willmake theperiod ofthesteady state motion 1r/3? (b)What frequency ofthefunction f(t)willproduce resonance? 9.Aparticle moves according tothelaw dzu du _..E+5E+6y=e s1n2t. (a)Solve foryasafunction oft. (b)What istheinput; theoutput? (c)Describe themotion. 10.InExercise 29A, 8,weaskqd youtoshow thatthemotion ofaparticle whose differential equation is%+ 2a%%+ by=0,isstable only ifa>0, b>0.Since theaddition totheequation ofafunction f(t)does notaffect thecomplementary function y,,,itfollows that a>0,b>0isalso a necessary condition forthestability ofthemotion ofaparticle whose differential equation is 5+259+ by=f(t) dt2 dt 'X Prove thatitisnotasufficient condition bysolving theequation ﬂ+5d"+5 -12‘ dtg E y_ e7 andthen showing thatthesolution y(t)-—->wast-—+9°. When thedamping orresisting factor ofasystem isnotnegligible and aforcing function f(t)isattached toit,thedifferential equation (28.63) forthehelical spring must bemodiﬁed toread[Seealso(29.37).] 2 (29.1) m%+T%+ku=f(t). Lesson 29B—Exercise 367 where risthecoefficient ofresistance ofthesystem: Use (29.7) tosolve thenext twoproblems. ll.A16-lb weight stretches aspring 1ft.Thespring isimmersed inamedium whose coefficient ofresistance is4.After thespring isbrought torest, a forcing function 10sin2tisapplied tothesystem. (a)Find theequation ofmotion. (b)What isthetransient motion; thesteady state motion? (c)Find theamplitude, period, andfrequency ofthesteady state motion. (d)What isthemagniﬁcation ratio ofthesystem? 12.A16-lb weight stretches aspring 6in.Itscoefficient ofresistance is2.The 16-lb weight isremoved, replaced bya64—lb weight andbrought torest. Att=0,aforcing function 8cos4tisapplied tothesystem. Find the steady state motion andtheampliﬁcation ratio ofthesystem. 13.In(29.4), let f(t)=m(A1 sinwlt-1-A2sinwgt+---+A,sinw,,t), sothatndifferent oscillations areimpressed onthesystem. (a)Find thesteady state motion. Hint. Usethesuperposition principle, seeComment 24.25; alsoExercise 19,6. (b)What isthemagniﬁcation ratio due totheinput mA1 sinw1t,to mA2 sinwgt,---,tom/1,, sinw,.tf Aglance atthedenominator ofeach magniﬁcation ratio term willshow that those terms with frequencies close towowillbemagniﬁed toamuch larger extent than those with frequencies farther away. Asystem ofthiskind thus actsasaﬁlter. Itresponds tothose vibrations with frequencies near woandignores those vibrations with frequencies notnear wo. 14-.InExercise 28D, 14,weintroduced thediscontinuous unitimpulse function 1f(t): 5.0§t§b, 0,t>b. Solve theequation dzu do _Ed" 25+ 211'f(t)) foryasafunction oft,where f(t)istheabove function andinitial conditions aret =0,y=0,y’=0.Hint. First solve withf(t)=1/b.Find y(b)and y’(b). Then solve theequation withf(t)=0andinitial conditions t=b, 2/=1/(5),J1//dl =y'(b)-l5.Solve problem 14if f(t) {e,0§t_l. 0. t>1. Hint. Seesuggestions given in14. 368 PROBLEMS LEADING 'roLINEAR EoUA'r1oNs orORDER Two Chapter 6 ANSWERS 29B 5.w=\/(.502 —2r2,thesame value ofwthatmakes theamplitude amaxi- mum, see(29.53); M(w)=1/2r\/woz -—r2. 5 .6.(a)y,=E sin(2t—a),where oz=Arctan§, 5\/41,-1r,2rad/sec. (b)1\/41. (c) _, F — _7.(a)y=ce‘cos(\/woz —r2t—|- 6)+ . with V(@102 -912)”+(2"-")2 1-2<wozandozgiven by(29.45). (b)First term onright of(a);second term onright of(a).(c)Same as(29.55). s.(5)5=5.(5)5=vs.-1 9.(a)y=c1e_2' —|—cge_3' ——326-(sin2t+3cos2t). (b)e-‘sin2t;solution y(t)asgiven in(a). (c)Each term iny(t)approaches zero ast—>w. The complementary function isnotoscillatory; theparticular solution, however, isdamped oscillatory since y—>0ast—>w. 10.y=c1e“2‘+ cge"3‘+ e‘—+wast—+ 90. -4: 11.(5)y=553-(sin4:+s5544:)+1150S1112:-455521). (b)First term in(a);second term in(a). (c)V65/13, 1rsec,1/1rcps. (d)8v65/65. 12.y,=sin4t;4. 13(a)y_ A1sin(w1t _011) +___+ A,.sin (w..t ——an) ,I P W V(~02—@112)” +(21191)? V(woz——491.2)’ +(2fw»)2 where a,;,i=1,---,n,isdeﬁned asin(29.45). 2 2O00 010,(b) . 7 \/(wo2 ~—w12)2 +(2rw1)"’ \/(woz —401.202 +(2Tw»)2 14. 1 _.%[l—e '(s1nt+cost)], 0§t§b. = —l y 27[{eb(sin b+cosb)—1}sint +{eb(cosb -sin5)-1}55541, 1>5. 15. 5-‘(1-cost), 0gzg1. 1/=e_'[sin1sint+ (cosl —1)cost] =e"'[cos (t——1)—cost], t>1. Lesson 30A SIMPLE ELECTRIC Cmcmr 369 LESSON 30. Electric Circuits. Analog Computation. ByNewton’s laws ofmotion wewere able tosetuparelationship among active forces inamechanical system. Analogous laws, known asKirch- hoff’s (1824-1887) laws, make itlikewise possible forustosetuparela- tionship among those forces which supply anduseenergy inanelectrical system. InLesson 30A below, westate oneofthese laws andapply itto asimple electric circuit. LESSON 30A. Simple Electric Circuit. Inthesimple electric circuit which wehave diagrammed inFig. 30.1, thesource ofenergy inthecir- cuitismarked E.Itmay beacell, battery, orgenerator. Itsupplies the energy intheform ofanelectrical ﬂow ofcharged particles. Thevelocity oftheparticles iscalled acurrent. However, theenergy source will produce thisﬂow only when thekeyatAismoved toB.The circuit is then saidtobeclosed. Theelectromotive force ofthebattery orother source ofA energy, usually written asemf, isdeﬁned B R asnumerically equal totheenergy sup- plied bythebattery orsource when one L unitcharge iscarried around thecomplete C circuit. Forexample, ifthree units of energy aresupplied byasource when one unit charge iscarried around thecomplete Figure 30.1 circuit, then itsemfisthree units. There are,fortheelectrical system, asinthemechanical one, different systems of units inuse. Intheoneweshall adopt, theunit ofemfiscalled avolt. The other three elements inthecircuit labeled R,L,andCareusers of energy. Innontechnical terms, thismeans that acertain amount ofenergy isneeded tomove theelectrical ﬂow ofcharged particles across these barriers. Weexpress theenergy each uses bygiving thevoltage drop across it.* From thephysicist, welearn that: (30.11) thevoltage drop across aresistor (Rinﬁgure) =Ri, thevoltage drop across aninductor (Linﬁgure) =L%%, thevoltage drop across acapacitor (Cinﬁgure) =éq, ‘The voltage drop across each element iseasily measured bymeans ofaninstrument called avoltmeter. Alloneneed doistoconnect onewire ofthevoltmeter tooneside oftheelement, another wire totheother side, andthen read how farapointer moves. 370 Pnonm-ms LEADING 'roLmmn EQUATIONS orOnnnn Two Chapter 6 provided : theresistance Roftheresistor ismeasured inohms, thecoefficient ofinductance Loftheinductor ismeasured in henrys, thecapacitance Cofthecapacitor ismeasured infarads, thecharge qinthecircuit ismeasured incoulombs, thecurrent iinthecircuit, which isdeﬁned tobetherateofchange ofthecharge q,orthevelocity ofq,i.e., (30.12) i=‘%, ismeasured inamperes. The resistor, asthename implies, resists theﬂow ofthecharged par- ticles, andthus energy isneeded tomove theparticles across it.The inductor’s jobistokeep therate ofﬂow ofthecharged particles asnear constant aspossible. Itthus opposes anincreas; oradecrease inthe current. The capacitor stores charged particles and thus interrupts the electrical ﬂow. When theaccumulated charges become toonumerous for itscapacity, thecharged particles leapacross thegap(that iswhen the spark occurs) andtheparticles then continue their course inthecircuit. Kii-ehhoﬂ"s second lawstates thatthesumofthevoltage drops ina closed circuit isequal totheelectromotive force ofthesource ofenergy E(t). Hence, by(30.11), . dz 1(30.13) R1+La +ﬁg=E(t). By(30.12), wecanwrite (30.13) as dzq dq1_ which isthedifferential equation ofmotion thecharge qinthecircuit asafunction ofthetime t. Toﬁnd thecurrent iinthecircuit asafunction ofthetime t,wecan either solve (30.14) forqandtake itsderivative, orwecandifferentiate (30.13) toobtain, with thehelp of(30.12), thedifferential equation d2i dz"1. <1 andthen solve (30.15) fori. Assume (30.16) E(t) =Fsin(wt+B);therefore %E'(t) =Fa:cos(wt+I3). Lesson 30A SIMPLE Etacrruc Cmcurr 371 Then (30.15) becomes dz‘ d‘1.(30.17) Ld—;+R;:+5@=Fwc0s(w1+)s), dz‘ Rd‘ 1. F#+ZF€+C7Jz=Twcos(wt+ﬁ). Itssolution, byanymethod youwish touse,is,assuming theroots ofthe characteristic equation areimaginary, u _ 0 i (30.18) 1=Ae""“"S111( z +as < 1'. > +"’l (R100)? +(1-0L..,2)=' < 1', > Let(Fig.30.211) (30.19) sin.1FCRwC' sin(wt+13)+(1—CLw2) cos(wt+6)]_ = 1—C'Lw2 _ V(RwC)2 —|—(1-—CLw2)2 ROJCcosa= —-—-—--L- V(RwC)2 +(1——CLw2)2 Then the1',part of(30.18) canbewritten as . FC(30.2) 1,,=-—-————"',i \/(RwC)2 +(1——CLw2)2 X[sin(wt+19)cosoz+cos(wt+)3)sina] = FwC \/(Rw€')2 +(1—(»'Lw2)2 Hence thesolution (30.18) becomes _,R,,L,, (x/4CL -R202[sin(wt+I9+01)]- < 2,, > +—--8sin(<»¢+ 0+0‘)~/<R~v>* +(1—CL~2>2 < '1', >- Thecurrent inthecircuit, therefore consists oftwoparts, adamped harmonic motion duetothe1'.part ofthesolution andasimple harmonic 372 Pnontans LEADING TOLINEAR EQUATIONS orORDER Two Chapter 6 motion duetothe1,,part. Asinthemechanical case, thepresence ofthe damping factor e_‘R/21')‘ causes thecurrent duetothe1',part ofthesolu- tion todieoutintime. [Ifwehadassumed realroots ofthecharacteristic equation of(30.17), instead ofim- “2 aginary ones, thecurrent duetothe /Q11”) 1',part ofthesolution would stilldie G“*0 2outintime. Seethesolutions for (Y~°‘ 1'CL“ each ofthedifferent cases inthecor- responding mechanical case, Lesson Rwc 29A.] The 1'.part ofthesolution is therefore called appropriately the Figure 30-211 transient current. The current equation (30.21) willthus beacom- plicated oneonly forthetime inwhich thetransient current iseffective. Thereafter thecurrent will bedetermined entirely bythe1',part ofthe solution. The 1',current hastherefore been named, also appropriately, thesteady state current. Comment 30.212. Asinthemechanical case, thefunction E(t) of d(30.14) oritE(t)of(30.15) iscalled theinput ofthesystem; thesolution ofeach equation theoutput oftherespective system. By(30.21) and (30.16), weseethat thesteady state current hasthe same frequency asthat oftheenergy source E(t), namely wrad/sec, but isoutofphase with it. The amplitude ofthesteady state current is,by(30.21), (30.22) A=\/(ROW _GLOW =\/(R2+(F1“)2.E_ The denominator ofthelast expression in(30.22) iscalled theim- pedance Zofthecircuit. When itsvalue isaminimum, theamplitude Aisamaximum. Toﬁndthevalue ofwthatwillmake Zaminimum, for ﬁxed R,C,andL,wedifferentiate theimpedance equation (30.23) z=,fR2+ -La)’ with respect towandsetdZ/dw equal tozero. The result is[square Zin (30.23) andthen differentiate with respect tow] (30.24) 0=2(i-L...)(-C%-L), Lesson 30A SIMPLE ELECTRIC Cmcurr 373 from which weobtain (30.25) .12= 1..=\/1/CL. Forthisvalue ofw,theimpedance Zofthecurrent willbeaminimum, theamplitude Awillbeamaximum, andasinthemechanical system, we saytheelectromotive force isinresonance with thecircuit. Substituting intheﬁrst equation of(30.22), this resonant value of wz=1/CL asgiven in(30.25), weobtain forthemaximum value ofthe amplitude, F(30.20) A_E- From (30.26) weobserve that when resonance occurs, themaximum value oftheamplitude Aisinversely proportional totheresistance R.Hence when Rissmall, themaximum value ofAislarge, andwhen Rislarge, themaximum amplitude issmall. The condition ofresonance therefore is always dangerous unless theresistance Rissufficiently large toprevent a breakdown ofthecircuit. And ifR=0,abreakdown isbound tooccur. IfweﬁxF,R,L,andw,then by(30.22), theamplitude Aofthesteady state current isafunction ofthecapacitance C.IfC=0,A=0,and ifCisadjusted sothat \/1/CL isequal tothefrequency toofE(t), then Awillbelargest. Hence byadjusting C,wecanmake theamplitude of thesteady state current small orlarge. Inapublic address system when wewant alarge ampliﬁcation ratio [this means, by(29.54), that wewant theamplitude Aofy,,toberelatively large], orinahome radio setwhen wewant alower ampliﬁcation ratio, weadjust thecapacitance Caccord- ingly byturning adial. Example 30.27. Acapacitor whose capacitance is2/1010 farad, an inductor whose coefficient ofinductance is5%henry, andaresistor whose resistance is1ohm areconnected inseries. Ifatt=0,1=0and the charge onthecapacitor is1coulomb, ﬁnd thecharge andthecurrent in thecircuit duetothedischarge ofthecapacitor when t=0.01second. Solution. Here E(t) =0,C=ﬁn, L=216,and R=1.Hence (30.14) becomes 1.1’ .1 4’ .1(2)ﬁat-§+17,%+305q=0, -at-§+20j§+10,100q=0. Itssolution is (b) q=e_‘°‘(c1 sin100t+C2cos1001). 374 PROnLEMs LEADING T0LINEAR EQUAr1oNs orORDER Two Chapter 6 Therefore (c) 1'=%%=—10e_1°'(c1 sin1001 —|—C2cos1001) +1-‘°‘(100¢, cos1001-1001,sin1001). Theinitial conditions aret==0,q=1,1'=0.Substituting these values in(b)and(c),weobtain 1ZC2, 0= —IOC2 +10061, C1= In(b)and (c)replace c1and C2bythese values. Then when t=0.01 there results (e) q(0.01) =e_°'1(0.1sin1 +cos1)=0.57 coulomb, 1'(0.01) =-101-°-1(0.1 sin1+cos1)+1-°~1(10 cos1—100sin1) =——76.9 amperes. Thenegative current indicates that thecondenser isdischarging, i.e., thecharged particles aremoving inadirection opposite totheonein which they moved when thecapacitor wasbeing charged. Example 30.3. Tothecircuit oftheprevious problem isadded a source ofenergy whose electromotive force E=50sin120t. Change the capacitance ofthecapacitor to2X10"3 farad. Att=0seconds, the switch isclosed. Ifatthat instant there isnocharge onthecapacitor and nocurrent inthecircuit, ﬁnd: 1.The equation ofmotion ofthesteady state current after theswitch isclosed. 2.Theamplitude ofthesteady state current. 3.Thefrequency ofthesteady state current. 4.The value ofthecapacitance which willmake theamplitude ofthe steady state current amaximum. d dSolution. Here EE(t) =E(50sin1201) =6000 cos120t. Using the ﬁgures forL_andRasgiven inExample 30.27 andofCasgiven above, thedifferential equation ofmotion (30.15) becomes 2- - 3 (a) %)%+g+ =6000 cosl20t, .12 .1 .at-Z+20It+1041=120,000 cos1201. Itssteady state solution is (b) 1',=11.5sin120t —21cos120t. Lesson 30B ANALOG COMPUTATION 375 ByComment 28.32, theamplitude ofthesteady state current is (c) A=\/11.52 +212=23.9. Thefrequency <11ofthesteady state current is120rad/sec, thesame as thefrequency ofthesource ofenergy E(t). By(30.25), theamplitude ofthesteady state current willbeamaximum ifChasavalue such that\/1/CL =w,i.e.,when 1 20 1 LESSON 30B. Analog Computation. Werecopy below thedifferen- tialequation ofmotion (28.63) ofamechanical system withthecoefficient ofresistance andforcing function terms added, andthedifferential equa- tions (30.14) and(30.15) ofanelectrical system. 2 (30.4) mag +r%+Icy=Fsinwt. d’q dq1_(30.41) LE; +Ra? +6q-E(t). 1 . 2 (30.42) L%+11%+=;%[E(t)]. When placed underneath each other inthismanner, thesimilarity in form ofthetwosystems isstriking. Itshould beevident toyouthatif inanelectric circuit, Fig.30.43(a), weinsert aresistor R=r,aninductor L=m,acapacitor C=1/k,andasource ofenergy E=Fsinwt(or R=' k=spring constant L=7'" jForcing function =Fsin(0)t) C,_1_ Dashpot whose 71 j jcoefficient of resistance isr (11) (b) Figure 30.43 —Fw coswt),thesolution qof(30.41) [or1'of(30.42)] willbethesame asthesolution yof(30.4). Bysolving theelectrical system, itisthen possible todetermine themotion ofacorresponding mechanical system, suchastheonepictured inFig.30.43(b). Since itisusually lessexpensive 376 PROBLEMS LEADING roLINEAR EQUATIONS orORDER Two Chapter 6 andeasier tosetupasimple electric circuit than itistoconstruct ame- chanical system, theimportance ofthisfortunate coincidence should be evident toyou. This method, which isnow well developed, ofcomputing themotion ofamechanical system from asimple electric circuit isknown asanalog computation. However, because ofthecurrent accessibility tohigh-speed digital computers, themost accurate and least expensive method atpresent of computing themotion ofamechanical system istousesuch acomputer. EXERCISE 30 1.Verify theaccuracy ofthesolution of(30.17) asgiven in(30.18). 2.Verify theaccuracy ofthesolution (b)ofExample 30.27. 3.Verify theaccuracy ofthesolution (b)ofExample 30.3. Intheproblems below, itisassumed, when notexplicitly stated, that thecoefficient ofinductance Loftheinductor ismeasured inhenrys, the resistance Roftheresistor ismeasured inohms, thecapacitance Cofthe capacitor ismeasured infarads, thecharge qisincoulombs, thecurrent 1' isinamperes andtheemfofthesource ofenergy isinvolts. 4-.Iftheemfi.e.,ifthesource ofenergy, ismissing from thecircuit, then the differential equations (30.14) and(30.15) become respectively (305) 1.!’i1+1a@+lq=0" .112 110 ’ (3031) L@+1zi‘+l'—0' 1112 1110*‘ ' (a)What isthenatural (undamped) frequency ofvibrations ofcurrent and charge? Hint. SetR=0. (b)Forwhat values orRwillthecharge andcurrent subside tozerowithout oscillating; forwhat values ofRwillthey oscillate before subsiding to zero? (c)Find thegeneral solutions forqand1'asfunctions oftimeifR2=4L/C. Towhat mechanical caseisthissituation comparable? (d)Find qand1'asfunctions oftimeifatt=0,q=qoand1'=0.Assume R’<4L/C. 5.Foracertain LRC electric circuit, L=Q,C= (a)Forwhat values ofRwillthecurrent subside tozerowithout oscillating after theemfisremoved from thecircuit; forwhat value ofRwillit subside tozerowith oscillations? (b)What isthenatural (undamped) frequency ofthesystem? 6.Acapacitor whose capacitance isl0'5 farad, aninductor whose coefficient ofinductance is10henrys, andaresistor whose resistance is3ohms are connected inseries. Att=0,1'=0andthecharge onthecapacitor is 0.5coulomb. Find thecharge andcurrent inthecircuit asfunctions oftime duetothedischarge ofthecapacitor. Lesson 30—Exercise 377 7.Ifaresistance ismissing from thecircuit, then by(30.14) dzq q(30.52) LE5—|—6-E(t). Equation (30.52) isthedifferential equation oftheharmonic oscillator for theelectric current andcorresponds totheforced undamped motion ofthe mechanical system, seeLesson 28D. (a)Solve forqand1asfunctions oftime ifE(t) =0andt=0,q=qo, 1'=0.What isthenatural (undamped) frequency ofthesystem? (b)Solve forqand1'asfunctions oftime ifE(t) =aconstant emfEand t=0,q=0,i=O. (c)Solve forqand1'asfunctions oftime ifE(t) =Esin wtandt=0, q=0,1'=0(two cases). What value ofwwillproduce (undamped) resonance? 8.(a)Find qand1'asfunctions oftime ifin(30.52) C=104, L=1, E(t) =100,andatt =0,q=0,1'= 0. (b)What isthenatural (undamped) frequency ofthesystem? (c)What isthevalue ofthecurrent when t=0.02sec? 9.(a)Find qand1asfunctions oftime ifin(30.52) C=10", L=1, E(t) =100sin50t andatt=0,q=0,1'=0. (b)What isthevalue ofthecurrent when t=0.02sec? (c)What isthemaximum value ofthecurrent? (d)What isthenatural (undamped) frequency ofthesystem? 10.Ifthecapacitance ismissing from thecircuit, then by(30.13), (30.53) L%:+R1=E(t). (a)Find 1asafunction oftifE(t)isaconstant emfEandatt=0,1'=0. What isthetransient current, thesteady state current? (b)Find 1'asafunction oftifE(t) =Esinwtandatt=0,1'=0.What is thetransient current, thesteady state current? 11.Find 1'asafunction oftifin(30.53) R=20,L=0.1,and (a)E(t) =10, (b)E(t) =100sin50t. 12.Aninductor ofLhenries, aresistor ofRohms, andacapacitor ofCfarads areconnected inseries toabattery whose emfisEvolts. (a)Find qand1'asfunctions oftime. Assume R2<4L/C. (b)What isthefrequency ofthetransient charge andcurrent? (0)Isthere asteady state charge, asteady state current? 2 Him.By(30.14), 11..differential equation isL53;+Rg11%q=E. 13.(a)Findqandias functions oftimeifin(30.14) and(30.15),L =1,R=5, C=10*‘, E(t) =50,andatt =0,when theswitch isclosed, q=0, 1'=0. (b)What isthefrequency ofthetransient charge andcurrent? (c)What isthesteady state charge? 14.(a)Find the steady state current if,in(30.15), L=%, R=5,C= 4X10-4, dE/dt =200cos1001, andifatt=0,when theswitch is closed, q=0,1=0. 378 PRosLEms LEADING roLINEAR EqUA'rIoNs orORDER Two Chapter 6 (b)What istheamplitude andfrequency ofthesteady state current? (c)Forwhat value ofthecapacitance willtheamplitude beamaximum? (d)What should thefrequency oftheinput E(t)beinorder that itbein resonance with thesystem? (e)What isthemaximum value oftheamplitude forthisresonant fre- quency? (f)What istheimpedance ofthesystem? 15.Find thesteady state charge andthesteady state current if,in(30.14) and (30.15), L=-213,R=20,C=10-4, E=100cos2001. Inregard tothesteady state current, answer allquestions (b)to(f)of14. 16.In(30.15), let E(t) =E;sin1011+ E2sinwgt—[--~-—|—E,sinw,,t, sothatndifferent frequencies areimpressed onanelectric system. (a)Show thatthesteady state current is E .(20.54) 1.=-——“’-’lC'———— sin<<».1+<1.)+---V(Rw1C')2 +(1-(3'I/1012)” +————-l"—""’—"C——— sin(11.1+11.).V(Rw,.C)2 +(1-——CLw,,2)2 where 01;,1'=1,---,n,isdeﬁned asin(30.19). Hint. Usethesuper- position principle, seeComment 24.25; alsoExercise 19,6. (b)Show that theamplitude ofthesteady state current duetotheinput E),sinoutis Ak= . Weproved inthetextthat A1.willbelargest when \/1/CL isequal to thefrequency wk.Hence byadjusting Cuntil \/l/CL =1.0),,wecanmake theamplitude oftheresponse oroutput duetotheinput E1,sinwit larger than theamplitudes duetotheother inputs. Theelectrical system willthus actasaﬁlter, responding tothose inputs whose frequencies arenear \/1/CL andignoring those inputs whose frequencies arefarther away. Iftheinputs, forexample, arecoming from different radio stations which arebroadcasting atdifferent frequencies, youtune your radio to oneofthem byturning adialandadjusting thecapacitance until the amplitude oftheoutput isgreatest forthatstation's input. Theampli- tude A1alsohasE1,inthenumerator. Hence forgood reception from station k,youwould want itsE),tobelarger, i.e.,more powerful, than theEofother stations andthefrequencies oftheother stations tobe nottooclose towk.Compare thisproblem with Exercise 29B, 13. ANSWERS 30 4-.(a)\/1/CL rad/sec. (b)Nooscillations ifR2Z4L/C, oscillations if R2<4L/C. (c)y=e‘R’/“(C1 +Cgt), critically damped case. Nora. Here y=qor1'. Lesson 30—Exercise 379 L ._ . 1 R2 <°‘>4=2q°\/fan?“ '“'“S‘“(\/E "mi”‘)' 8=‘Arctan‘ -1;i =dq/dt. 5.(a)Rg40,nooscillations; R<40oscillations. (b)40rad/sec. 6.q=ﬁe-3‘/2° sin(100! —|—8)approximately, where 6=Arctan(2000/3) ap- proximately; i=—-ﬁ;e_3‘/2° sin(100t +5)+50e-3‘/2° cos(100t +8). 7.(a)q=qocos\/1/CLt;i =—\;—i)isin\/1/CLt,\/1/CLrad/sec. (b)q=CE(1 —cosvl/GL1), 1'=dq/dt. (c)q— CE (sinwt w\/C'Lsin\/1/CLt), waé1/\/CL;'1-cL@2 " 1'=dq/dt, EC’ 1 E l=—-—' ————t-—— C’/Lt os——t, =1/\/CL; q 2sin\/ai 2V c CL w i=dq/dt; w=1/\/CL rad/sec. 8.(a)q=Thu —-cos100t),i =sinl00t. (b)100rad/sec. (c)0.909 amp. 9.(a)q=71§(sin 50t—§sin100t);i =§(cos 50t—cos100t). (b)i(0.02) =0.638 amp. (c)max =ﬁamp. (d)100rad/sec. 10.(a)i=%(1-—2-1“/L);i¢ =-—ge_R'”‘;i, = _ E , _(b)1.=ﬁg’; (Rsinwt—Lwcoswt+Lwe RM‘), ._ ELw -mil. "'R2_|_L2,_,,2 e ' —L (R't—Lot) 1,-R2+L2w2 sinw wcsw . ll.(a)i=}(1-—e-2°°‘). (b)i=-f-‘H4 sin50t—-cos50t+e'2°°‘). 12.(a)q=q,+EC,where q,isthesame as1',in(30.18), .dq13 E > (b)\/4CL —R2C2/41rCL cps. (c)q,=EC’.There isnosteady state current. Intaking thederivative ofq, theconstant ECvanishes. 13.(a)q=—10“3e-2-5‘(0.125 sin99.97t +5cos99.97t) +0.005, i=dq/dt =0.500e_2-5‘ sin99.97t. (b)99.97/21r =15.9cps. (c)0.005. Forashort time thecharge onthecapacitor willoscillate about thisﬁgure andapproach thisﬁgure ast—>°°._ 14.(a)11,=§5(sin100t+ 4cos100t). (b)-3%\/T7, 100/21r cps. (c)C’=2X10-3 farad. (d)100x/5 rad/sec. (e)2/5amp. (f)5x/T7 ohms. 380 Pnonusms LEADING T0LINEAR EQUATIONS orORDER Two Chapter 6 15.(a)q,_=5X10_3(sin 200t+2cos200t), i,=cos2008 —2sin200t. (b)\/5,200/21r cps. (c)5X10-4. (d)200x/5. (e)5amp. (f) 20%. LESSON 30M. MISCELLANEOUS TYPES OF PROBLEMS LEADING TO LINEAR EQUATIONS OF THE SECOND ORDER A.Problems Involving aCentrifugal Force. When abody is whirled inacircle attheendofstring, aforce, directed toward thecenter ofthecircle, must beexerted toprevent thebody from ﬂying off;the faster therotation, themore powerful theforce. Since this force isdi- rected toward thecenter ofthepath, ithasbeen called thecentripetal force orcentral force. And since thebody remains initspath, there must beanoutward force intheopposite direction equal tothecentral force. This force iscalled thecentrifugal force. Ithasbeen proved that thecentripetal force required tohold amass minacircular path of radius r,moving with alinear velocity vis 2 (30.6) C.F.= Hence thisformula must also give thecentrifugal force ofthemass m. The linear velocity voftheparticle isv=1'd0/dt, where 0isthecentral angle measured inradians through which theparticle isrotated. Substi- tuting thisvalue ofvin(30.6), weobtain __ m 2_ 2 where wistheangular velocity oftheparticle. With thehelp of(30.61), solve thefollowing problems. 1.Asmooth straight tube rotates inavertical plane about itsmid-point with constant angular velocity w.Aparticle ofmass minside thetube isfreeto slide without friction. (a)Find thedifferential equation ofmotion oftheparticle. Hint. There aretwoforces acting ontheparticle attime t,seeFig.30.62. (b)Solve theequation withl =0,r=10,dr/dt =vo. (c)From theintroductory remarks, itisclear that iftheparticle istoofar from 0orifdr/dt istoogreat, theparticle willﬂyofffrom anendofthe tube; forcertain values ofrand dr/dt, itwillnot. Find values ofthe initial conditions roand vo=dr/dt sothat theparticle will execute simple harmonic motion. Write theresulting equation ofmotion for these values. Can youidentify it?Draw theﬁgure. 2.Solve problem 1,ifthetube rotates inahorizontal plane about avertical axis. Assume att =0,1‘ =0anddr/dt =vo. 3.Solve problem 1,ifatt =0,r=0,dr/dt =0. Lesson 30M B.Rotmno Booms 381 C.F.=mrw2 5r‘/0 0=0 att=0 mg mgsin0 Figure 30.62 B.Rolling Bodies. Newton’s ﬁrst lawofmotion states that abody atrest ormoving with uniform velocity will remain intherespective state ofrest ormotion unless aforce acts onit.Wesaythebody has inertia, i.e., itresists having itsstatus changed. Similarly, abody at rest orrotating about anaxis with aconstant angular velocity willre- main intherespective state ofrest orrotation, unless atorque ora moment offorce acts onit;fordeﬁnition oftorque, see(30.63) below. Inthiscase wesaythebody hasrotational inertia, alsocalled moment ofinertia. Bydeﬁnition, themoment ofinertia Iofaparticle is (30.621) I=mx2, where misthemass oftheparticle located xunits from theaxisofrotation. Inthecalculus, youwere taught how tocalculate themoment ofinertia of different bodies. Forexample, forasolid cylinder ofradius randmass m rotating about anaxiscoinciding with theaxisofthecylinder, I=mr2/ 2. Itisasiftheentire mass ofthecylinder were concentrated atadistance r2/2 units from theaxis. Wedeﬁne thetorque ormoment offorce Lasfollows, seeFig.30.631. (30.63) L=2:times thecomponent oftheforce Facting at right angles tothelinejoining theaxis ofrota- tion and thepoint Pwhere Fisbeing applied; :1:isthedistance between theaxisandP. O x P / Component ofFat right angles toOP Axis ofrotation ‘Fl 1toplane ofpaper Figure 30.631 382 Paosnnus LEADING 'roLINEAR Eouxrrons orORDER Two Chapter 6 Finally, ithasbeen proved that corresponding tothelawF=mass X acceleration governing thelinear motion ofabody, thelawgoverning the rotational motion ofabody isgiven by _d20__ dw where aistheangular acceleration ofthebody, wisitsangular velocity, and0isthecentral angle through which thebody hasrotated from 0=0. With thehelp ofequations (30.62l) to(30.64), solve thefollowing problems. 4-.Acordiswound afewturns around asolid cylindrical spool ofmass mand radius r.Oneendofthecordisattached totheceiling. SeeFig.30.65. At t=0,thespool, which isbeing heldagainst theceiling with axishorizontal, isreleased. Cord attached here _A Ceiling t=0’y=0‘0=0 F P F y=re r r0 P ‘"'47 Figure 30.65 (a)Ifgravity istheonlyacting force, ﬁndthedifferential equation ofmotion ofthespool. Hint. ByNewton’s law,mass Xacceleration ofbody must equal thenetforce acting onthe‘body. These forces areFandmgas shown inFig.30.65. By(30.63), (30.64) andthefactthat I=mrz/2 forasolid cylinder rotating about itsaxis, wehave mrzd20FT =I0! =7 When thecylinder hasrolled through acentral angle 0sothatthepoint ofthespool initially atAisnowatB,thedistance yfrom theceiling isr0. Therefore, 2 2 2 at, aaaa1.11, ”="’»a.?=’aa' amt? HenceF =(-m/2)(d2y/dt’). (b)Solve thediﬁerential equation foryasafunction oft.Remember at t= 013/ = = 5.Answer questions (a)and(b)ofproblem 4ifthere isaresisting force dueto friction andairof(m/80) (dy/dt). (c)What isthelimiting velocity? Lesson 30M D.Bnnnmo orBalms 383 6.Att=0,asolid cylinder ofradius randmass misplaced atthetopofan incline andreleased, Fig.30.66. Assume itrollswithout slipping andthata frictional force Factstooppose themotion. B L=Ia; 8=7‘0 sI5"‘AZF \s=0, 0=0 O, mgsin a. Figure 30.66 (a)Find thedifferential equation ofmotion. Hint. The only difference between thisproblem and4isthatmgisreplaced bymgsin0:. (b)Solve thedifferential equation. Remember att=0,0=0,ds/dt =0. C.Twisting Bodies. When aspring isstretched, aforce results, proportional totheamount ofstretch, that tries torestore thespring to itsoriginal natural length. Similarly, when ahanging wire istwisted by rotating abobabout itasanaxis, where thebobisrigidly attached toit atoneend,atorque ormoment offorce results thattries torestore the wire toitsoriginal position. This torque Lis,inmany cases, proportional totheangle 0through which thebobisturned. By(30.64), therefore, 2 (30.67) 1%’=-1.0, where kiscalled thetorsional stilfness constant. Thenegative sign isnecessary, because when 0isturning clockwise, thetorque actscounter- clockwise; hence torque and0have opposite signs. 7.(a)Solve (30.67) for0asafunction oftime ifthetorque isequal inmag- nitude totheangle 0,i.e.,k=1. (b)Ifthebobreturns toitsequilibrium position attheendofeach §second, ﬁndthemoment ofinertia ofthebobwith respect tothewireasanaxis. Assume themass ofthewireisnegligible. D.Bending ofBeams. We consider abeam with thefollowing properties. 1.Itisrelatively longincomparison with itswidth andthickness. 2.Every cross section isuniform. 3.Thecenter ofgravity ofeach cross section liesonastraight line,called theaxisofthebeam. Itisthelinejoining (0,0) to(L,0) inFig.30.7. 384 PROBLEMS LEADING TOLINEAR Equxrrons orORDER Two Chapter 6 Ifabeam merely rests onsupports atitstwoends, itiscalled asimple beam anditissaid tobesimply supported attheends. Ifabeam is supported only atoneend, asforexample, when itisembedded inmasonry atoneendandhangs freely attheother end, itiscalled acantilever beam. InFig.30.7, wehave drawn asimple beam oflength Lftwith rectangu- larcross sections whose centers ofgravity lieinthegeometrical center of ((,,(,,A‘xY/';;_j"4.*_;1};;;;;;;/'2/*2; //_ Axis ¢/// A§=.o) (Ln)CA Figure 30.7 therectangle. (However, beams may have other shaped cross sections, as long asallcross sections areuniform andthecenter ofgravity ofeach lies onastraight line.) Wemay look onsuch abeam ascomposed ofﬁbers parallel totheaxis ofthebeam, each ofwhose length isLft.When a load isdistributed along asimple beam, asagdevelops sothat theﬁbers ononeside ofthebeam arecompressed andﬁbers ontheother side are ”’;;4 '/ /..‘/// ’ Q Figure 30.71 stretched, Fig. 30.71. Itfollows, therefore, that somewhere between the two sides, aneutral surface exists that isneither stretched norcom- pressed (shaded area inFig. 30.71), i.e.,itretains itsoriginal length L. Theintersection ofthisneutral surface with avertical plane through the xisofthebeam iscalled theelastic curve ofthebeam. Itisthecurve ilpining (0,0) to(L,0) inFig.30.71. 1Ithasbeen proved inmechanics thats(st (30.72) Ma)=FR5. where : M(x)isthebending moment atany cross section A,asunits from oneend ofthebeam. The bending moment atAis deﬁned asthealgebraic sum ofallthemoments offorce Lesson 30M D.BENDING orBmms 385 acting ononly one side ofAabout anaxis through the center ofthecross section A,marked CD intheﬁgure. (For deﬁnition ofamoment offorce, see(30.63) above.) Iisthemoment ofinertia ofthecross section Aabout its center axis CD (for deﬁnition ofmoment ofinertia, see (30.62l) above). Ristheradius ofcurvature oftheelastic curve ofthebeam. Eisaproportionality constant, called Young’s modulus or modulus ofelasticity. Itisdependent onlyonthematerial ofwhich thebeam ismade. The radius ofcurvature isgiven bytheformula R=[1+<y'>’1=*”/y". Itssubstitution in(30.72) gives (30-73) M(iv)=E11/"[1 +(2/')2l_3'2 =EIy"[1 -%(y')2 +“s§(1/’)‘ —-'-l- Since thebending isusually slight, y’isvery small. Hence itisnotunrea- sonable toassume that wecommit asmall error ifin(30.73) weneglect (1/)2 andhigher powers ofy’.Equation (30.73) thus simpliﬁes to (30.74) M(w)=Ely". which isthedifferential equation oftheelastic curve ofthebeam. Weshall arbitrarily assume that anupward force gives apositive moment andthat adownward force gives anegative moment. With thehelp of(30.74), solve thefollowing problems. 8.Ahorizontal beam oflength 2Lftissimply supported atitsends. The weight ofthebeam isevenly distributed andequals wlb/ft. (a)Find theequation oftheelastic curve. Hint. SeeFig.30.75. Thetotal weight ofthebeam is2Lwlb.Therefore theupward force ateach end Lw Lw %—_| P(I,°) (L0)(0.0) ' ' ’ (21-.0) wx -I 7(2L -x)w Figure 30.75 isLwlb.Since thebeam isuniform, wecanconsider theweight ofthe beam from (0,0) toP(a:,0) asconcentrated atitsmid-point (:2:/2, 0). Hence thedownward force atthismid-point iswa:lb.Thebending 386 PROBLEMS LEADING 'roLINEAR EQUATIONS orORDER Two Chapter 6 moment M(22)atPistherefore—remember thebending moment isthe algebraic sum ofallmoments offorce acting ononesideofthecross section Awhose axisgoesthrough P- 2 M(a:)=(Lw)a:-(wx) =Lwa:- Substitute thisvalue ofM(x)in(30.74) andsolve. Theinitial conditions are2:=0,y=0;1:=L,y’=0.(Note there isalsoathird initial condition x=2L,y=0.However, thethree arenotmutually inde- pendent. Useofanytwoofthethree willresult inasolution which satisﬁes thethird condition. Verify thisstatement.) (b)What isthemaximum sag? Hint. The maximum sagoccurs when 2:=L. (c)Show that thesame bending moment atPresults, iftheforces tothe right ofPwere used. Hint. Thebending moment atPduetotheforces ontheright is Ma)=Lw(2L -1)-[(2L-:2:)w] Simplify theright side. 9.Ahorizontal beam oflength 2Lftissimply supported atitsends andcarries aweight Wlbatitscenter. Iftheweight ofthebeam isnegligible compared toW,ﬁndtheequation oftheelastic curve andthesagatthecenter. See Fig.30.76. Two cases must beconsidered. K H’. K W2 2 2 T P(1,0) (L0) (L10) P(1.0) (0,0) (211.0) (0,0) (211.0) at-L W W 0%xéL Léxé2L Figure 30.76 Case 1.IfPistotheleftofthemid-point, theonly active force totheleft ofPisW/2.Hence thebending moment atPis (a)M(a:) =%,a:which canbewritten asQWL ——<}W(L —2:), 0§0:<L. Case 2.IfPistotheright ofthemid-point, then theactive forces totheleft ofPareW/2upward andWdownward. Hence thebending moment atPis (b)M(a:) =g:2:—W(a: ——L)which canbewritten asQWL +§W(L —2:), L<2:§2L. Cases 1and2cantherefore betreated asoneifwewrite (c) M(x) =QWL =F=}W(L —x), where itisunderstood that theminus signistobeused when 0§:c<L andtheplussignwhen L<2:§2L. When a:=L,(a),(b),and(c)are Lesson 30M D.BENDING orBmms 387 10. ll. 12.thesame. Hence thesolution obtained byusing (c)isalsovalid when 2:=L. Initial conditions are2:=0,y=0;:0=2L,y=0.(Athird initial condition :2:=L,y’=0isnotindependent oftheother two. Verify that itsatisﬁes thederivative ofthesolution.) Solve problem 9,iftheweight ofthebeam isnotnegligible andiswlb/ft. Hint. Follow alltheinstructions given in8and9.Asin9there willbetwo cases, onewhen Pistoleftofcenter, theother when Pistoright ofcenter. Thebending moment atPduetotheforces totheleftofPare 2 .'!~I(:z:)=(wL+%):c -% =wL2:—§w:z:2 -—}W(L—z)+<}WL, cg2:<L, ifPistoleftofthecenter, and W w:c2M(I) = it'—T— -—L) =wL:c-gm’+§W(L-1)+;WL, L<1g2L, ifPistoright ofthecenter. When :7:=L,both bending moments arethesame. Thus both cases can becombined ifyoutake M(:t) =wL:c —§w:r2 =F§W(L —zc)+QWL, where itisunderstood that theminus signistobeused when 0§2:<L andtheplussignwhen L<:0:§2L. Ahorizontal beam oflength 30ftissimply supported atitsends andcarries aweight of360lbatitscenter. Iftheweight ofthebeam isnegligible, ﬁnd theequation oftheelastic curve foreach halfbeam. What isitssag? Solve independently. Check your results with solutions given inproblem 9. Asimply supported horizontal beam oflength 2Lcarries aweight Wlb attached toitatadistance 2L/3 from oneend. Assume theweight ofthe beam isnegligible. Find theequation oftheelastic curve. Hint. Theend ofthebeam closer totheweight nowsupports 2W/3 lb;theother endsup- ports only W/3 lb.Two cases willbeneeded asin9and10,oneifPisto theleftofW,theother ifPistotheright ofW.Thebending moment atP using forces totheleftofPare M(a:)=%:r, 0§a:<%» ifPistotheleftofW,and 2W 2L 2LM(I) =3‘-Z—W($—?)! ? <$ ifPisto theright ofW.§2L, Initial conditions are1=0,y=0;:0:=2L,y=0.Note also, since the elastic curve iscontinuous at:4:=2L/3 andhasatangent there, thatwhen 21=2L/3, thevalue ofyandthevalue ofthederivative y’foreach ofthe twocurves must bethesame. These conditions areknown respectively as thecondition ofcontinuity ofthecurve andthecondition ofcontinuity 388 PROBLEMS LEADING T0LINEAR EQUA'rIoNs orORDER Two Chapter 6 oftheslope. Youwillneed tousethese facts inorder toevaluate some of theconstants ofintegration. Ahorizontal beam oflength 2Lftandofuniform weight wlb/ftisembedded inconcrete atboth ends. Find theequation oftheelastic curve andthe maximum sag. Take theorigin atoneendofthebeam. Here inaddition to theusual moments found inproblem 8,there isanadditional moment of force ateach endacting tokeep thebeam horizontal, i.e.,themasonry at each endprevents thebeam initsimmediate neighborhood from sagging. Callthisunknown moment offorce M.Theinitial conditions are1:=0, y=0;:c=0,y’=0;:c=L,y’=0;z=2L,y=0;x=2L,y’=0. There areﬁvesetsofinitial conditions. Useofthree, saytheﬁrstthree, will enable youtoevaluate Mandtheconstants ofintegration. Verify thatthe resulting equation satisﬁes theother twoinitial conditions. Solve problem 13,ifthebeam alsosupports aweight Watitscenter. Hint. Here, inaddition totheusual moments found inproblem 10,there isamo- ment Mateach end. Asinproblem 10,there aretwocases tobeconsidered, onewhen Pistotheleftofcenter, theother when Pistotheright ofcenter. Itwillbeeasier totreat each caseseparately instead ofcombining them as wedidin9and10.Initial conditions are:7:=0,y=0;2:=0,y’=0; 2:=L,y’=0;2:=2L,y=0;:r=2L,y’=0.Note thatthecondition :7:=L,y’=0,applies toeach case, since thecurve iscontinuous at:1:=L. There isonemore condition than youneed. However, itisnotindependent oftheothers. Verify thattheoneyouomit satisﬁes thesolution. (0,0) x (x,0) 2L-x(2L,0) y 2L-I > 2 P(x.y) Q(2L—x)w Figure 30.77 Acantilever beam oflength 2Landofuniform weight wlb/ft isembedded in concrete atoneend. Find theequation ofitselastic curve andthemaximum deﬂection. SeeFig.30.77. Inthiscase, itwillbeeasier toconsider moments offorce totheright ofP.Since thebeam isuniform, there isadownward force atthecenter ofPQ. And since thisistheonly acting force tothe right ofP,thebending moment atPis Ma)=—w(2L -1) =-(2L_1)’. Substitute thisvalue in(30.74). Initial conditions area: =0,y=0;:=0, y’=0. Acantilever beam oflength 2Landofnegligible weight supports aload of Wlbatitscenter. (a)Find theequation ofitselastic curve, thedeﬂection atitscenter, andits maximum deﬂection. Hint. SeeFig.30.78. There aretwocases tobe Lesson 30M D.BENDING orBEAms 389 considered. Take moments toright ofP.Initial conditions area:=0, y=0;:2:=0,y’=0.When Pistotheright ofcenter, there areno forces totheright ofP.Hence thebending moment M(x)=0atP. Youwillalsoneed tousethefactthatwhen 2:=L,thesolution yand theslope y’forthecase0§:1:<Lmust agree respectively with the solution yandtheslope y’forthecaseL<z§2L. (0,0) (x,0) (L,0) (2L,0) (0,0) (L,0) (x,0) (2L,0) P P W Figure 30.78 (b)Find themaximum deﬂection andtheequation oftheelastic curve ifthe weight Wwere placed attheendofthebeam. Hint. Here there isonly onecasetoconsider, namely, P-totheleftofW.Theonly force tothe right ofPcontributing tothebending moment M(zt)atPistheweight W. 17.Solve problem 16(a) iftheweight ofthebeam isnotnegligible andiswlb/ft. Hint. There willbetwocases asin16.Thebending moment atPwillbe thesumofthebending moments given in15and16.Andremember when 2:=L,thesolution yandtheslope y’must agree forthetwocases. 18.Ahorizontal beam oflength 2Lisembedded inconcrete atoneendandis simply supported attheother endwithboth endsatthesame level. Aweight Wissuspended atitsmid-point andthebeam itself weighs wlb/ft. Find the equation oftheelastic curve. Take theorigin attheembedded end. Hint. Weneed twocases, Case 1when Pistotheleftofthemid-point; Case 2 when Pistotheright ofthemid-point. SeeFig.30.79. Take moments to theright ofP.CallFtheunknown upward force at(2L,0). Initial condi- tions are:2:=0,y=0;2:=0,y’=0;:0=2L,y=0.And remember, when 1=L,thesolution yandtheslope y’must agree forboth cases. 2L—x M=F(2L—r) 2L_x M=F(2I--I) 2 —2—— (0.0) (E0) (11.0) (0.0) (13.0) (1,0)(211.0) P P (L_x) W(2L -X)!!! W _x)w Figure 30.79 19.Aspring board, ﬁxed atoneendonly, maybeconsidered asacantilever beam. Itisdesired thatitsmaximum deﬂection be1ftwhen a240-lb mansteps on theend. Iftheboard is20ftlongandweighs 5lb/ft, ﬁndthevalue ofthe constant EI. Hint. Theformula formaximum deﬂection isthesumofthe maximum deﬂections given in15andl6(b). Andremember inthese formulas Lisone-half thelength oftheboard. 390 Pnonmms LEADING T0Lmmn Eqnnrons orORDER Two Chapter 6 1. 2. 3. 4 5. 6 7. 8 9. 10. ll.ANSWERS 30M. 2 (a)m%=mrwz —mgsin wt. ul —wt ml —ut2 ——- -— . (b)r=roe ~28 +(wv;w2 g)e 26 -l-2:2 smwt 2 — . .=rocoshwt—|—— smhwt+#S111wt. (c)ro=0,vo= Resulting equation is =_9_-=L- r20,2S111wt 20,2sin0. Inpolar coordinates itistheequation ofacircle with center at(g/4:02, 1r/2) andradius g/4:02. U0 lot —ut r-2w(e e). r=If;(2sinwt—e"'+e"""). 2<1 dz<a>mg,-2?=m@—F, if=st.<1»)y=4:”/3. 3dzy 1dy_ <a>5E2"+E5-"~ (b)y=9e0og(e-" 12°-1)+80gt,v=—s0ge-" 12°+80g. (0)80g. d2s md2s 3dzs(a)mEt;=mgsina—F, F=5;i§» 5a§=gsina. (b)-S=(gsin<1)?/3. (a)0=61cos(\/1/1z+ a.(b)1=1/(6412). (a)Ely=53(4142-Z3-8L3),ogItg2L.24 (b)sag=5wL4/24EI rt. Ely=%[3La:2q=(L-a:)3-eL%+ L3],0§1g2L; Ely=lg(x2—3L2), 0;:c§L; W WEIy=T§(z2—3L2)+—6(L——a:)3, L§x§2L; sag=WL3/6E1 rt. Ely =sum oftheresults obtained in8and9;sag=sum oftheresults obtained in8and9. Ely =30:2:(:c2 ——675), 0§a:§15, =30a:(x2 —675) +60(15 ——a:)3, 15§a:§30; sag=202,500/EI ft. Lesson 30M—Answers 391 W1 2 212.Ely =8T(92: —20L ),0§:1:§2L/3; 3 Ely=%(9¢’-20L’) --l;1<1-?§) ,%;2;2L. 13.Ely”=M+I/war:- M=—L2w/3; Ely=%(4La:-J-4L’)=-5“-2@2(2L-@)2; sag=wL“/24111 rt. 14.Moment M=-—§wL2 —%; 2 EI;/"=(wL+5vK)x-—%+M, 0§a:§L,' w 3 4 22 W a 2 _EIy=—-(4L:2: -2: —4La:)+—(2:2: —-3L:c), 0§:2:§L,24 24 ,, W 1012Ely =wL+? :c—T—W(:c—L)+M, L§:t§2L, Ely=£1(4L¢“-1‘-41,21”) +Ll;[28-4(:r:-L)3-3L:|:2], Lg1§2L, sag=(wL4+WL3)/24EI. 1s.Ely=3[l6L4-s2L%-(2L-¢)‘1 =%(8La:3 —24L2x2 -—2:4). Maximum deﬂection =2wL4/EI. 16.(a)Ely=%[L3-313%-(L-:c)3] =%(:c3—-3L:c2), 0§a:§ L; Ely=gm“ -31,2»), L§1g2L. Deﬂection atmid-point WL3/3 EI;maximum deﬂection 5WL3/6EI. (b)Ely =%(:23-—6La:2), maximum deﬂection 8WL3/3EI. 11.EIy=3(sL¢“-24L'*’¢’ -1?4)'+Z($3-am’), 0g2:gL;24 6 E11,=$1(8L:c3_24L’¢’ -2:4)+lg(L3-3L2:c), Lg1;g2L. Deﬂection atmidpoint: (17wL4 +8WL3)/24EI ,'deﬂection atendpoint: (12wL4 +5WL3)/6E1. 392 PROBLEMS LEADING T0Lmnxn Equurons orORDER Two Chapter 6 5W18. F=£1011 +W ' Ely"=——W(L _y)-g(2L-y)’+F(2L-2:); Ely=3(10La:3 -12L2a:2 -21‘)+1'(1113-18Lx2),4s 90 0§1§L; Ely"=~g(2L-y)’+F(2L-1), Ely=1%(10La:3 -12L2a:2 -22‘) +g(16L3_48L2z+30La:2-sf),Lg1g2L. 19.140,000. Chapter 7 Systems ofDifferential Equations. Linearization ofFirst Order Systems LESSON 31. Solution ofaSystem ofDifferential Equations. LESSON 31A. Meaning ofaSolution ofaSystem ofDiﬁerential Equations. Inalgebra itisfrequently necessary tosolve asystem of simultaneous equations ofthetype (a)2:v+3y=5, (b):v2—3y=2, (c):c+3y—2z=l5, a:——y=15; :l:+y2=5; x—y+z=7, 3:c+2y-—z=12. Similarly, itisfrequently necessary tosolve asystem ofdifferential equa- tions ofthetype d d(d) x;f+y’;§+2w—3y=3l, dzx d dW+2y%+37’:-3x=e‘, where :0andyaredependent variables andtisanindependent variable. InLessons 33and34,wediscuss andsolve numerous physical problems which give risetosuch systems. Asolution ofanalgebraic system oftwoequations isapairofvalues of2: andysuch thatthispairsatisﬁes both equations. Forexample, x=10, y=-5isasolution of(a),since thispair ofvalues satisﬁes both equa- tions. Analogously wesaythat thepair offunctions x(t), y(t), each de- ﬁned onacommon interval I,isasolution ofthesystem (d),ifthispair satisﬁes both equations identically onI,i.e.,ifineach equation of(d)an identity results when 2:isreplaced by:c(t), ybyy(t), andtheir respective derivatives by:c’(t), y’(t), etc. The extension ofthemeaning ofasolution ofasystem ofthree ormore equations should beapparent. 393 394 Srsrsms. LINEARIZATION orFmsr ORDER Srsrsms Chapter 7 LESSON 31B. Deﬁnition and Solution ofaSystem ofFirst Order Equations. Deﬁnition 31.1. The pair ofequations (31.11) %=rm/,1), %=r.<x,y,1>, where f1andf2arefunctions of2:,y,t,deﬁned onacommon setS,iscalled asystem oftwo ﬁrst order equations. Asolution of(31.11) will then beapair offunctions x(t), y(t), each deﬁned onacommon interval I contained inS,satisfying both equations of(31.11) identically. Ageneralization ofthistypeofsystem isgiven inthefollowing deﬁnition. Deﬁnition 31.12. Thesystem ofnequations <31-13> %=r.<y.,y.,---,y.,1>, d '3? =f2(i|/111/21' ''21/mt): d 'ay?n' =.f1l(ylry2: '''1it/ﬂat): where f1,''',f..areeach functions ofyl,1/2,---,y,,,t,deﬁned onacom- mon setS,iscalled asystem ofnﬁrst order equations. Deﬁnition 31.14. Asolution ofthesystem (31.13) isasetoffunc- tions y1(t), y2(t), ---,y,,(t), each deﬁned onacommon interval Icon- tained inS,satisfying allequations of(31.13) identically. Comment 31.141. InLesson 62,youwillﬁndacriterion, Theorem 62.12, which gives asuﬂicient condition fortheexistence anduniqueness ofasolution ofthesystem (31.13) satisfying theninitial conditions, (31-15) 3/1(lo) =111, 3/2(to) =112.'''1y.(1o) =fl»- Comment 31.16. InLesson 62B, weshow how anonlinear differen- tialequation oforder greater than onecanbereduced toasystem ofﬁrst order equations. If,therefore, wecandevise methods forﬁnding solutions ofthesystem (31.13), then theoretically anynonlinear differential equation canbesolved. InLesson 34,infact, wesolve certain special types ofsecond order nonlinear differential equations byreducing them toasystem of twoﬁrst order equations. Comment 31.17. Solutions ofsystems ofﬁrst order equations will notingeneral beexpressible explicitly orimplicitly interms ofelementary functions. Only afewvery special ﬁrstorder systems willhave such solu- Lesson 31B SOLUTION orAFmsr Oansa SYs'rEM 395 tions. Even thissimple looking pair ofﬁrst order equations, d$ __ g3 _ Tl?—e’dz_x cannot besolved interms ofelementary functions. Inlater lessons, we shall show youhow apair ofﬁrstorder equations may sometimes be solved bymeans ofseries methods, numerical methods, andbyamethod known asPicard’s method ofsuccessive approximations. Wewish toim- press onyouthattheexamples ofﬁrstorder systems which wehave solved below areofavery special kind, artiﬁcially designed toenable ustoob- tainsolutions interms ofelementary functions. Example 31.18. Solve theﬁrstorder system dz: t d(8.) it-=5-5» %==%-s {B9£0,t#0. (Note thateach euation isoftheseparable type.) '31 Solution. From theﬁrstequation, weobtain a 2 (b) %=*5+cu,1“=131”+0.. andfrom thesecond, provided y950, <<=> logy=-§+02'.y=¢ze'”'- Thepairoffunctions deﬁned by(b)and(c)isasolution ofthesystem (a). Example 31.19. Solve theﬁrstorder system d1?_ 2: QQ___$2—Z/(8.) -a?—26 , dt—-it 11950. (Note thattheﬁrstequation hasnoxoryinit.) Solution. Solving theﬁrstequation in(a),weobtain (b) x=em+cl. Substituting (b)inthesecond equation of(a),there results dy _y__e“+2c1e2' +cl’ (°) E+1_ 1 ’ which isaﬁrstorder linear equation iny.Itssolution, byanymethod 396 SYSTEMS. LINEARIZATION orFmsr ORDER Srsrsms Chapter 7 youwish tochoose, is (d) yt=/(e“ +2c1e2‘ +(:12)dt=ﬁe“+c1e2‘+ c121+c2, ll=(1fe“+ 6162’ +6125+62)l_1»t 5*0- Your canverify that thepair offunctions deﬁned by(b)and (d)satisﬁes both equations of(a)andis,therefore, asolution ofthesystem. LESSON 31C. Deﬁnition and Solution ofaSystem ofLinear First Order Equations. Aspecial type ofﬁrst order system isonein which thefunctions f1(x,y,t) andf2(:c,y,t) of(31.11) arelinear in:0andy. This means that each equation ofthesystem hastheform <31-2) ‘f,—’f=mow+my+1.0). 3'5=f2(t)x+my+12(1). Apair ofequations ofthis type iscalled asystem oftwo linear ﬁrst order equati0ns.. Note that atandyboth have theexponent one, but that nosuch restriction isplaced ontheindependent variable t.Ageneral- ization ofthistype ofsystem isgiven inthefollowing deﬁnition. Deﬁnition 31.21. The system ofnequations d<s1.22> %=f..<1>y.+/..<1>y.+---+r..<0y.+0.0). 95’,-2=r..<0y.+r..<0y.+---+r..0>y.+Q.<1>, %=r..<1>y.j-4..<0y. +---+r..<1>y..+Q.<0. iscalled asystem ofnlinear ﬁrst order equations. Comment 31.23. InLesson 62Cyouwillﬁndacriterion, Theorem 62.3, which gives asuﬁicient condition fortheexistence anduniqueness ofa solution ofthesystem (31.22) satisfying theinitial conditions, (3124) I/i(io) =111, 1/2(lo) =112,'''.y»(l0) =ﬂy- Comment 31.24. Nostandard method isknown ofﬁnding asolution interms ofelementary functions, ifoneexists, ofageneral linear ﬁrstorder system (31.22). If,however, allcoeﬁicients f,~_,-(1), i=1,---,n;j= 1,---,n,areconstants, then standard methods ofsolution areavailable. These methods arediscussed inLesson 31D, where thesystem (31.22) with constant coeﬁicients, isincluded inthelarger class ofsystems of Lesson 31C SoLo'rroN orALINEAR FIRST ORDER Srsrnu 397 linear equations with constant coeﬁicients oforder greater than orequal toone. Intheexamples solved below, where thecoefﬁcients arenotcon- stants, weagain impress upon youthefactthat they have been artiﬁcially selected toyield elementary functions forsolutions. Example 31.25. Solve thelinear ﬁrst order system d d(a) Ff= 2:ct—-2:, Fit/= 2yt+:l:. (Note thattheﬁrstequation hasnoyinit.) Solution. Theﬁrstequation in(a)istheseparable type discussed in Lesson 6C.Bythemethod outlined there, weobtain thegeneral solution (b) :1:=c1e‘a_‘. Substituting (b)inthesecond equation of(a),there results d _ (°) %=21?!-I"61¢‘: i» which isaﬁrst order equation, linear iny.Itssolution, bythemethod of Lesson 11B, is (d) ye-'2 =cl‘/e"‘dt =——c1e_' +C2, 2/=e‘z(¢2——cw“)- You canverify that thepair offunctions deﬁned by(b)and (d)satisﬁes both equations of(a)andistherefore asolution ofthesystem (a). Example 31.26. Solve thelinear ﬁrst order system Q- =1@___=»—1/.(a) dt_2e’ dt_ 1 Solution. The solution oftheﬁrst equation is (b) x=ea‘+cl. Substituting b)inthesecond equation of(a),there results ( @_e2‘—2/+c1 dy2/_e"+c1<°> .1.-"—r‘—' a+r"".—' The solution of(c),bythemethod ofLesson 11B, is (d) 1/1=/<e"+c.>d1. y=Ge"+0.1+0.):-K Thepairoffunctions deﬁned by(b)and(d)isasolution ofthesystem (a). 398 Srsrnus. LINEARIZATION orFmsr ORDER SYSTEMS Chapter 7 LESSON 31D. Solution ofaSystem ofLinear Equations with Constant Coeﬂieients bythe Use ofOperators. Nondegenerate Case. Thepairofequations (31~3) f1(D)fv +91(1))?! =711(1). f2(D)%'+ 92(D)1/ =h2(¢), where Distheoperator d/dt andthecoefficients of:1:andyarepolynomial operators asdeﬁned inLesson 24A, iscalled a.system oftwolinear differential equations. Ageneralization ofthistype ofsystem isgiven inthefollowing deﬁnition. Deﬁnition 31.31. Thesystem ofequations (31-32) P11(D)y1 -l"P12(D)Z/2 +---+P1n(D)yn =h1(t)» P21(D)?/1 +P22(D)3/2 +'''+P2n(D)yn =712(1)» ---¢-------¢----~-----Q---¢.-- Pn1(D)yl +Pn2(D)l/2 "1'‘''+Pnn(D)1/n =h»(l). where Distheoperator d/dtandthecoefficients of1/1,212»''‘13/»arepoly- nomial operators, iscalled asystem ofnlinear diﬂerential equations. Deﬁnition 31.321. Asolution ofthelinear system (31.32) isasetof functions y1(t), y-y(t), ---,y,.(t), each deﬁned onacommon interval I, satisfying allequations ofthesystem (31.32) identically; thesolution isa general oneiffin addition, thesetoffunctions y1(t), ---,y,,(t)contains the correct number ofarbitrary constants; seeTheorem 31.33 below. Examples ofsystems oflinear equations are (a) (2D+3):c+(5D—l)y=e‘, (D—1):c—|—(3D+1)y=sint; (b) (D’+3D-1)1+y=6+t2. (D+2)1— (D2+D)1/=1; (0) (3D’+1)rv+D321—(D+1)z=1’+2. (D—-1)r+(D2+l)1/+(D2—2);!=6‘, Dx—Dy— (D3+5)z=2t+5. Systems oflinear equations with constant coeﬂicients lend themselves readily tosolutions bymeans ofoperators and,iftgO,byLaplace trans- forms. Although weshall conﬁne ourattention primarily toasystem of twolinear equations andtouch brieﬂy onasystem ofthree linear equations, theextension ofthemethod ofsolution toalarger system oflinear differen- Lesson 31D Somrrron orALINEAR SYSTEM BYO1>E1u.'roRs 399 tialequations with constant coefficients willbemade apparent. Inthis lesson weshall solve alinear system bytheuseofoperators; inLesson 31H bymeans oftheLaplace transform. Since polynomial operators with constant coefficients obey alltherules ofalgebra summarized in24.523, themethod weshall beable tousefor solving asystem (31.32) willbesimilar tothatusedinsolving analgebraic system ofsimultaneous equations. There are, however, two important differences between thetwosystems. 1.Theoperator symbol Ddoes notrepresent anumerical quantity such as2,\/3, etc. Itrepresents adifferential operator, operating ona function. Hence theorder inwhich operators arewritten isimportant. 2.Solutions ofalgebraic systems donotusually have arbitrary constants; general solutions ofsystems oflinear differential equations usually do. ByDeﬁnition 31.321, ageneral solution, inaddition tosatisfying the system, must contain thecorrect number ofsuch constants. The relevant theorem needed inthis connection forthetwo equation system (31.3) isthefollowing. Theorem 31.33. Thenumber ofarbitrary constants inthegeneral solu- tiona:(t), y(t)ofthelinear system (31.3) isequal totheorder of (31-34) f1(D)92(D) —91(D)f2(D). I"‘0vid@df1(D)92(D) -91(D)f2(D) 5*0- Theproof ofthetheorem hasbeen deferred toLesson 31E. Comment 31.341. Ifthedifference in(31.34) iszero, thesystem is called degenerate. Weshall discuss thedegenerate caseinLesson 31F. Comment 31.35. The quantity alby —(12171, where a1,G2,bl,172 areconstants, appears sofrequently inmathematical literature that it hasbeen given aspecial name. Itiscalled adeterminant.* Ade- terminant isalsousually written as 111112 1. a‘b’ b b ' ence 5 a * G .blb21 ,blb2 12 21 Since (31.34) hasthesame form asadeterminant, weshall alsorefer toit asadeterminant, andwrite (31-36) f1(D) 91(1)) f2(D) 92(1)) ‘The solution ofthepair ofequations alx+bly=cl,G33!+byy-03is:4:= (C1173 —69171)/(ﬂlbg —a¢b1), y=(ale; —Ggﬂj)/(fllbg —Gaby). Note that the quantity (dlbg -Ggbl) appears inthedenominator ofboth equations andthus de- termines whether asolution exists; seeLesson 63A. Hence thename determinant.IEf1(D)92(D) —91(D)f2(D)- 400 Svsrnus. Lrnnnarzxrron orFmsr Onnna Srsrnms Chapter 7 Observe thatthemembers oftheleftsideof(31.36) arethecoeﬁicients of(31.3) written inthesame relative position inwhich they appear there andthateach term ontheright sideisacross product ofthese coeﬂicients inadeﬁnite order with aminus signbetween them. Weshall therefore refer to(31.36) asthedeterminant ofthesystem (31.3). Using this terminology, wecansay,byComment 31.341, thatthesystem (31.3) is degenerate ifitsdeterminant (31.36) iszero. Fortheremainder ofthisLesson 31D, weconsider onlynondegenerate systems. Theusual procedure followed tosolve thealgebraic system (a) 2:r+3y=7, 3:0—2y=4, istomultiply theﬁrstby3,thesecond by—2,andthenaddthetwo. In thiswayweeliminate 2:,thusobtaining 13y=13, y=1. Toﬁnd2:,wecanagain start with (a)andeliminate y,orwecansubstitute y=1ineither ofthetwoequations. Byeither method, weﬁndas=2. This pairofvalues x=2,y=1satisﬁes both equations andistherefore asolution of(a). Wefollow anidentical procedure insolving thesystem (31.3). Multiply- ingtheﬁrstbyf2(D), thesecond by—f1(D) andadding thetworesulting equations willeliminate anandyield alinear differential equation iny which canbesolved byprevious methods. Substituting thisvalue ofy ineither ofthetwogiven equations willenable ustoﬁndx(t). [Orwecan eliminate yinthegiven equations andsolve fora:(t).] Theunfortunate feature ofthisstandard method isthatinmultiplying each equation bya polynomial operator, weusually raise theorder ofthegiven equations and thus introduce superﬂuous constants inthepairoffunctions a:(t), y(t). Itthen becomes necessary, ifthepairistobeageneral solution of(31.3), todetermine how these constants arerelated: usually atedious task. We shall show byexamples howtoeliminate such superﬂuous constants and thusobtain thegeneral solution of(31.3). Later weshalldescribe amethod which willimmediately givethecorrect number ofconstants inthepair offunctions :c(t),y(t),andhence willimmediately yield ageneral solution of(31.3)—see Lesson 31E. Example 31.361. Solve thesystem da: d(a) 2a—x+%+4y=1, dz_Q__ E dt--t 1. Lesson 31D Somrrron orALmmn SYSTEM BYOPERATORS 4-01 Solution. Inoperator notation, with D=d/dt, wecanwrite (a)as (b) (21)—1)w+(D+4)y=1, D2:-— Dy=t—1. Following theprocedure outlined above, wemultiply theﬁrst equation byD,thesecond by(D—|—4)andaddthetwo. There results (c) (3D2 +3D)x =D(1) +(D+4)(t——1)=4t——3. Thegeneral solution of(c),byanyoftheprevious methods discussed, is (d) z(t)=cl+c2e_‘ +§t2—§t. Substituting (d)inthesecond equation of(b),weobtain (e) D(c1 +c2e" +-§t2—?;t)——Dy=t—1, which simpliﬁes to (f) Dy=——cze" +%—;- Integration of(f)gives itsgeneral solution, (2) y(t)=ere“+%2-—tt+@3- ByComment 31.35, the determinant of(b)is(2D —1)(—D) — D(D +4)=——3D2 —3Dwhich isoforder two. Hence, byTheorem 31.33, thegeneral solution of(b)must contain only twoarbitrary con- stants. Butthepairoffunctions :c(t),y(t)asgiven in(d)and(g)respec- tively hasthree. Toﬁndarelationship among thethree constants, weuse thefact, seeDeﬁnition 31.321, that thesolution ofasystem ofequations isasetoffunctions which satisﬁes each equation ofthesystem identically. Since y(t)wasobtained bysubstituting x(t)inthesecond equation of(b), weknow that thepairoffunctions x(t), y(t)willsatisfy thisequation identically. Hence wesubstitute (d)and(g)intheﬁrstequation. Making these substitutions, there results (11)(2D—1)(¢1+¢¢e"+§l2 —fit)+(D+4) (¢2e"'+§ —%i+c3)=1- Performing theindicated operations in(h),weobtain . 7 <1)~6—c1+4c3= 1,4c3=<=1+7, c3=°%,i—- Hence ifc3=(cl+7)/4,(h)willbeanidentity int.Substituting this value in(g),weobtain thecorrected function 2 (i) y(t)=62¢"+%—gt+%- 402 Srsraus. Lmmmrurron orFmsr Osman Srsrmrs Chapter 7 Thepairoffunctions :c(t), y(t)deﬁned by(d)and(j)nowcontains the correct number oftwoarbitrary constants. Youcanverify thatthispair offunctions satisﬁes both equations of(a)andis,therefore, byDeﬁnition 31.321, itsgeneral solution. Example 31.37. Solve thesystem §'§_.g-d (3 —:5+%+l/=02 dy_ _ Solution. Inoperator notation, wecanwrite (a)as (b) (D-1)@=+(D+1)y=0, (2D+2)::+(2D—2)y=t. Multiplying theﬁrst equation in(b)by(2D+2),thesecond by —(D —-1)andadding thetwo,there results (<=) [(2D+2)(D+1)—(D—1)(2D —2)]?!=—(D —1)¢| 8Dy =t-—1, 2 In/=§—§» 1/(¢>={;-§+¢.. Substituting thisvalue ofy(t)intheﬁrstequation of(b),weobtain (d) (D-1)¢+(z>+1)[{%-§+¢,]=0, which simpliﬁes totheﬁrstorder linear equation 1’1(e) (D—1):c=—E+-5-01. Thegeneral solution of(e)is 2 <0 we=g+§+c.+ere‘- Thedeterminant of(b),byComment 31.35, is(D—1)(2D —2)— (D+1)(2D +2)=—8D which isoforder one. Hence byTheorem 31.33, thepairoffunctions x(t),y(t)asgiven in(f)and(c)respectively should have only onearbitrary constant. Butthepairhastwo. Toﬁnd therelationship between theconstants weproceed aswedidinthepre- vious example. Substituting (f)and(c)inthesecond equation of(b)(we have already used theﬁrst), there results 2 2 <8)2(D+1>[{5+§+c.+¢.e‘]+2<D— 1>[§§—§+c.]= r, Lesson 31D Sourrrou orALINEAR Srsrrm BYOrmwrons 4-03 which simpliﬁes to I(h) 2,:-5 +202$] =t,cget=0. Equation (h)willbeanidentity intonly ifCg=0.Substituting this value in(f)weobtain thecorrected function <0 x<¢>=§+i+c16s1' Thepairoffunctions :z:(t), y(t)deﬁned in(i)and(c)respectively is,by Deﬁnition 31.321, thegeneral solution of(a). Example 31.38. Solve thesystem (=1) (D+3)@=+(D+1)y=e‘, (D+1)w+(D— 1)y=¢- Solution. Multiplying theﬁrstby—(D —1),thesecond by(D+1) andadding thetwo,weobtain (bl [—(D2 +2D—3)+(D2+2D+1)]!!!=1+ti 4:1:=1+ t, x(t)=ﬁt—|—1). Substituting thisvalue of:z:(t)intheﬁrstequation of(a),there results <0)<1>+s>(§+§)+<1>+1>y=¢‘. <1>+1>y=e'—1—&¢- The general solution of(c)byanymethod youwish tochoose is z (<1) yo=6.6-‘+3—=1—it- The determinant of(a)is(D+3)(D —1)—(D+1)”=--4which isoforder zero. Hence, byTheorem 31.33, thepairoffunctions :c(t),y(t) should have noarbitrary constant. Butthepairasgiven in(b)and(d) hasone. Weknow thepairsatisﬁes theﬁrstequation of(a)identically, since weused ittoﬁndy(t). Wetherefore substitute :c(t)andy(t)inthe second equation of(a).There results t1 _,e‘1(6) (D-l'1)<z+Z)+(D—1)(616 "l-5-‘Z—i'5)=h ¢ ¢ which simpliﬁes to (f) t—2c1e_‘ =t. 404 Srsrrzms. Lrumnrzsrron orFrnsr Onnnn Srsrmrs Chapter 7 Hence (f)willbeanidentity int,only ifcl=0.Substituting c1=0 in(d)gives thecorrected function <g> ya)=-1—it ByDeﬁnition 31.321, thepairoffunctions a:(t), y(t)deﬁned by(b)and(g) isthegeneral solution of(a). Example 31.39. Solve thesystem <1’ d(a) E’;-4x+if=0, dx dzy _ Solution. Inoperator notation wecanwrite (a)as (b) (D2-4)vv+ Dy=0, (—4D)rv +(D2+2)?!=0- Multiply theﬁrstequation by(4D), thesecond by(D2 —4)andaddthe two. There results (<1) [4172+(D2+2)(D2 —4)]y=0, (D‘+2D’~8)y=0- The general solution of(c)is (d) y(t)=¢,e‘/5‘ —|—c2e"/E‘ +c3cos2t+c4sin2t. Substituting (d)inthesecond equation of(b)gives (e) -—4D:c +2c1e‘/E‘ +2c2e“/-5' ——403cos2t——4c4sin2t +20165‘ +26,6-‘/5‘ +2c3cos2:+26,sin2:=0, D1:=cle‘/E‘ +c2e"‘/5‘ —ﬁescos2t—£04sin2t, 2 _ .x(t)=%?ole‘/E’ —%—C28‘/5'-—icgS1I12t+104cos2t+c5. The determinant of(b),byComment 31.35, is(D2 —4)(D2 +2)- D(—4D) which isoforder four. Hence byTheorem 31.33, thepair of functions a:(t),y(t)should have fourarbitrary constants. Butthepairas given in(d)and(e)hasﬁve. Youcanverify thatthesubstitution ofx(t), y(t)intheﬁrst equation of(b)—we have already used thesecond—will beanidentity ifc5=0.Hence thispairoffunctions, with 05=O,isthe general solution of(a). Lesson 31E AnEQUIVALENT TRIANGULAR SYSTEM 405 LESSON 31E. AnEquivalent Triangular System. The standard method outlined above forsolving thesystem (31.3) canbetedious andtime consuming, ifbyitsusethepair offunctions thus obtained hasmore than therequired number ofconstants. Amethod therefore which would im- mediately give thecorrect number ofconstants inthepair offunctions should indeed bewelcomed. Weshall now develop such amethod. Forconvenience werecopy thesystem (31.3), (31-4) f1(D)1v -|'91(1))?! =111(1). f2(D)Iv +92(D)?l =712(1)- From it,weobtain anewsystem inthefollowing manner. Weretain either (meoftheequations in(31.4). Letussayweretain theﬁrst. Thesecond is then changed asfollows. Multiply theoneretained, here theﬁrst, byany arbitrary operator lc(D) andaddittothesecond. The new system thus becomes (31-41) f1(D)$ + 91(D)Z/ =111(1), lf1(D)k(D) '1'f2(D)]I +[!J1(D)k(D) -1'g2(D)l!/ =k(D)h-1(1) +112(1)- Thenewsystem (31.41) isequivalent totheﬁrstsystem (31.4) inthesense thatapairoffunctions :c(t),y(t)which satisﬁes theﬁrstsystem willalso satisfy thesecond system, andconversely, asolution ofthesystem (31.41) willsatisfy thesystem (31.4). ByComment 31.35, thedeterminant of(31.4) is (31-42) f192 —91.7.21 andthedeterminant of(31.41) is (31-43) f191l<? +f192 —fithk —f291 =f192 -"91f2- Hence weseethatthedeterminants ofboth systems arethesame. There- foretheorder ofthedeterminant ofouroriginal system (31.4) must be thesame astheorder ofthedeterminant oftheequivalent system (31.41). Letusassume momentarily—we shall show youlater how todoit— that byalways retaining oneequation inasystem andchanging theother inthemanner described above, wecanobtain anewsystem, equivalent to (31.4), intheform (31.44) F1(D):z: =H1(t), F2(D)Il= -l"G2(D)l/ =H2(l)~ Hence thedeterminant ofthesystem (31.44) isthesame asthedeter- minant ofthesystem (31.4) andtheorders oftheir determinants arealso thesame. 406 Srsrnms. LIN!-:An1zArroN orFins-r ORDER Srsrams Chapter 7 Wecallsuch anequivalent system, i.e.,oneinwhich acoeﬁicient ofreor ofyiszero, anequivalent triangular system. Butnow note. Ifwe solve thesystem (31.44) forac,using theﬁrstequation in(31.44), itsgen- eral solution x(t)willhave asmany arbitrary constants astheorder of F1(D). Substituting thissolution inthesecond equation of(31.44) will result inalinear equation inywhose order isequal tothat ofG2(D). Hence, insolving thisequation fory,weshall introduce additional con- stants equal totheorder ofG2(D). Therefore thetotal number ofarbi- trary constants inthepairoffunctions :r(t),y(t)of(31.44), ifsolved ﬁrst foratandthen fory,willequal thesum oftheorders ofF1(D) andG2(D). But, byComment 31.35, thedeterminant of(31.44) isF1(D)G2(D) whose order isexactly thesum oftheorders ofF1andG2. (Remember inmulti- plying differential operators, theorders areadded just asifthey were exponents, forexample D2D3 =D5.) Wehave thus shown that forthe system (31.44) thepair offunctions :e(t), y(t), ifthesystem issolved ﬁrst forx(t)and then fory(t), will contain thecorrect number ofconstants (since thepair satisﬁes each equation identically), andthat thisnumber equals theorder ofthedeterminant of(31.44). Buttheorder ofthedeter- minant of(31.44) isthesame astheorder ofthedeterminant oftheoriginal system (31.4) from which itcame. Wehave therefore notonly proved thatasolution of(31.44) willalsobeasolution of(31.4) andcontain the correct number ofarbitrary constants, buthave atthesame time proved Theorem 31.33. Iftheoriginal system isalready inthespecial form (31-45) (8) fi(D)w =111(1), 92(D)!l =112(1), orinthetriangular form (31~45) (b) f1(D)1 =111(1), f2(D)11 +yz(D)y =112(1), then thepair offunctions ::;(t), y(t)obtained bysolving thesystem will contain thecorrect number ofconstants. Insolving (b),itisessential to obtain :c(t)ﬁrstandthen y(t). Ifyousolve ﬁrstforybyeliminating 2:, superﬂuous constants may beintroduced. Example 31.46. Solve thesystem (=1) (D+1)w =1. D2y =e2‘. Solution. The general solution oftheﬁrst equation in(a),bythe method ofLesson 11B, is (b) :v(t)=t—1-1-c1e_‘. Leann 31E ANEQUIVALENT TRIANGULAR Srsrau 4-07 Thegeneral solution ofthesecond equation in(a)is (0) 2/(1)=11¢”+621+63- Youcanverify, byTheorem 31.33, thatthepairoffunctions in(b)and (c)hasthecorrect number ofthree constants andistherefore thegeneral solution of(a). Example 31.47. Solve thesystem (a) (3D2 +3D):c =4t—3, (D-—1):c—D2y =t2. Solution. Thegeneral solution oftheﬁrstequation in(a),byany oftheprevious methods discussed, is (b) x(t)=cl+c-ye" +§t2—§t. Hence (c) (D——1)a:=——c2e" +§t—;7;—c1—c2e" —§t2+-}t =—-§—cl—2c2e" -1--131t—-§t2. Substituting (c)inthesecond equation of(a),there results (d) D2y=—-§—c1—2c2e_‘ +-1;-,1t—17¢’. Integrating (d)twice, weobtain thegeneral solution (e)y(t)=0.+at-+%)I’++11“—at‘—20.6-‘. The pair offunctions deﬁned in(b)and (e)isthegeneral solution of(a). Youcanverify, byTheorem 31.33, thatthispairoffunctions contains the correct number offourarbitrary constants. Before proceeding tothegeneral system, weconsider onemore special type, namely oneinwhich acoefficient of:0oryisaconstant. Hence the system isoftheform (31-48) f1(D)@ + 111/=111(1), fz(D)w +a2(D)y =112(1)- Inthiscaseitwillbepossible toobtain anequivalent triangular system in onestepbythefollowing procedure. Retain theequation which contains the constant coe_ﬂicient—in (31.48) itistheﬁrst—and change thesecond by multiplying theﬁrstby——g2(D)/k andadding ittothesecond. Example 31.49. Solve thesystem (8) (3D— 1)fv+42/=1. D:c—Dy=t—-1. 408 SYSTEMS. LINnAnIzA'rIoN orFmsr Onnan Svsrmrs Chapter 7 Solution. Following theprocedure outlined above wecopy theﬁrst equation, andobtain asecond equation bymultiplying theﬁrstbyD/4 andadding ittothesecond. Wethus obtain theequivalent triangular system (b) (3D-1)w+42/=1. i}(3D2+3D)x =€(¢)+¢—1=¢-1. Thegeneral solution ofthesecond equation in(b),byanyofthemethods previously discussed, is (0) 11(1)=61+02¢"+ﬁt’—Zit- Substituting thisvalue of:1:intheﬁrstequation in(b),weobtain (<1) (317—1)(c1+02¢“+£1’—it)+4y=1, which simpliﬁes to @ yw=ﬂ{l+w*+%#-Q Thedeterminant of(1.),by(31.36), is (31)-1)(—D) -40=-301-so, whose order istwo. Note thatthepairoffunctions :c(t),y(t)contains the correct number oftwoconstants andistherefore thegeneral solution of(a). Example 31.5. Solve thesystem (11) (D+4)x+Dy=1. (D—2):c+y= t2. Solution. Wecopythesecond equation andobtain anewﬁrstequation bymultiplying thesecond by—D and adding ittotheﬁrst. Wethus obtain theequivalent triangular system (b) (——D2 +3D-1-4):: =—2t +1, (D——2):c+y= t2. The general solution oftheﬁrst equation in(b)byany ofthemethods previously discussed is (0) w(1)=ere“+ere"—é+2- Substituting (c)inthesecond equation of(b)gives _ 5w <D—a@W+ew-§+Q+y=a y(t)=—2c1e'“ +3c2e_' +t2—-t+ Lesson 31E ANEQUIVALENT TRIANGULAR Srsram 409 The determinant of(a)is(D+4)—-D(D —2)=—-D2 +3D+4, which isoforder two. Note that thepair offunctions :c(t), y(t)contains thecorrect number oftwoconstants, andistherefore thegeneral solution of(a). Example 31.51. Solve thesystem (a) (D+1):c+y=e‘, (D2—|—1)x+(D-1)y =t. Solution. Wecopytheﬁrstequation, andchange thesecond bymulti- plying theﬁrst by—(D —-1)andadding ittothesecond equation. We thus obtain theequivalent system (b) (D+1)w+y=6‘. 2;» =-(1)-1)¢‘+¢=¢, :c(t)=%- Substituting intheﬁrst equation of(b),thevalue of:1:asfound inthe second equation, there results <0) (1>+1>§+y=e‘. y<»>=e‘-§—§- The determinant of(a),by(31.36), is(D+1)(D —1)—(D2-1-1)= D2-—1-—D2-1=-2, which isoforder zero. Note that thepair offunctions :c(t), y(t)contains thiscorrect number ofzero constants, and istherefore thegeneral solution of(a). Weshall now show youamethod bywhich youcanreduce ageneral system toanequivalent triangular one. Wedemonstrate themethod by anexample. Consider thesystem (31.52) (D3-D2+2D+1);»+(D4+302+1)y=0, (D—-2)::—|— (D3—3)y=O. Ofthefourpolynomial operators in(31.52), concentrate ontheoneof lowest order. Intheabove example itisD-—2.Retain theequation in which itappears, multiply theequation by—D2 andaddittotheﬁrst. Theresulting equivalent system is (D2+21)+l):c+(—D5+1)‘+602+1)y=0, (D—2)x+ (D3—3)y=0. Note that bythisprocess, wewere able toreduce theorder oftheco- efficient ofreintheﬁrst equation from three totwo. The polynomial operator oflowest order inthisnewsystem isagain (D—2).Wethere- 410 Srsrams. LINEARIZATION orFrnsr Onnan Srsrnms Chapter 7 fore again retain theequation inwhich itappears, multiply itby--D andaddittotheﬁrst. Wethusobtain theequivalent system (4D-1-1):c+(—D5 +6D2_+ 3D+1)y=0, (D—2)a:-1- (D3 —3)y=0. Note that theorder ofthecoeﬂicient ofxintheﬁrst equation hasbeen reduced from twotoone. There arenow twopolynomial operators with thesame lowest order. Wecanretain either one.Because thecoeﬂicient ofDin(D—2)isone, itwillbefound easier toretain this equation. Multiply itby-4andaddittotheﬁrst equation. Wethus obtain the equivalent system 9a:+(-—D5-41>“+602+so+13);,=0, (D-2)x+ (1)3-3)y=0. Thesystem isnow oftheform (31.48). Therefore, retaining theﬁrstequa- tion, multiplying itby~—(D -2)/9 and adding ittothesecond will ﬁnally give theequivalent triangular system 9x+(—D5 -4D3 —|—6D’ +3D+13)y =0, [(D5+4D3 -co’ -30- 13)-D_9I3-+0“-3]y=0. Inthemanner described above, itisalways possible toreduce theorder ofthecoefficient ofonepolynomial operator inthesystem tozero. When that point hasbeen reached, thesystem willbeintheform (31.48). One more step willthen give therequired equivalent triangular system. Comment 31.53. Useoftheabove method willalways enable youto obtain anequivalent triangular system. Useofalittle ingenuity may at times enable youtoobtain itsooner. Forexample, ifthegiven system is, orifinthecourse ofyour work, itbecomes, D3:c +(D2—2)y=e‘, -20% +(D+3)y=41+2, then retaining theﬁrst, multiplying itbytwoandadding ittothesecond willgive youanequivalent triangular system immediately, even though D3isnotthepolynomial operator oflowest order. Example 31.531. Solve thesystem (9-) (D+1)1v +(D—|—l)1/=1, D2x- Dy=t—-1. Lesson 31E ANEQUIVALENT TRIANGULAR SYSTEM 4-ll Solution. Here there arethree polynomial operators ofthesame lowest order. Itwillbefound easiest toretain thesecond andchange the ﬁrstbyadding thetwoequations. There results theequivalent system (11) (D2+D+1)11+ 1/=1» D2:c—Dy=t—1. The system isnow oftheform (31.48). Wetherefore copy theﬁrst equa- tionandchange thesecond bymultiplying theﬁrstbyDandadding itto thesecond. Wethusobtain theequivalent triangular system (c) (D”+D+1):c+y=t, (D3+2D’+D)Cl2 =1. Ageneral solution ofthesecond equation is 2 (d) :c(t)=c1—|—c2e_' +c3te" +15—2t. Substituting thisvalue of:c(t)intheﬁrst equation of(c),weobtain (e) y(t)=1—(D2+D+1)(vi+cw“+cite“ +g—21)l =t—(c2e"‘ —2c3e"‘ +c;,te_‘ —|—1—c2e"' +c3e_‘ -¢.a"+¢- 2+¢, +626-‘+¢3a-‘+§ -2:). 1/(1)=-[C1+02¢"+ca(-F‘ +16")+§—21—1]- The determinant of(a), by(31.36), is(D+1)(—D) +(D+1)(D2) which isoforder three. Note that ourpairoffunctions :c(t), y(t)contains thecorrect number ofthree constants, andistherefore thegeneral solution of(a). - Example 31.54. Solve thesystem (11) (2D—1)w+(D+4)y=1.‘ Dx-— Dy=t—.'1. Solution. Here allfour polynomial operators in(a)areofthesame order one. Wecantherefore retain either equation of(a).Itwill, however, befound easier toretain thesecond and change theﬁrst byadding the second toit.Wethus obtain theequivalent system (b) (3D——1):c+ 4y= t, D:c—Dy=t—-1. 4-12 Svsraus. LINEARIZATION orFmsr Onnnn Srsrams Chapter 7 Thesystem isnow oftheform (31.48). Wetherefore retain theﬁrstequation andchange thesecond bymultiplying theﬁrst byD/4 andadding itto thesecond. There results theequivalent triangular system (0) (3D—1)r+4y= 1. (3D2+30);»=4:-3. The general solution ofthesecond equation in(c)is (d) a:(t)=cl+c2e" —|—§l2—-§t. Substituting thisvalue ofacintheﬁrstequation of(c),weobtain (e) (3D—1>(c.+er‘+it’—so—1=-41). which simpliﬁes to2 <f> ya)=ca-1+‘-6-§:+ The pair offunctions (d)and (f)arethesame asthose obtained inEx- ample 31.361 byusing thestandard method ofelimination. Note howmuch easier theabove method is.Note toothat weobtained thecorrect number oftwoconstants immediately. Example 31.55. Solve thesystem (P1) (D—1)w+(D+1)2/=0. <1>+1>x+(D— 1>y=§- Solution. Weretain theﬁrstequation andchange thesecond bymulti- plying theﬁrst by——1andadding ittothesecond. Wethus obtain the equivalent system (b) (D—l)w+(D+ 1)1/=0. 2z—— 2y=%- Retain thesecond equation andchange theﬁrstbymultiplying thesecond by(D+1)/2 andadding totheﬁrst. There results theequivalent triangular system (o we =@%1—) 1=1<1+1). 21:—2y= Integrating theﬁrst equation in(c),weobtain (<1) 2x—3+1+2c w<¢>—‘—2+1+¢ _8 4 ’ _16 8 ' Lesson 31F DEGENERATE CAsE 413 Substituting (d)inthesecond equation of(c),weobtain 2 2 (e) 2:/=‘§+§-§+2c. 1/(t>=f—6—-é-+¢. These arethesame functions obtained inExample 31.37. Note howmuch easier theabove method isover theprevious one. Note, too,that weob- tained thecorrect number ofoneconstant immediately. Example 31.56. Solve thesystem (*1) (D+3)$+(D+1)?!=6'. (D+1)r+(D-" 1)y=1- Solution. Weretain thesecond equation andchange theﬁrst bymulti- plying thesecond by—-1andadding ittotheﬁrst. Wethus‘ obtain the equivalent system (b) 2:1: —|—2y=e‘—t, (D+1)=v+(D— 1)y=1- Retain theﬁrst equation andchange thesecond bymultiplying theﬁrstby -—(D —-1)/2andadding ittothesecond. There results theequivalent triangular system (c) 2x+2y=e‘—t, u =—ﬂD-Imh-o+z=§a+n. From thesecond equation of(c),weobtain (<1) 11(1)==l:(1+ 1)- Subtracting thesecond equation in(c)from theﬁrst, weobtain a1 ‘3t1(9) 21/=¢""§—§' !l(1)=e§—'Z—Z' These arethesame functions :c(t),y(t)obtained inExample 31.38. Note that bythis method weobtained thecorrect number ofzero constants immediately. LESSON 31F. Degenerate Case. f1(D)g2(D) —g1(D)f2(D) =0.An algebraic system ofequations may bedegenerate. Insuch cases thesystem will have nosolutions orinﬁnitely many solutions. Forexample, the systems (8)2w+31/=5, (b)2r+32/=5. 2a:—l—3y=7; 4x+6y=7, 414 SYSTEMS. LINEARIZATION orFmsr QRDER Srsrnus Chapter 7 have nosolutions. Ontheother hand, each ofthesystems (c)2:c+3y=5, (d)2a:+3y=O, 4a:+6y=10; 4z+6y=0, hasinﬁnitely many solutions. Note that inallfour examples, thedeter- minant formed bythecoefﬁcients of:1:andyiszero. Note alsothat there arenosolutions, when intrying toeliminate :1:ory,theright sidedoes not also reduce tozero. There areinﬁnitely many solutions when thelight sidereduces tozero. Similarly, wecallthesystem oflinear differential equations (31.4) de- generate whenever itsdeterminant f1(D) 91(D) f2(D) 92(1)) iszero. Asinthealgebraic system, there willbenosolutions if,intrying toeliminate :1:ory,theright sideofthesystem isnotzero; there willbe inﬁnitely many iftheright sideiszero.I=f1(D)92(D) '"01(D)f2(D) Example 31.6. Show that thesystem (a) Dz—Dy=t, D2:—Dy=t2, isdegenerate. Find thenumber ofsolutions ithas. Solution. By(31.36), thedeterminant of(a)is—D2 —(-—D2) =0. Hence thesystem isdegenerate. Since theright sidedoes notreduce to zerowhen weeliminate :cory,ithasnosolutions. This example corresponds to(a)atthebottom ofpage 413. Example 31.61. Show thatthesystem (a) D2:—Dy=t, 4D:z: -—4Dy =4t, isdegenerate. Find thenumber ofsolutions ithas. Solution. Thedeterminant of(a)is—4D2 —(—4D”) =0.Hence thesystem isdegenerate. Itsright side, however, reduces tozerowhen weeliminate :1:ory.Inthiscase there areinﬁnitely many solutions ofthe t2 system. Forexample, thepair, ac=5+c,y=5cisasolution; thepair 2l Z l _ _ , :0=5+cl,y=5—5+Cgisasolution. [Ineither equation, deﬁne x(t)arbitrarily, andsolve fory(t). This pairoffunctions willalsosatisfy theother equation.] This example corresponds to(c)attopofpage. Lesson 31G Srsrsns orTHREE LINEAR Eouxrxons 4-15 Example 31.7. Show that thesystem (Q) (D+1)=v+(D+1)y=0. (D—1)w+(D— 1)y=0. isdegenerate. Find itssolutions. Solution. The determinant of(a)is ‘bl‘DH D+1l=<1>+1><1>—1>—<1>+1><1>—1>=o.D-—1D—1 Hence thesystem (a)isdegenerate. By(24.21), wecanwrite (a)as (0) (D+1)(w+y)=0, (D-—1)(:v+ y)=0. Letu=:0+y.Therefore (c)becomes (D+1)u=0,and(D-—1)u=0. The solution oftheﬁrst equation isu=c1e"; thesolution ofthesecond equation isu=c2e'. Hence thesolution ofthesystem (c)is (d) x+y=ole" and x+y=cge‘. However, ifc1960andc29-60,then thepairofsolutions of(d)isin- consistent, i.e.,there arenofunctions a:(t)andy(t)thatwillsatisfy them simultaneously. Ifhowever, c1=C2=0,then theinﬁnitely many func- tions, such that (e) w+2/=0orw(l)=-2/(1) willsatisfy thesystem (c). LESSON 31G. Systems ofThree Linear Equations. Wediscuss brieﬂy thesystem ofthree linear equations: (31-71) f1(D)w +91(17):! +h1(D)z =k1(l), f2(D)w +92(1))?! +712(1))? =k2(¢). fs(D)w +93(9)?! +ha(D)Z =ks(l)- Itsdeterminant isdeﬁned as (31-72) f1(1)) 91(1)) h1(D) f2(D) 92(1)) h2(D) Efiqzha +f29ah1 +fagihz —fiﬂahz 13(0)g.<1>>ha(D) "“"1"”_’*‘”'“' The method ofﬁnding ageneral solution ofthesystem (31.71) follows thesame rules outlined previously insolving atwo-equation system. The number ofconstants inthegeneral solution of(31.71), i.e.,inthesetof 416 SYs'rEus. Lrnnanrzarxon orFmsr ORDER SYSTEMS Chapter 7 functions :c(t),y(t),z(t)satisfying (31.71), asinthecaseofatwo-equation system, must equal theorder ofthedeterminant (31.72), provided this determinant isnotzero. Theusual standard method forsolving thesystem (31.71) istoeliminate oneofthevariables, sayz,inthesame manner weeliminate thisvariable inanalgebraic system ofthree equations. Thesystem (31.71) canthus bereduced tothesystem (31-73) F1(D)¢ +G1(D)y =11(1), F-.»(D)w +G2(D)y =12(13)- Finally byeliminating, intheusual manner, onevariable from thesystem (31.73), sayy,weobtain thesingle linear equation (31.74) F(D):c =k(t), which wecansolve forx.Substituting thisvalue ofa:ineither equation of (31.73) willenable ustosolve fory.Substituting both values 2:andy inanyoneoftheequations in(31.71) willenable ustosolve forz.Since inthisprocess, weusually must multiply byanoperator inorder toeffect thedesired eliminations, thenumber ofconstants inthesetoffunctions :c(t), y(t),z(l)willgenerally bemore than required. Wethen have togo through thetedious process ofﬁnding what relationship exists among the constants bysubstituting thethree functions x(t),y(t),z(t)ineither ofthe twoequations of(31.71) which were notused toﬁndz(t). Since thethree functions must satisfy each equation in(31.71) identically, wecanthus determine from theequation arelationship among theconstants, justas wedidinthetwo-equation system. Fortimately itisalways possible, inexactly thesame manner described forthetwo-equation system, toreduce thesystem (31.71) toanequivalent triangular system oftheform Fz(D)=v +Gz(D)y =7620), F3(D)1¢ +G3(D)y -l"Ha(D)(Z) =753(0- Ifsolutions arethenobtained intheorder a:(t),y(t),z(t)starting withthe ﬁrstequation in(31.75), thissetoffunctions willcontain thecorrect number ofconstants required inthegeneral solution ofthesystem (31.71). Example 31.76. Reduce thefollowing system toanequivalent tri- angular oneanddetermine thenumber ofarbitrary constants needed inthe general solution. (=1) (D-2)w+ y— z=i, -—:c+(2D+l)y+ 2z= 1, 2:c+ 6y—|-Dz=0. Lesson 31G Srsrsus orTHREE LINEAR Eqmrrrons 417 Solution. Here wecanretain any oneofthethree equations. We decide toretain theﬁrstequation andchange thesecond equation bymulti- plying theﬁrstoneby2andadding ittothesecond. Wechange thethird equation bymultiplying theﬁrst byDandadding ittothethird. There results theequivalent system (b) (D—-2)x+ y—z=t, (2D -—5)::+(2D +3)y =2t+1, (D2—2D+2)x+ (D+6)y =1- Weretain theﬁrstandthird equations andchange thesecond bymultiplying thethird by-2andadding ittothesecond. Wethus obtain theequiva- lentsystem (C) (D—2)I+ 1/—z=t, (—2D’ +6D—9)@- 9y =2:-1, (D2—2D+2)x+(D—|—6)y =1. Weretain theﬁrstandsecond equations andchange thethird bymultiplying thesecond by(D+6)/9andadding ittothethird. Wethus ﬁnally obtain theequivalent triangular system (d) (D-—2)w+ 1/—z=t. (-202 +6D-9)@-9y =2:—1, (-21)“ +3D’+9D—36):: =12¢+5. Solving thethird equation for:1:willgive three constants. Substituting this value of:1:inthesecond equation willgive anequation inyoforder zero. Hence thesolution y(t)willcontain nonew constants. Substituting the solutions :t(t), y(t)intheﬁrst equation willgive anequation inzthat is also oforder zero and hence thesolution z(t)willnotcontain any new constants. Wewillthus have three arbitrary constants inthesetoffunc- tions a:(t),y(t),z(t). Thedeterminant of(a),by(31.72), isalsooforder three, andhence only three constants areneeded inthegeneral solution of(a).Wesee,therefore, thatbyusing thismethod thesetoffunctions obtained willcontain thecorrect number ofconstants. The extension ofthestandard method ofsolution tosystems oflinear equations oforder higher than three should now beapparent toyou. To determine thenumber ofarbitrary constants inthegeneral solution, you willneed toknow theorder ofthedeterminant ofthesystem. Forthis information werefer youtoanybook ondeterminants andmatrices. If, however, you solve thesystem byﬁrst reducing ittoanequivalent tri- angular one, your solution will always contain thecorrect number of constants. 418 Srsrnms. LINEARIZATION orFIRs'r ORDER SYSTEMS Chapter 7 LESSON 31H. Solution ofaSystem ofLinear Differential Equa- tions with Constant Coeﬁicients byMeans ofLaplace Transforms. Asystem oflinear differential equations with constant coefficients can also besolved bymeans ofLaplace transforms. Unlike theoperational method, however, theLaplace transform method canbeused only ifinitial conditions aregiven andtheinterval over which thesolutions arevalid is 0§t<oo.Ontheother hand, theLaplace method hasanadvantage over thepreceding operational oneinthat itwill immediately yield a particular solution satisfying given initial conditions without thenecessity ofevaluating arbitrary constants. Although wehave conﬁned ourattention toasystem oftwodifferential equations with two dependent variables, theextension ofthemethod to alarger system willbeapparent. Example 31.8. Solve thesystem dz d(a) é-4:0+%=0, d dz—4if+;,,%+2y=0. forwhich a:(0) =0,x’(0) =1,y(0) =-1,y'(0) =2. Solution. InLaplace notation, wecanwrite (a)as—see Comment 27.24- (b) L[=v”(i)] —4Llrv(t)] +Ll!/’(i)l =0, —4L[¢'(t)l +Lly”(l)l +21-[y(t)] =0- By(27.33), ordirectly from (27.29) and(27.31), (b)becomes (6)82L[=v(i)l —¢’(0) —8w(0) —4L[Iv(t)l +8L[y(i)l -1/(0)=0- —48L[w(l)] +4w(0) +821/[y(t)] -y'(0)—82/(0)+2I/[y(t)] =0- Inserting theinitial conditions in(c)andsimplifying theresulting equa- tions, weobtain (<1) (82—4)I-[ml + SI-[2/l=1—1=0. -—4sL[:z:] +(s2+2)L[y] =2-—s. Just aswith operators, wenow solve (d)asifthey were ordinary algebraic expressions inL[a:] and L[y]. Multiply theﬁrst equation in’(d)by4s, thesecond bys2—-4,andaddthetwo. The result is (e) (s4+2s2—8)L[y] =—s3 +282—|—4s--8. Hence __—s3+2s2+4s-—8__1 1+\/5 1--\/2__8(s—2)_(f) Lilli-' (82+4)(s2-2) "_5[8+\/§+8_\/5 824.4 Lesson 31H SOLUTION orALINEAR SYSTEM BYLAPLACE Tnmsronms 419 Referring toatable ofLaplace transforms, weﬁndthat s) mH~%*%=¥@@s+\/2-, -vs L_is=1__. wva1s_v§ ——8s . 2008 2t]= r S1112i]=8 ' ' Therefore, by(f)and(g), (11)y(t)=an+\/in-~/5‘ +(1-\/a)u/7* -scos2t+ssin2:1. Again aswith operators, twomethods areavailable forﬁnding a:(t). Wecanstart with (d)again andeliminate y(t)orwecansubstitute (f) ineither equation of(d)andsolve forL[:c(t)]. Using thislatter method, weobtain by(f)andtheﬁrstequation in(d) . -“+2’+4 -8_(1) (82 — —0. - _ (—m%+e) _ <—m°)LM'o—ﬁQaw1ow—a‘wihw—m =I;[f:£+fi.‘?§-4(:-i.i)l~Referring toatable ofLaplace transforms, weﬁndthat (1.)T—;‘L[(2 -\/QM‘ +(2+\/in-*5‘ -4cos2:-4sin2:1 equals theexpression ontheextreme right of(j).Therefore, by(j)and(k), (1):c(t)=T—;[(2-\/§)e~’§‘ +(2+\/§)e-‘/5‘ -4cos2:-4sin2:]. Example 31.81. Solve thesystem (seeExample 31.5) w §+a+%=t %%—2x+y=12, forwhich a:(0) =2andy(0) =—1. Solution. InLaplace notation (a)canbewritten as (b) L[1=']+41-lwl+Ll!/'1=Llll, Llrv’l—2L[¢vl+Llyl=LE2]- 4-20 Srsrsns. LINEARIZATION orFIRST ORnER SYSTEMS Chapter 7 By(27.33) ordirectly from (27.29), (b)becomes (0) 81-[w(t)l —16(0)+4Llx(l)l +sl-[y(t)] -1/(0)=11(1), 81-[w(i)l -w(0)—21-11(1)] +I-[y(t)] =L02)- Referring toatable ofLaplace transforms, weﬁndthat 1 2(<1) L(1)=g. La’)=3-5- Substituting these values andtheinitial conditions in(c),andsimplifying theresulting equations, weobtain (8) - '?»l\'Jao>--+!°(8+4)Ll$(i)l +8L[1/(1)1 =+1, (8-2)LlI(l)l +Ll!/(1)1 = Multiplying thesecond equation in(e)by—-sandadding ittotheﬁrst, there results (r) (8+4)L(x) +(-82+2s)L(:c) =1+1-g-28. Solving (f)forL[x(t)] gives _23—’— +2_<+na2-a+a— :g(s2 _s_38i _“ 8s2(s __s4)(s _Z§i:_3L‘:_?.=_~’L__L 11 s2(s 4) 8s 2s?+8(s-4) Referring toatable ofLaplace transforms, weﬁndthat 5_i _.l __i 1_14¢_i.G‘)Ll§l*ss’ Llfl‘ 282’ Lise i_8(s--4) Hence, - §_£ L1.4¢_i__1_ in .(‘l Lia 2+s“l*ss 2s2+8(s-4) Therefore by(g)and(i) - _§_iH41 Toﬁndy(t), wesubstitute (g)ineither equation of(e).You canverify that (k) y(t)=t’—t+-1—12-“- NOTE. These arethesame results youwould have obtained ifyouhad inserted theinitial conditions in(c)and (d)ofExample 31.5 where we solved thissame problem bymeans ofoperators. Verify it. Lesson 31--Exercise 421 EXERCISE 31 Solve each ofthefollowing systems ofequations. dx_ 2dy_1.dt--—:2:, Ft--—y. 9..Qd _2.?:=3e‘, =:2:+y. a. E-‘We§_,Q..s. d3.i=12¢,-=ya’. d d4.E€=2t, —=3a:+2t, F:=:c—|—4y+t. £-§“ ..._.L_—y- 5.%:§=e‘, 6.%;Z=x+sint, %=t—y. 7.%=3:c+2e3" Z—:+%—3y=sin2t. 8.:—f=y, %%=-—:c+2y. 9.3%-1—3:r:—l-2y=e', 41-s%§+sy=s¢. 2 2 l0.%T;€+4:c=3sint, %%—%+y=2cost. 11.:1ul:—4:v—2%?+y=t, 2%—|—:c+%jg-=0. 12.2%:+%-¢=e‘, 33-f+2%‘,!+y=¢. 2 2 l3.%;-+x—‘fTg—y=—cm2t, 2l%—%—y=0. 14.%-%=1-:, Z-§+2g%=4e‘+¢. d2 4’ a al5.F:—x+%+y =0,7’§+2@+%+2y -o. dz: dy _ dz: dzy dy _z16-J+Et~+t/-1, $+$+$+$+!l—i- Verify that each ofthefollowing systems 17-20 isdegenerate (D=d/dt). Find solutions ifthey exist. 17.D:c+ 2Dy =e‘, Dz+2Dy =t. 18.Dz—Dy=e‘, 3Dx —-3Dy =3e‘. 19.(D2—1)x+ (D2-1.)y=0. (D2+4)!"+(D2+4)?!=9- 422 Srsrsns. LINEARIZATION orFrnsr ORDER SYSTEMS Chapter 7 20.(D—2)l-l-(D-2)y =1. (D-I-3)1-I-(D+3)y =1- (D—-2)1+ 31/- ——1+(D-I-3)!/+ 2::+ 39y+(D—4)z 22.Solve each ofthefollowing systems ofequations. (R)(D-l)w -—w+(D—-3)y ——w+ y+(D—2)Z (11)(D-l)1 (C)(D- (d) D:c— y+z Solve each ofthefollowing systems ofequations bythemethod ofthe3w+2(D+1):;2y+(2D-3) 2):: -31+(D+2)y -211+(D—3)Z 0,= 0, =0.-—w+(D—1):I/-2:+ (D-—1)z Laplace transform. da:23.5-y-:, Zd -7’:=1.-(0)=2.1/(0)=1- 24.3:—:+3x+2y =e‘, 42:—~3-ji:+3y =3t,:c(0) =l,y(0) =--1. dz: dy25.21?-'l'E—4y -1, d-+7f—sy =1’,-(0)=2,1/<0)=—2dz dy_ ___26.Tt2—E—1 l,Z0.0.0. 0, 0,0. 2.21.Reduce thefollowing system toanequivalent triangular one How many arbitrary constants should thegeneral solution contain ‘:,lf+2%=4-e‘+w. 1(0)=0.1/<0)=0.170) = 08MI-IO--RR8|-ANSWERS =l-|- 61, y=cge"‘.31 =—3e"‘ -1-c1,y=§e"‘ —61—|—62¢‘- |- =—%(¢’+ C1),1/=626"”- Lesson 31-Answers -1-23 2" 95" 7. Ii999° I-II-Is 1-In-I5*’? I-IIP 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.12-l-61, 1/=13-1-12+ 3611-I-C2, t“.+its+(601+1)t2'+ (01+4¢2)t+ca- e'+c1,y=t-1e‘ +c1+c2t'1. cle‘—§(sint+ cost), y=t-—1+c2e"'. 3.'2t 2 2t(2:+c1)e3', y=-i1l‘-:;-°‘-’5- -é“(a¢’+ 3617+21+C2). c1te‘+ age‘, y=c1te‘+(c1—|—cg)e‘. cle‘/3 —|—ego"/3 —6t,y=——2c1e‘/3 —c2e"/3 —|—ﬁe‘—|—9t+9. sint—|—c1sin2t—|—02cos2t, 1}cost —§c1cos2t+icgsin2t+cge‘+04c". c1sint+62cost+cge‘+c4e", (c1—262)sint+(2c1—|—62)cost ——3c;;e' +c4e“ —|—t+2. c1e'+ cge" +-Z-te'-1-1,y=t—2—(Q+c1)e‘ —3c2e" —-§-te‘. c1e'+62sint+03cost -—-f5(3 cos2t+4sin2t), cle‘+(C2—c3)sint+(cg—|—c3)cost ——-f*5(2 cos2t+sin2t). c1e"‘ +cge‘/2 +2e‘+2t, 2 —c1e_l —|—C52em+2e’+ti—t+c3. 5ce"2‘, y=-—3ce"2‘. :v=t2+t—1, y=—t. Nosolutions. Inﬁnitely many. Deﬁne :c(t)arbitrarily andsolve fory(t). Inﬁnitely many; a:(t)=——y(t). Nosolutions. Onesuch equivalent triangular system is (D—2)w+ ,81/—- (2D—-5)w+(D+9):; (D2-121)+2s)¢ General solution should contain three arbitrary constants. 1 31 61¢ 2: 361: 3:(a):v=c1e, y=c2e ——§e, z=c3e -3-e —cge .H8128 x2 HQSSQHQHHH y= 3: z=t, 2t+1, -—10t —2. (b)a:=c1e', y=L261 e'+ c2e_', z=-—-3-;e'+ L?e"'+ 4:303‘/2. (c):1:=4c1e2‘, y=3c1e2‘+ 5c2e'2‘, z=——6c1e2‘ -2c2e"2‘ +c3e3‘. (d)x=c1+cge‘,y=-—c1+(c2t+ c3)e‘, z=—c1+(Cgi—-62+c3)e'. x=§re‘+1}e"‘, y=-Q-e‘-ﬁe" —t. 2,=122.6:/3 _1216-¢/a _ y=_19e:/3 +1216-us +9,+9+.}e:_ 7 1 _ 92; 2; t 1 +1’ y‘46'4“ +2+4+cazo§_»3:-Z=Ze2¢_4te2:+,2 a:=§e_‘—-159/2+2e'+2t, 4-z 5:12 1121/=——§e —-ée +2e+§—t—|-1. 4-24 Srsrsns. LINEARIZATION orFmsr ORDER Srsrsms Chapter 7 LESSON 32. Linearization ofFirst Order Systems. Lettheacandycomponents ofthevelocity ofaparticle begiven by (32.1) $5=ray). $5-,’=gay). where thefunctions f(a:,y) andg(:c,y) areassumed tohave continuous partial derivatives forall(z,y). This system oftwoﬁrst order equations may, inmany problems, bedifficult tosolve. What wedoinsuch cases istoﬁndtheﬁrst three terms oftheTaylor series expansion off(z,y) and g(:c,y), and then eliminate theparameter tbetween them. The Taylor series expansions ofthese functions about thepoint (0,0) are,seeLesson 38, (32-11) %=f(w.y) =an+aw+at-y+031112+airy+cry”+---. 5%=o(w.y) =be+b1r+ 1121/-l" bar’+My+115112+---- Using only theﬁrstthree terms ineach expansion, anddividing thesecond bytheﬁrst, weobtain §11_ bo+b1$+ (>21/_ (3212) drv"at+rm:+1121/ The solution ofthisequation willgive what iscalled aﬁrst approximation tothepath oftheparticle.* Theequation (32.12) isnow ofthetype with linear coeﬂicients discussed inLesson 8.Ascommented in8.26, theresult- ingimplicit solution of(32.12) willfrequently besocomplicated astobe oflittle practical useindetermining thepath oftheparticle. Weshall nowshow howimportant information astothecharacter oftheparticle’s motion canoften beobtained more easily from thedifferential equation (32.12) itself than from itscomplicated implicit solution. InLesson 8B,weshowed how byatranslation ofaxes, itisalways possi- bletotransform adifferential equation oftheform (32.12) with linear coefficients (assuming these represent nonparallel lines) tooneoftheform (alflf+bly)dzc+(1122+bgy)dy=0withhomogeneous coeﬂicients. All weneed doistotranslate theorigin tothepoint ofintersection ofthetwo nonparallel lines represented bythese linear coefficients. Hence weshall assume inwhat follows that thisshift oforigin hasbeen made. Wethere- foreneed consider only equations oftheform Q_111$+bi?! Q b_1(32.13) dz-————a2x_,_bzy! ax95b2and G223+bgyre0. ‘Itispossible fortheﬁrstapproximation tobevery wide oftheactual path ofthe particle because theomitted terms arerelevant andimportant. Lesson 32—Case 2 LINEARIZATION orFIRST ORDER SYSTEMS 425 Tosave writing, wedeﬁne D=a,b; -D102. Note that Disthedeterminant <11bl (12 D2 whose elements arethecoeﬂicients ofxandyin(32.13). Weconsider separately each ofthecases resulting from different values ofa1,a2,b1,b2. Inallcases, dy/dx ismeaningless attheorigin (0,0). Hence, inallcases, nosolution willlieonthispoint. However, every other point intheplane isanordinary point, including points inaneighborhood oftheorigin, nomatter how small. Hence, byDeﬁnition 5.41, theorigin is, inallcases, asingular point. For convenience, however, weshall say “solutions through theorigin." Thisexpression istobeinterpreted tomean solutions through points inaneighborhood oftheorigin, butnotthrough the origin itself. Case 1.a1=b2=0andbl=a2.Inthiscase (32.13) simpliﬁes to d(32.15) i= .1-s0, whose solutions are (32.16) y=ex, x750. Itisafamily ofstraight lines through theorigin (0,0). Thus forthis special case, thefamily ofintegral curves of(32.13) iseasily drawn. Case 2.b1=——a2. Inthiscase, (32.13) becomes Q_ala:—112]]_ (3217) dz“12¢+bzy Itssolution bythemethod ofLesson 7B,is (32.18) a1a:2 —2a2xy —b2y2 =c. Since b1=—-a2, weobtain by(32.14), (32.181) D=albg +(Z22, which isalsothediscriminant ofthequadratic equation (32.18). From analytic geometry weknow thefollowing. 1.IfD<0andcsf0,(32.18) isafamily ofellipses with center atthe origin; ifc=0,only thepoint (0,0) satisﬁes theequation. Butthe 426 SYSTEMS. LINEARIZATION orF1Rs'r ORDER SYSTEMS Chapter 7 point (0,0) isexcluded since, by(32.17), dy/da: ismeaningless there. Hence thefamily ofintegral curves of(32.13) will resemble those shown inFig.32.l9(a). J’ \, k_% /\’ @\ Figure 32.19 2.IfD>Oandc¢0,(32.18) isafamily ofhyperbolas with center at theorigin; ifc=0,(32.18) degenerates intotwostraight linesthrough theorigin which aretheasymptotes ofthefamily ofhyperbolas. Hence thefamily ofintegral curves of(32.13) willresemble those shown in Fig.32.19(b). For this special Case 2,therefore, thefamily ofintegral curves of (32.13) canalsobeeasily drawn. Case 3.General case, a,b; 9-6(12111, withnoother restrictions placed on a1,a2,bl,b2.LetP(z,y) beapoint onanintegral cm've of(32.13), y P(x’ y 1‘%\°QG B B c y y.P(x.y) I ' xy, -74)‘ J’'90’ yl y OO A A y—10" <-> C an Figure 32.2 Lesson 32—Case 3 LINEARIZATION orFmsr ORDER SYSTEMS 427 Fig.32.2(a). The slope ofthetangent lineCPtothecurve istherefore y’. Assume thislinedoes notgothrough theorigin O.Draw OBparallel to CP. Itsslope isalso y’.From Fig. 32.2(a), weseethat (tan {AOB = y’=AB/ac; therefore AB=xy’) (32.21) OC=AP——AB=y——xy’. Iftherefore OCliesabove theasaxis, then y—xy’>0.Further weknow from thecalculus that ify”>Oatapoint Pofacurve, then thecurve inaneighborhood ofPisconcave upward andatangent atPliesbelow thecurve. Wehave thus proved that ifatapoint P(z,y) onanintegral curve, y">0andy—xy’>0,then thetangent lineatPseparates curve andorigin inaneighborhood ofP. Asimilar conclusion canbeproved ify"<0andy—xy’<0,see Fig. 32.2(b). If,however, y”andy—xy’have different signs atP(z,y), then origin andanintegral curve near Pwilllieonthesame sideofthetangent line atP.Forexample, inFig. 32.2(a), draw atPanintegral curve which is concave downward sothaty”<0. Bydifferentiation ofouroriginal differential equation (32.13), weobtain after simpliﬁcation, b— b<32-22> y"=,i,,‘j—,,—,f‘g§,-,1‘. (y—-1'» By(32.13), thedenominator ofthefraction in(32.22) isnotequal tozero. Hence itisalways positive since itissquared. Itsnumerator by(32.14) isD.Iftherefore (32.221) D>0,y"and(y—any’)have thesame sign, D<0,y”and(y—-xy’)have opposite signs. Because of(32.221) andtheremarks after (32.21), wecannowassert thefollowing. Comment 32.23. Ify—xy’950,i.e.,ifthetangent lineatapoint Pofanintegral curve of(32.13) does notgothrough theorigin, andif Dof(32.14) >0,then curve andorigin, inaneighborhood ofP,are separated bythetangent atP;ifD<0,then curve and origin, ina neighborhood ofP,lieonthesame sideofthetangent atP. Ifthetangent atapoint Pofanintegral curve of(32.13) goes through theorigin, then itisevident from Fig. 32.2(a) or(b),that OC=0and therefore by(32.21), that y—say’=0.This means that if(32.13) has rectilinear solutions (i.e., straight linesolutions) through theorigin, these solutions must satisfy theequation y—-my’=0.Forexample, thestraight linesolutions through theorigin which wefound inCase 1,namely y=cx, satisfy thisrequirement. Forthen y’=candy—-xy’=ca:—xc=0. 428 SYs'rEns. LINEARIZATION orFmsr ORDER SYs'rEus Chapter 7 The question wenow askisthefollowing. Arethere, forthegeneral case, rectilinear solutions of(32.13) which gothrough theorigin? The answer is:allthose solutions which satisfy theequation y—xy’=0. From (32.13) weﬁnd, after simpliﬁcation, 2231 _,__52!/2+(112—b1)wy —a1w’_ (3I) y xy_ 1121?+52?! Hence thelocus ofthose points (z,y) which willmake thenumerator of (32.23l) zerowillmake y—xy’=0.Such lociwillthen berectilinear solutions of(32.13). Toﬁndthem, weset (32.24) b2]/2 +(a2-—bl):cy —al:c2 =0. LetuscallAthediscriminant of(32.24). Then (32.25) A=((12--bl)2 +4alb2 =a-22-—2a2bl +bl2+4:(l1b2 =(122+211251 +1112—4112171 +441152 =(112-1-b1)2 +4(a1b2 -02171)- By(32.14), wecanwrite (32.25) as (32.26) A=(<1,+bl)’+4D. From algebra, weknow thatif (32.27) (a)A>0,then (32.24) willhave tworeal, distinct factors, say 01¢+by)(¢r +dy)=0. (b)A=0,then (32.24) willhave onerealrepeated factor, (c)A<0,then (32.24) hasonly imaginary factors. Incase (a),there willthus betworectilinear integral curves of(32.13) through theorigin. These twolines willseparate theplane intofourregions inwhich allother integral curves of(32.13), forwhich y—my’960,will lie. Incase (b),there willbeonly onerectilinear integral curve through the origin. This linewilldivide theplane into tworegions inwhich allother integral curves of(32.13), forwhich y-—xy’¢0,willlie. Incase (c),there willbenorealrectilinear solution of(32.13). By(32.26), wenotefurther thatwhen D>0,AisB-180>0-B111when D<0,Amay take onanyofthethree values in(32.27). Wehave then four possibilities toconsider inthegeneral Case 3:onewhen D>O; three when D<0. Case 3-1. Dof(32.14) >0,[and therefore Aof(32.26) >0].[No'rE. IfD>0andalsobl=-a2, then thespecial Case 2applies; see2after (32.l81) ofthat case.] ForthisCase 3-1weknow from theremarks above, that (32.13) hastwo rectilinear solutions, each ofwhich isafactor of Lesson 32—Case 3-1 Lmmmzzmon orFmsr ORDER SYSTEMS 429 (32.24). Allpoints onthese twolines willmake y—xy'=0.Allother integral curves forwhich y—my’;£0,willlieinthefour regions made by thetwo rectilinear solutions. And since D>0,weknow byComment 32.23, that atangent linedrawn atanypoint ofanintegral curve willsepa- rate curve and origin. The integral curves will thus have thegeneral appearance ofthose shown inFig. 32.281. The rectilinear solutions are asymptotes oftheintegral curves. Theorigin itself, however, isasingular point. The curves, while nothyperbolas, willhave their general appear- ance. Example 32.28. Discuss, without solving theequation, thecharacter ofthesolutions of 4::-—y I2. _ w y5?; Solution. Comparing (a)with (32.-13), weseethat a1=4,bl=-1, a2=2,b2=1.By(32.14), D=4+2=6>0.Hence thisCase 3-1 applies. There should therefore be,andasweshall now show, there are twofactors of(32.24) each ofwhich isarectilinear solution of(a). Sub- stituting in(32.24) theabove values ofa1,a2,bl,b2,weobtain W f+%w%#=Q o+em—o=u Hence thetworectilinear solutions of(a)are (c) y+4:z:=0, y—:z:=0. 0.0) XI l Figure 32-281 430 Srsrmas. LINEARIZATION orFmsr Onnaa Srsraus Chaptgf 7 [Verify that thefunctions deﬁned in(c)aresolutions of(a).] They are shown inheavy linesinFig.32.281. Note thatatangent drawn atapoint ofanonrectilinear integral curve separates curve andorigin. Case 3-2. Dof(32.14) <0.[NOTE. IfD<0andalsobl=—a2, thenthespecial Case 2applies; see1after (32.l81) ofthatcase.] Forthis Case 3-2,asremarked earlier, Aof(32.26) may take onanyofthethree values in(32.27). Weshall therefore need toconsider each ofthese three possibilities separately. Before doing so,however, itwillbenecessary for ustohave additional information. Wetherefore digress momentarily in order toobtain thisinformation. Thefamily ofsolutions of(32.13) hasslope y’.Letyl’betheslope of anisogonal trajectory family which cutsthisgiven family inapositive <lCa, measured counter clockwise from y’toyl',Fig.32.3. Y y‘=slope ofgiven family yi=slope ofisogonal trajectory family O X Figure 32.3 Bya.formula. inanalytic geometry, therefore, I_ I (32.31) t3.Xl0l= Solving (32.31) fory’,weobtain _yl’-—tana _(32.32) y’-i———i1+(tan“)3”, Replacing y’in(32.32) byitsvalue asgiven in(32.13), there results yl'—tana _alx+bly_ (32-33) 1+(tall '-‘>311’ 112$-l-1723/ Thesolution of(32.33) foryl’is ,_(a2tana+al):v+(b2tana+bl)y , (3234) yl_(a2—altana)x+(bg—bltana)y which hasthesame form as(32.13). Wehave thusproved thatthediffer- ential equation ofafamily ofisogonal trajectories, making an{ozwith thegiven family ofsolutions of(32.13), hasthesame form as(32.13). Lesson 32—Case 3-2(a) Lrusmrzxrron orFmsr ORDER Srsrnms 431 The determinant D1,whose elements arethecoefficients of:0andyin (32.34), is (32.35) D1=(a2tana—|—al)(b2 —bltana)—(b2tana—|—bl)(a2 —altanoz) =(albg —a2bl)(tan2 oz—|—1)=D(tan2 a+1)=Dsecz a, where Disgiven by(32.14). Since sec2a>0,weseefrom (32.35) that DIandDhave thesame sign. ByCase 2,thefamily ofsolutions of(32.34), which remember isan isogonal family oftrajectories ofthefamily ofsolutions of(32.13), willbe ellipses orhyperbolas with center attheorigin if bg123.11 (2+bl= -(G2 "-G1138.11 (1), i.e.,if __112‘l'bl_(32.37) tana--————a1__b2 They are,byCase 2,ellipses ifD;<0orequivalently, asnoted above, ifD<0. Comment 32.371. Wehave thus proved that ifD<0andaisan angle which satisﬁes (32.37), then thefamily ofisogonal trajectories which cuts thefamily ofintegral curves of(32.13) inthisangle a,measured from theintegral family counterclockwise totheisogonal trajectory family, is afamily ofellipses with center attheorigin. Ifoz;-50,each integral curve willtherefore approach theorigin inonedirection andrecede inﬁnitely in theother direction. Theorigin itself isasingular point. If0:=O,then the isogonal trajectory family coincides with theintegral family and the integral family isthus alsoafamily ofellipses asinCase 2.[Note that if oz=0,then tanor=Oandby(32.37), a2=—-bl asinCase 2.] Wearenow ready toexamine each ofthethree possibilities under Case 3-2. Case 3-2(a). Dof(32.14) <0andAof(32.26) >0.Inthis case, there willbe,by(32.27) (a),tworectilinear solutions of(32.13) through the origin. These lines willseparate theplane into four regions. Foranon- rectilinear integral curve of(32.13), wenow know thefollowing facts. 1.Itcannot cross either oftherectilinear solutions. 2.ByComment 32.23, origin andintegral curve, inaneighborhood ofa point Ponit,lieonthesame sideofthetangent atP. 3.Ifozisanangle which satisﬁes (32.37), then thefamily ofisogonal tra- jectories cutting theintegral family of(32.13) inthisangle a,measured 4-32 Svsrrms. Lrunanrzxrron orFmsr Osman Srsraus Chapter 7 from theintegral family counterclockwise totheisogonal trajectory family, isafamily ofellipses, withcenter attheorigin, satisfying (32.34). Inaddition, ithasbeen proved that 4.Anintegral curve willapproach theorigin inadirection tangent tooneoftherectilinear solu- (°’) tions; seeExercise 32,12. 5.Aradius vector toapoint P moving along anintegral curve away from theorigin willap- Figure 32.38proach coincidence with thesec- ondrectilinear solution. Thegraphs oftheintegral curves willtherefore resemble those shown in Fig.32.38. Thetwoheavy lines aretherectilinear solutions. Example 32.39. Discuss, without solving, thecharacter ofthesolu- tions of 2 4 <a>Solution. Comparing (a)with (32.13), weseethat al=2,bl=4, a2=1,bl=——1.By(32.14), D=-2—-4=-—6<Oand by(32.26), A=(1+4)’+4D=25-—24=1>0.Hence this Case 3—2(a) applies. Since A>0,there should be,andasweshall nowshow, there aretworectilinear solutions of(a),each ofwhich isafactor of(32.24). Substituting in(32.24) thevalues ofal,bl,a-2,bl,weobtain (b) 2/’+31y+2w’=0. (y+2w)(1/+Z)=0~ Each ofthelinesy+2x=0andy+as=0aresolutions of(a).(Verify it.) They areshown inheavy lines inFig. 32.391. Allother integral curves must lieinside thefourregions formed bythese twolines. From (32.37), weﬁndtana=5/3, sothat aisapproximately 59°. With thisvalue oftanoz,(32.34) becomes y1,= (§+2)$+(-§-l-4)y =J3'“9=+:7;?/ =11$-l-71/, (1"1s“)%+(""1 —25°)?! —'.7§$ "2531'! “‘7$ —23? which has,asitshould, thesame form as(32.17). Hence by(32.18), its solution is (c) 11¢”+14:cy+23y’=c. Lesson 32-Case 3-2(b) Lrnmarzuron orFrnsr Onnan Svsrans 4-33 Arotation oftheaxes through anangle ofapproximately 65°18’ will eliminate themyterm and(c)willbecome, approximately, (d) 26222 +8172=c, which represents, asitalsoshould, afamily ofellipses. Some ofthese ellipses areshown inFig.32.391. Each nonrectilinear integral curve of 3' _x Y x Figure 32.391 (a)willcutthisfamily attheconstant angle oz=59°,measured counter- clockwise from theintegral curve totheellipse. Wehave shown apart of onesuch integral curve intheﬁgure. Case 3-2(b). Dof(32.14) <0andAof(32.26) =0.Inthiscase, there willbe,by32.27(b), only onerectilinear integral curve through the origin. This linewillseparate theplane into tworegions. Thesame com- ments wemade intheprevious lesson inregard toanonrectilinear integral curve willapply here. These are: 1.Itcannot cross therectilinear solution. 2.ByComment 32.23, theorigin andeach integral curve inaneighbor- hood ofapoint Ponit,lieonthesame sideofthetangent atP. 3.Iforisanangle which satisﬁes (32.37), then thefamily ofisogonal trajectories cutting thegiven family ofsolutions of(32.13) inthisangle oz,measured from theintegral family counterclockwise tothetrajectory family, isafamily ofellipses, with center attheorigin, satisfying (32.34). 4.Anintegral curve willapproach theorigin inadirection tangent tothe rectilinear solution; seeExercise 32,13. 5.Aradius vector drawn toapoint Pmoving along theintegral curve away from theorigin will approach coincidence with therectilinear solution. 434 Srsrnms. Lrnaxnrzxrron orFmsr Oanaa Srsrr-ms Chapter 7 The graphs oftheintegral curves will thus resemble those shown in Fig. 32.4. Figure 32.4- Example 32.41. Discuss, without solving, thecharacter ofthesolu- tions of ,__rv+3y_(a) y—x__y Solution. Comparing (a)with (32.13), weseethat al=1,bl=3, a2=1,b2=——1. By(32.14), D=-1—3<0.By(32.26) A= (4)2—l-4(——4) =0.Hence this Case 3—2(b) applies. There should be, and asweshall now show, there is,only onerectilinear solution of(a). Substituting in(32.24) thevalues ofal,a2,bl,blgiveabove, weobtain (b) 1/”+2wy+w”=0. (y+w)2=0- Hence theliney=-—:vistheonly rectilinear solution of(a). Wehave leftittoyouasanexercise, toﬁndthetrajectory family ofellipses aswe didinExample 32.39, andtheangle aof(32.37) which each integral curve makes with these ellipses. Draw some ofthese ellipses and anintegral curve. Itsgraph should resemble acurve ofFig. 32.4. Case 3-2(c). Dof(32.14) <0,Aof(32.26) <0.Inthis case by (32.27) (c),there arenorectilinear solutions of(32.13). And if0:isanangle satisfying (32.37), then byComment 32.371, thefamily ofisogonal trajec- tories which cuts thegiven family ofsolutions of(32.13) inthisangle a measured from theintegral family counterclockwise tothetrajectory family, isafamily ofellipses with center attheorigin. Iftherefore a=0,the family ofsolutions oftheoriginal equation (32.13) coincides with thetra- jectory family andthus arealsoellipses with center attheorigin. Hence ifoz=O,every integral curve encircles theorigin. If,however, oz¢0,then each integral curve of(32.13) willcuttheisog- onal trajectory family ofellipses, which areintegral curves of(32.34), in Lesson 32—Case 3-2(c) Lrnnxmzxrron orFmsr ORDER Svsrsms 435 thisconstant angle oz.Weshall now prove forthisCase 3—2(c), that when an9'50,anintegral curve willalsoencircle theorigin. The equation ofthefamily ofstraight lines through theorigin is (a) 1/=M; y’=m- Replacing minthesecond equation of(a)byitsvalue y/2:obtained from theﬁrst equation, wehave y'=1n=-Z-. Let0betheangle ofintersection ofalineof(a)andanintegral curve of (32.13), measured from theintegral curve totheline, Fig. 32.42. Then, byaformula inanalytic geometry, 2g__<11w+b1y _ g_ (.)tan0= rw+'>21/ = +(“’ we“‘.y zazx-l-bzy _ a1+b2);+a2 O"Q‘.-to AmQQQR‘-QN +/-\ By(b)wecanwrite (c)as __bzmz "l"(412""bi)?" "a1_ (d) tan0—blmz -l"(411-1-52)?" -l"<12 Y Xl\!>°°&la0 P, P(x.y) "1023 y,=alx+ bly '3 ¢zr+bzy O X Figure 32.4-2 Anintegral curve of(32.13), therefore, cutseach lineof(a)attheangle 0 given by(d). [For each lineof(a),there isadistinct mand therefore by (d),adeﬁnite angle 0.]Thenumerator of(d)is (9) bzmz +('12"-bi)?" -"<11- Itsdiscriminant is ff) (112—b1)2 -l"411152, which isexactly thesame asthediscriminant Aof(32.24) [see(32.25)]. Since byourassumption forthisCase 3—2(c), A<0,itfollows thatthe 436 Srsrsns. Lrnnxarzxrron orFmsr Oanaa Srsrams Chapter 7 expression in(f)islessthan zero. Hence, asnoted in(32.27) (c),there are norealvalues ofmwhich willmake (e)zero. Since (e)isthenumerator of(d),thismeans that theangle 0in(d)isnever equal tozero. Anintegral curve of(32.13), since itcuts every straight linethrough theorigin at anangle 0950,must therefore en- circle theorigin. Hence inboth cases, a=0and a¢0,anintegral curve of(32.13) willgocompletely around theorigin. Intheﬁrst case, itwillenclose the origin intheform ofanellipse asin Fig. 32.19(a); inthesecond case in (<1) (b) theform ofspirals asshown inFig. 32.43(a) and(b).Inboth cases the Fiﬂllre 32-4-3 origin isasingular point. Example 32.44. Discuss, without solving, thecharacter ofthesolu- tions of dy____2:c—y+1_ la) 75* w+y (Nora. This equation wassolved inExample 8.25. Itsimplicit solution canbefound there.) Solution. Atranslation oftheaxes sothat theorigin isat(——§, §), seeExample 8.25, transforms (a)totheform dy_—2a:—l-y_ ‘bl at-TH Comparing (b)with (32.13), weseethat al=-2, bl=1,a2=1, b2=1.You canverify that Dof(32.14) <0and Aof(32.26) <0. Hence thisCase 3—2(c) applies. By(32.37) (c) tana=—§, oz=-—33°41'. With thisvalue oftana,(32.34) becomes (d) I: _833‘l‘?/_ ”‘ <1—ax+(1+%>y —-@+5@/ Itssolution by(32.18) is (e) 8:02-2:vy—|—51/2=c. Lesson 32—Case 3-2(c) Lrnamrzxrron orFrasr Onnan Srsrsms 437 You canverify that (e)has, asitshould, only imaginary factors. Arota- tion oftheaxes through anangle _ ofapproximately 73°10’ willtrans- X form (e)to Y (f) 4.722’+8.37;’=c, which represents afamily ofellipses. X Afewofthese ellipses areshown in Fig.32.45. Each integral curve must cuteachellipse attheconstant angle —33°4l’, measured from integral curve toellipse. Wehave alsoshown inthisﬁgure part ofonesuch inte- _ gralcurve. F'g“"e 32'“ Comment 32.46. Position oftheParticle asaFunction ofTime. Intheabove discussions ofthepath ofaparticle, weeliminated theim- portant element oftime. Hence, although wehave shown thegeneral appearance ofthepath, intheabsence ofanequation connecting theposi- tionoftheparticle with time, wecannot know where theparticle isatany instant. Toobtain this information, wemust change equation (32.13) back tothesystem from which itcame, namely d d(32.47) i=112$+bay, %=alx—|—bly, andthen solve itbythemethods ofLesson 31E. Inoperator notation, we canwrite (32.47) as (32.48) (D—a2)a: —bgy=O, —al:c +(D—bly) =0. Following themethod outlined inLesson 31E, weretain theﬁrst equation andchange thesecond bymultiplying theﬁrstby(D——bl)/b-2 andadding ittothesecond. There results theequivalent triangular system (32.49) (D—a2):c —bzy=0, (D-b)(D—) __..].-..Thesecond equation in(32.49) reduces to (32.5) [D2——(a2+bl)D +02171 —-alb2]x =0. Thecharacteristic equation of(32.5) is (32.51) m2—(a2+bl)m —|—azbl —albg =0, 438 Srsrnns. Lrnnxarzxrron orFrasr Oannn Srsrans Chapter 7 whose roots are ./*~—}(32.52) ml="2+bl+(“E “)2+‘mlb’. 02-l"b1"‘V(112*b1)2 +411152 2 Hence thesolution of(32.5) is (32.53) x=cle"“‘ +c2e""',m2= where mlandmghave thevalues given in(32.52). Note that thechar- acter oftheroots mland mg,and therefore thesolution av,depends on whether thediscriminant (a2——bl)2 +4alb2 ispositive, zero, ornegative. Substituting thisvalue ofxintheﬁrstequation in(32.48), weobtain (32-54) bay=(D—¢l2)(¢1¢m‘t —|—Czemﬂ) =mlcle'"“ -l—m2c2e""' —a2cle’”“ —a2c2e""' ___ __ mlt __ ‘mg! -—¢1(m1 a2)e +cz(m2 a2)¢ - Hence 1 1.. m(32-55) 1/=El61(m1 —¢12)6 ‘l-l-¢2(m2 "-412)‘? atl- ByTheorem 31.33, thepairoffunctions a:(t),y(t)asgiven in(32.53) and (32.55), have thecorrect number oftwoarbitrary constants. Thevalue of theconstants clandC2willdepend ontheposition oftheparticle att=0. Thesolutions :c(t), y(t)of(32.53) and(32.55) willthen give theposition of theparticle atlater times t,see,forexample, Exercise 32,11. EXERCISE 32 Discuss without solving thecharacter ofthesolution ofeach ofthefol- lowing differential equations. (First determine towhich ofthedifferent cases each belongs.) Draw rectilinear solutions where these exist. Draw arough graph ofanintegral curve. 1.g ..35 ..gg ..gg dy5.8;=1.2:-1 2:--31-———~iI 7 31-}-y _2: 3y —32:—10y 8' _4:c+y-4 9.—2x—l—y =2:2:-l—y_ 3a:—l—ydy 8? dy Z15 dyH dyH Qdz:=3a:—l—y_ 2:c—l-y =-—2a:—l—y_ 1+1; =1iL.32:-—y _,,._y =7??? it—?! =3! Lesson 32—Exercise 439 11.Assume theoriginal system from which theequation ofproblem 4above came is dz dy_5-—21+1/. ,75—4:v+u. andthattheinitial conditions are:c(0) =1,y(0) =2.Find theparametric equations ofthepath andthevalues of2andywhen t=1. 12.Prove statement 4ofCase 3—2(a). Hint. Follow theproof given in Case 3—2(c). 13.Prove statement 4ofCase 3—2(b). Hint. Follow theproof given in Case3—2(c). ANSWERS 32 1.Case 1. 2.Case 2;family ofhyperbolas. 3.Case 2;family ofellipses. 4.Case 3-1;y=42:,y=—-x. 5.Case 3—2(a); y=0.732;, y=-2.732. 6.Case 3—1;y =1.3:, 1/=--2.31. 7.Case 3—2(c). 8.Case 3—2(b); y=2:. 9.Case 2;family ofellipses. 10.Case 2;family ofhyperbolas. ll.:c(t)=§e"3‘—|- §e2‘, y(t)=—-§e"3‘—l- 1-fez‘; 2:=4.45, y=17.71. Chapter 8 Problems Giving Rise toSystems ofEquations. Special Types ofSecond Order Linear andNonlinear Equations Solvable byReduction toSystems LESSON 33. Mechanical, Biological, Electrical Problems Giving Rise toSystems ofEquations. LESSON 33A. AMechanical Problem—Coupled Springs. Two masses mlandmgrestonafrictionless, horizontal table. They arecon- nected toeach other andtotwoﬁxed supports bythree unstretched springs Sl,S2,S3with respective spring constants kl,kg,k3,Fig.33.1(a). ——>s, m‘ s, m’ s, + (<1)kl kz ks Ol-equilibrium O2=equilibrium position ofml position ofml S1 ' S (b) 1'.kl k2: Ik3 |-~=-l.57““"-_l_w$5 Figure 33.1 [For themeaning ofaspring constant, see(28.61).] The masses canbe displaced from their equilibrium position byholding, forexample, ml, moving mgtotheright andthen releasing both. Figures 33.1(a) and (b) show theposition ofthetwomasses before andafter adisplacement. We wish toﬁndtheequation ofmotion ofeach mass. 44-0 Lesson 33A AMncnxnrcn. PROBLEM——-COUPLED Srnmos 441 Denote byOltheposition ofmlandbyO2theposition ofm3,when thesystem isatrest. Letxlbethedisplacement ofmlfrom itsequi- librium orrestposition Ol,and2:2thedisplacement ofm2from itsequi- librium position O2,attime t.Hence when mlisatmlandm2isat(I32, Fig. 33.1(b), spring Slhasbeen elongated adistance xlfrom itsequi- librium position and spring S2hasbeen elongated adistance :02—xl from itsequilibrium position. Forexample, ifacl=1footand:03=3feet, then spring S3hasbeen elongated only 1}foot, i.e.,adistance x3——xl. Therefore therestoring forces acting onml,with thepositive direction to theright, aretwoinnumber. 1.Aspring force duetoSlacting totheleft. ByHooke’s law, which youwillﬁndinExample AofLesson 28C, thisforce is—Iclxl. 2.Aspring force due toS2acting totheright. ByHooke’s law, this force isk2(a:2 xl). (Remember thespring S3wants torestore itself toitsoriginal sizeand willtherefore pull onmltotheright and on m3totheleft.) Hence thedifferential equation ofmotion ofmlis d2 ml -1% Z —-lC1I1 +lC2(.’lIg -Z1). Asimilar analysis shows that therestoring forces acting onm3are twoinnumber. 1.Aspring force duetoS3acting totheleft. ByHooke’s law, thisforce IS—lC2(IlZ2 —IE1). 2.Aspring force duetoS3acting totheleft. Remember S3hasbeen squeezed adistance :02anditwants torestore itself toitsoriginal size. Therefore, byHooke’s law, thisforce is—k3:z:3. Hence thedifferential equation ofmotion ofm3is d2 mgm? = —lC2(1U2 —$1) *-lC3Z2. Thesolution ofthesystem oflinear differential equations (33.11) and (33.l11) willgive theequations ofmotion xl(t) andx3(t) ofthecoupled springs diagrammed inFig. 33.1. If,inaddition tothespring forces, aforcing function Fl(l)isattached tomland aforcing function F3(t) isattached tom3asshown inFig. 33.121, then thepair ofequations (33.11) and (33.l1l) becomes 2 <33-12> m.dy,?+16.».+k2(w1-22>=mo. (1212 "12W" +763962 —|—k2(@2 —$1)=F2“)- 44-2 Pnonnmasz Svswams. Sracln. 2NDORDER EQUATIONS Chapter 8 m s, '"‘F.s, ’F.S8 +—> O1 O2 Figure 33.121 And ifweassume that because offriction orbecause thesystem is . . . . dilil .immersed 1namedium, there ISadamping force 1'1Wacting onmland dadamping force 1-2%acting onm2then thesystem (33.12) becomes 2 <33-122) m.9%+11‘1’j—;+701311+km.—32>=F10). 3” <1m2"a*t% +13% +793112 +k2(f'32 "‘931)=F2(t)- Ifallspring constants have thesame value k,then (33122) simpliﬁes tothelinear system 2 (33.13) m1% +r19%+216$,-la,=F1(t), dz dmgﬁt? +T2"-% +2,0112 _"10131: F2(t). Inoperator notation, wecanwrite (33.13) as (33.14) (m1D2 —|—1‘1D —|—2k):t1 -—101:2 ==F1(i), **k$1 +(m2D2 +T21) —|—270112 =F2“)- Wecan eliminate 2:2bymultiplying theﬁrst equation of(33.14) by (m2D2 -l-r2D +2k), thesecond bykand adding thetwo. Wethus obtain (33.15) {m1m,D* +(mm+m2r,)D3 +[rm+2k(m1 +m2)]D2 +[Mn+r.>1D+3k’}w1=<m2D’+120+2k>F.(¢) +kF2<¢>, anequation which isnow linear inx1.Itsgeneral solution canbeob- tained bytheusual methods outlined inprevious lessons. When 1:1has been determined, thegeneral solution forx2canbeobtained from the ﬁrst equation in(33.14). Note that thegeneral solution of(33.15) for x1(t) willcontain fourconstants. Substituting thisvalue ofx1(t) inthe ﬁrstequation of(33.14) andsolving itfor:c2(t) willaddnonewconstants. ByTheorem 31.33, thecorrect number ofconstants required inthegen- Lesson 33A—Exe1-cise 443 eralsolution ofthesystem (33.15) isfour. Hence thepairoffunctions x1(t), a:2(t) willcontain thecorrect number ofconstants. Comment 33.16. The characteristic equation oftheleftside of (33.15) setequal tozero isofthefourth degree. Hence, asweremarked inComment 18.26, these roots may notbeeasy toﬁnd, especially ifthey areimaginary. Remember these roots areneeded inorder towrite the complementary function y,,ofthegeneral solution of(33.15). Ifhowever m1=mg,i.e.,ifboth masses arethesame, and1'1=r2,then theelimina- tionofx2in(33.14) willresult in (33.17) [('m1D2 +r1D +2k)2 —/021211 =("HD2 +T11)+2k)F1(t) —|—kF2(i)- Butnow ifwesettheleftsideequal tozero, itscharacteristic equation is (m1m2 +rlm—|—2lc)2—k2 =0which factors into (33.18) (m1m2 —|—rlm+Ic)(m1m2 +rlm+3k)=0. Itsroots canthus bereadily obtained bysetting each factor equal tozero. Comment 33.19. Iftheforcing functions F1(t) and F2(t) areiden- tically zero andifr1g0,r2g0,then asweshowed inLessons 28A and29A, themotion ofthesystem isstable. Iftheforcing functions are notidentically zero, then themotion ismuch more complicated and a breakdown ofthesystem may occur ifthephenomenon ofresonance is present, seeExercise 33A,1(b). (Ifyou have forgotten themeaning of resonance, reread Lessons 28D and29B.) EXERCISE 33A 1.(a)Solve thesystem (33.13) if‘M1=m2=m;r1=1'2=0;F1(t) = F2(f) =0. (b)The system in(a)hastwofrequencies. These arecalled normal frequencies. (Insimple harmonic motion, asystem had only one natural frequency.) Find thenormal frequencies of(a). Resonance occurs ifaforcing function, impressed onthesystem, haseither of these normal frequencies. (c)Find thesolution ifatt=0,mgisheld ﬁxed andm1ismoved tothe right Aunits andreleased. Hint. Initial conditions aret =0,:1=A, Z2=0,(1231/df =0,(1132/dl =0. 2.(a)Solve thesystem given byequations (33.11) and(33.11l) ifm1=1, mg =2,101 =1,102 =2,163 =3. (b)What areitsnormal frequencies? 3.Ifyouwrite thesystem ofproblem 1(a)inoperator notation, youwillobtain (a) (mD2 +2k):c1 ——16922 =0, -—kx1 —l—(mD2 +2k)Z2 =0. Since each differential equation ofthesystem islinear, itisreasonable to 4-4-4 Pnonnsmsz Srsrsms. SPECIAL 2NDOnnsn EQUATIONS Chapter 8 assume thatitssolutions willhave theform 2:1=c1e“‘, xg=cge"". Show thatthesubstitution of(b)in(a)yields thesystem (mwz —|—2k)c1 ——Iccg=0, —kc1 -l-(mwz —|—2k)cg =0. Note that (c)isobtainable from (a)byreplacing Dbyw,11byc1,xgbycg. Thesystem (c)willhave nontrivial solutions forc1andcg(i.e., solutions other than c1=0,cg=0)only ifthedeterminant mwz+211 -—k =0, 2 -——k mw +2k seeTheorem 63.42. Expand thedeterminant andsolve theresulting equa- tion, called thecharacteristic equation, forw.Note that theimaginary part ofthesolution forwisthesame asthenormal frequencies found in l(b). You willhave thus proved that theimaginary partofthesolution ofw,where wisobtained bytheprocess outlined above, willgivethenormal frequencies ofthesystem. Usethemethod outlined inproblem 3toﬁndthenormal frequencies ofthe system ofproblem 2.What isthedeterminant? Compare with answer to 2(b). (a)Show that when 2:1=zzg,thesystem ofcoupled iprings solved in problem 1(a)hasonly theonenormal frequency \/Ic/m rad/unittime; that when 1:1=——zg, ithasonly theonenormal frequency \/3k/m rad/unit time. (b)Show that iftheinitial conditions aresochosen that 03=0.1=0in theanswer toproblem 1(a), then 2:1=Iﬁld thesystem willvibrate with only theonenormal frequency \/k/m; ifchosen sothat c1= cg=0,then 2:1=—a:g andthesystem willvibrate with only theone normal frequency \/3k/m. (c)Lety1=3:1+zgandyg=2:1-—zg.Show that thesolution of1(a) becomes respectively y1=C1sin\/k/mt+C'g cos\/k/mt, yg=C3sin\/3k/mt+C4 c0s\/3k/mt. Note that each equation hasonly onenormal frequency. The ﬁrst equation (where 1/1=2:1+232)corresponds toamotion where thetwo masses move incoordination totheleftandtotheright with thesame amplitude; thesecond equation (where y1=2:1-—zg)corresponds toa motion where thetwomasses move inopposite directions with thesame amplitude. These newcoordinates y1andygwhich replace 2:1andxg arecalled normal coordinates. They areuseful insimplifying certain problems inthetheory ofvibrations. (a)Solve thesystem (33.13) ifm1=mg=2,k=32,r1=rg=0and I"1(t) =Fg(t) =4cost. (b)In(a),replace F1(t) andFg(t) by4coswt.What values ofwwillproduce resonance? Two masses m1andmgareconnected bytwosprings S1andSgwith respec- tivespring constants k1andkgasshown inFig.33.191. After thesystem is brought torest, themasses aredisplaced from their equilibrium position. Lesson 33A-Exercise 445 8. 9. lo. 11. l.Assume nodamping forces andnoforcing functions. (a)Find thedifferential equations ofmotion ofthesystem. (b)Find theequations ofmotion ofthesystem ifk1=kg=kandm1= mg=m. (c)What arethenormal frequencies ofthe system? S1kl Find thenormal frequencies ofthesystem ofproblem 7bymeans ofthemethod out- lined inproblem 3.What isthedetermi- nant? Compare with answer to7(0). (a)Usethemethod outlined inproblem 382kz toﬁndthenormal frequencies ofthe i x2=0,eq11i|ib1-gum system shown inFig.33.191, i.e.,with mz position ofm, m194mgandk19*kg.Seeproblem 7. (b)Write thenormal frequencies ifm1= Figure 33.191 mg=mZ x1=0,equilibrium ml position ofm1 Solve thesystem ofFig.33.191 ifk1=1,kg=2,m1=1,mg=2,andif, after thesystem is‘brought torest, themass mgisheldﬁxed, themass m1 isdisplaced Aunits downward andboth masses released. Hint. Att=0, x1=A,zg=0,d:c1/dt =0,dazg/dt =0. Inthesystem illustrated inFig.33.191, assume thatthemass m1isdamped byashock absorber whose coefficient ofresistance israndthat aforcing function F=F0sinwtisattached tom1. (a)Show that thedifferential equations ofmotion are ('m1D2 +TD-l-k1-l-k2)$1 —162$: =F0Sinwt, —-76211 +(M2132 —|—k2)fI=2 =0- Hint. Seeanswer toproblem 7(a). Modify ittoinclude thedamping factor andtheforcing function. (b)Byeliminating xg,show that [(m1D2 +1'D+k1+k2)(m2D2 +k2)—1622111 =Fo(k2 -—mzwz) Billwi- Remark. Thegeneral solution ofthislastequation for2:1will,asusual, be thesumofacomplementary function 2:1,andaparticular solution 2:1,. It canbeshown thatthemotion duetothe2:1,,partofthesolution isatransient oneandsubsides with time. If,therefore, kgandmgaresochosen that kg=mgwz, thentheright sideoftheequation iszero. Hence :01,=0,andthe motion duetothe0:1,,partofthesolution isalsozero. This means thatif kg=mgwz, themass m1willintime approach aposition 2:1andremain there. Hence, themass m1willnotvibrate. This result haspractical use when itisnecessary thatthevibrations ofaninstrument benegligible. ANSWERS 33A (a)2:1=c1sinmt-1- cgcosmhk 03sinmt +c4cos\/W t, zg=c1sinmhk cgcosmt —c3sin\/Wt —c4cos\/31::/Wm t. (b)\/k/Wm rad/unit time; \/W rad/unit time. 446 Pnonnsmsz Srsrams. Srncmt. 2m)Ono!-:11 Eouzrrrons Chapter 2. 4-. 6. 7. 8. 9. 10.(c)2:1=4;(cos\/k/mt—|— cos\/3k/mt), xg=3(cos\/k/mt —cos\/3k/mt). (a)2:1=c1cos(\/131+ 531)—|—cgcos(\/4-.-13¢ +8g), zz=111.6931 cos(\/E 1+3,)-1.1932cos(\/$131+ 32)]. (b)\/I33rad/unit time,\/H6rad/unit time. (.02+3 -2 -2 2...’+5 (a)2:1=c1sin4t+ cgcos4t+ c3sin4\/3t+ c4c0s4\/3t+ -fgcost, xg=c1sin4t—l- cgcos4t—c3sin4\/3t —c4cos4\/3t+ 1350051- (b)co=4rad/unit time andw=4\/3 rad/unittime. 3’ _(8.) m1-£1 =-——k1I1 —|—f62($2 --$1), d2 "I2ﬁ =—7¢2(I¢2 —11)- (b)a:1=c1sin t+ cgcos t+ casm t +c.1cos t. zcg=1.6201 sin t—|—1.62cg cos t _0.6233sinx/ -0.6231mi 1. (c)w= andw = mw2+21. -1. —k mwz +k (*1)miwz +kl+k2 —-kz —k2 m2w2 +k2 2__1k1+k1 Q) 1‘/(k1+k= kl)_4k1kg_w—\/ 2( m1 +mg 5:2 m1 +mg mung (b) _ 1kl-2k2:|:\/I012-l~4Jt22_ "’_ _ 2 .1=§[<3—\/5)cos(‘-/2-2}‘-[?)¢+ <3+\/:3) cost] 1»2:1=A(0.2l1 cos0.518t +0.789 cos1.932t), zg=A(0.289 cos0.518! —-0.289 cosl.932t). Lesson 33B ABIOLOGICAL Pnosusm 44-7 LESSON 33B. ABiological Problem. There isaconstant struggle forsurvival among different species. One species survives byeating members ofanother species; asecond bypreventing itself from being eaten. Certain birds liveonﬁsh. When, forexample, there isanabundant supply ofﬁsh, thepopulation ofthis bird species ﬂourishes and grows. When these birds become toonumerous andconsume toomany ﬁsh, thus reducing theﬁshpopulation, then their own bird population begins to diminish. Asthenumber ofbirds decreases, theﬁshpopulation increases. Andastheﬁshpopulation increases, thebirdpopulation starts increasing, etc., inanendless cycle ofperiodic increases anddecreases intherespective populations ofthetwospecies. _ Problems concerning theriseandfallofthepopulations ofinteracting species have been studied extensively bybiologists andmathematicians. The theoretical mathematical results which weshall develop inthislesson agree fairly accurately with actual population trends ofcertain interacting species!‘ Weshall consider theproblem ofdetermining thepopulations oftwo interacting species: aparasitic species Pwhich hatches itseggs inahost species H.Unfortunately fortheHspecies, thedeposit ofaneggina member ofHcauses thismember's death. Foreach 100ofpopulation: LetHbdenote thenumber ofbirths peryear oftheHspecies, H4denote thenumber ofdeaths peryear oftheHspecies, ifnoP species were present, i.e.,H4denotes H’snatural death rate, P4denote thenumber ofnatural deaths peryear ofthePspecies. Wenowmake thefollowing assumptions: 1.That H1,Hg,andP4areconstants. Hence ifattime t,a:isthepopu- lation oftheHspecies, ythepopulation ofthePspecies, then intime At (a)H1,%Atistheapproximate number ofbirths ofthehostH, (b) ——H,1 %Atistheapproximate number ofnatural deaths ofthe host H, (c) —P,1 %Atistheapproximate number ofnatural deaths ofthe parasite P. 2.That thenumber ofeggs peryear deposited bythePspecies result- inginthedeath oftheHspecies isproportional totheprobability that themembers ofthetwospecies meet. Since thisprobability depends on theproduct myofthetwopopulations, weassume that intime At, (d) —k:cyAt *V.A.Kostitzin, Mathematical Biology, London, G.G.Harrap &Co.Ltd. (1939); Vito Volterra, LesAssociations Biologiques anPoint dcVue Mathématique, Paris, Hermann dzCo.(1935). 448 Paonnsmsz Srsrsms. SPECIAL 2NDOansa Equxrrons Chapter 8 istheapproximate number ofdeaths ofthehostHduetothepresence of theparasite P,where kisaproportionality constant. Hence, also, since each such death ofHimplies thelaying ofaneggbyP, (e) km;/At istheapproximate number ofbirths ofP. Therefore intime At,theapproximate changes intheHandPspecies arerespectively (33192) Ax=Hbﬁm -H,1T3—6At ~k:cyAt, Ay=k:cyAt -P1-3’-At.100 Intheﬁrstequation of(33.192), leth=(H1,-—H4)/100, assumed >0, i.e.,histhenetnatural annual percentage increase ofthepopulation ofH. Inthesecond equation of(33.192), letp=P,,1/100, i.e.,pisthenatural annual percentage death rate ofP.Then (33.192) simpliﬁes to (33.2) Ax=h:cAt —k:z:yAt, Ay=k:z:yAt -pyAt. Note thatwehave used thewords “approximate changes” inconnec- tionwith thesystem (33.2). Itisconceivable thatinsome time intervals Population/A“-.,f Time Figure 33.21 fartobelieve that thesystem (33.22) Q2dt QdtAt,nobirths ordeaths take place; in other time intervals, oneormore births ordeaths occur. Agraph ofthepopula- tionofaspecies asafunction oftime is therefore notacontinuous curve. Itwill show sudden jumps ordrops asinFig. 33.21. However, ifthe population is large, itisconvenient toassume that there isacontinuous curve which willap- proximate thepopulation trend. Further, oneneed notstretch one’s credulity too ofequations =hm-—kxy, =hwy-zw, which result, when in(33.2), wedivide eachequation byAtandletAt—>0, willgive good approximations forlarge populations and long periods of time. Lesson 33B AB1o1.oc1c.u. Pnonum 44-9 The equations in(33.22) form asystem oftwo nonlinear ﬁrst-order equations which cannot besolved bythemethods thus fardiscussed. In later chapters weshall describe other methods bywhich anapproximate solution ofthesystem (33.22) may beobtained, ifthenumerical values ofh,k,andpareknown, seeExercise 39,5. Itispossible, however, tosolve thesystem (33.22) asanimplicit func- tion ofac,ifwedivide thesecond equation bytheﬁrst. Making this division, weobtain mm %=%§%§ g~Q@=@-an Integration ofthesecond equation in(33.23) gives (33.24) logy"—Icy=Icx——log:21"—logc, from which weobtain (33.25) logcyhx’ =Icy+lax, cy":c" =e""e'“, cy"e"“’ =x_"e"”. Ifatt=0,thepopulation ofthehost species is:00and thepopulation oftheparasitic species isyo,then thelastequation in(33.25) becomes (33.26) c=x0"’y0"‘e"“°+"°), which isthevalue ofthearbitrary constant inthesolution. Aswehave frequently mentioned, implicit solutions areusually oflittle value. Here itwould bedifficult indeed toplotagraph of(33.25) showing thepopulation yasafunction of1:. Although wecannot atthisstage solve thesystem (33.22) foracandy asfunctions oftime, nevertheless certain important useful information canbeobtained from it.Suppose, forexample, that atacertain moment oftime, thepopulations ofourtwointeracting species are h(mm) @=%, y=F Then by(33.22) dy/dt =0and dx/dt =O.Since therate ofchange of thetwo populations atthat moment iszero, there arenoincreases or decreases. Letuscallthese populations theequilibrium populations of:1: and yrespectively. Let (mm) x=%+X, y=%+K where Xand Yrepresent therespective deviations oftheequilibrium 4-50 Pnonums: Srsrnus. Srncnu. 2NnORDER Eouzmons Chapter 8 populations of:1:andy.Substituting (33.28) in(33.22), weobtain (33.29) %(%+X)=h<%+X)-—lc<%+X)c+ Y), %(%+Y)=k(%+X)(%+ Y)~P(%+ Y)-These equations simplify to (33291) %=——pY-kXY, %’=hX+kXY. Ifweassume that thedeviations Xand Yaresmall, wemay, without serious error, discard theXYterms in(33.291). These equations then become dX dY Butnowwerecognize that thesystem ofequations isexactly thesame asthose discussed inLesson 32under theheading “Linearization ofFirst Order Systems. ”Dividing thesecond equation in(33.292) bytheﬁrst, wehave dY hX($3.293) 55_-W- Itssolution, obtained bysolving itdirectly, orbyrecognizing that this differential equation comes under Case 2ofLesson 32where itssolution isgiven in(32.18), is ($3.294) hX2+pY2=¢. Both pandharepositive quantities. (Remember histhenetnatural percentage increase ofthehost popu- Y lation, assumed >0,andpisthenatu- 0, ralpercentage death rateofthepara- ” sitepopulation.) Iftherefore h=p, (‘/'%,0) (33294) istheequation ofafamily Xofcircles; ifh¢p,itisafamily of ellipses. Wehave thus shown that if the populations oftwo interacting species remain stationary over long Figure 33.295 periods oftime, asthey doinmany cases, then thegraph oftheir popu- lation deviations from equilibrium isacircle oranellipse, Fig. 33.295. Note how themathematical equation (33294) supports ouropening Lesson 33C Courmax ELECTRICAL Cmcurrs 4-51 remarks. When thedeviation Xofthehost species equals @ andis thuslargest, thedeviation Yoftheparasite species iszero, i,e.,thepara- sitepopulation isnormal. Butnow there aremore hosts available in which theparasites canlaytheir eggs. Thepopulation oftheparasite species thus begins toincrease, while that ofthehost species decreases. Butastheparasites gettoonumerous, they killmore hosts, until the parasite deviation ismaximum andthehostdeviation iszero. When this point isreached, there arerelatively somany parasites around, thatthey begin killing offsome ofthehost’s normal population. Asthehost’s population decreases below normal, theparasitic population, because there arenow insufficient hosts, must also decrease, until ﬁnally apoint is reached when itsdeviation iszero andthehost’s deviation hasitslargest negative value ——\/¢7h. Thehost's population isnowtoolowtosupport anormal parasitic population. Theparasitic deviation therefore begins todecrease below zero, thusallowing thehost’s deviation toincrease, etc., inanendless cycle ofperiodic increases anddecreases. LESSON 33C. AnElectrical Problem. More Complex Circuits. InLesson 30,wediscussed asimple electric circuit inwhich thecharged particles moved inonepath. Acircuit inwhich thecharged particles can move indifferent paths will,asweshall show below, giverisetoasystem ofdifferential equations. Toenable ustowrite differential equations with correct signs, weshall adopt thefollowing convention. Ifabranch ofthe circuit contains asource ofenergy E,weshall indicate byanarrowhead that thedirection ofthecurrent isfrom thesidewhich welabel +,tothe sidewhich welabel —-.Inevery other branch ofthecircuit, weputin arrowheads arbitrarily. Itisonly essential that thearrowheads remain ﬁxed inaproblem. 5.§?> Example 33.3. Setupa , system ofdifferential equa- ‘ ‘ 'V\/\' l L2 qt C1tions forthecurrent inthe + i2 electric circuit shown inFig. 33.31. _ Solution. By Kirch- hoﬂ"s ﬁrst law, thealge- braic sumofthecurrents at anyjunction point iszero. Figure 33_31 Ifwecallthecurrent whose arrowhead isdirected toward ajunction point positive and thecurrent whose arrowhead isdirected away from ajunction point negative, then at thejunction point marked 1inFig. 33.31,Q Q7:®I'M-9 (9,) ’I:"“’ll1-i2=0, 452 Paontamsz Srsrams. SPECIAL 2m)Onnan EQUATIONS Chapter 8 And atjunction point 2, (bl 'i1+i2-"i=0, i=‘i1+'i2- Applying Kirchhoff’s second lawtotheleftcircuit, weobtain (c) Ri+L2%-+%,q= E(t). Ifnosource ofenergy ispresent inaclosed circuit, then hissecond law states that thealgebraic sum ofthevoltage drops iszero. Applying this lawtotheright circuit, weobtain . 1 d (Q) R1l1+aT1q1—L2ﬁ‘=0- Differentiating (c)and (d)andreplacing i(= dq/dt), wherever itappears byitsvalue in(a),wehave <12" .1"d‘ 1..dE<e> J (1%. dz, 1._ L2-.1?-R‘E?_E“ —.°- Thelinear system (e)canbesolved for2'1and2'2bythemethod ofLesson 31D orE.The current iwillthen bethesum ofthese twocurrents. Comment 33.311. Weobtained twoequations intheabove problem byusing theleftandright circuits shown inFig. 33.31. Itispossible to obtain additional equations byusing other circuits. Each ofthese, how- ever, will notbeindependent ofthetwo obtained in(c)and (d). For example, wecould, ifwehadwished, taken thecircuit going completely around theborder andhave thus obtained 121+R.i.+51;q.+§q=E- This equation, however, isthesumof(c)and(d). Example 33.32. Setupasystem ofdifferential equations forthecur- rents intheelectric circuit shown inFig,33.33. Solution. ByKirchhoﬁ’s ﬁrst law, thealgebraic sum ofthecurrents atanyjunction point iszero. Hence atthejunction points correspondingly marked inFig.33.33, wehave (3) @3 ii—'52'*i4=0, i4=i1_'i2i @152‘-is-‘i3=0, is=i2—ia; @1ia+ie—i1=0. ie=i1—ia- Lesson 33C Courmx E1.ec'r1uc.u. Cmcurrs 453 Atjunction point Q),2'4+2'5-—2'6=0.Andweseefrom thesums ofQ), @,and@,thatthiscondition issatisﬁed. s= s>G) 5I'v\/\. |( 2mm+ qi 1,4 R. W‘: is@ lei" ®-—i—<.-3° is R3 L8 '\/\/\¢ WW‘ Figure 33.33 Applying Kirchhoff’s second lawtotheupper left,theupper right and thelower circuit respectively, weobtain thelinear system . 1 1 1 (bl R111+@:q1+'(Eq4+(IT6qe=E) J. . . 1 L2$+ R212-F R515 -aq4= 0, 1 . d . —(T640 —R5’l»5+ I/3%-F R313 =0- Differentiating (b)andreplacing d/dtbyDanddq,,/dt by2),,there results . . . 1. dE' (0) R1D1'1+(%;11+6%4‘l4+6T;1o=TlZ' . . . 1.LgD2‘lg +R3DZg +R5D'l.5 —6:14 =O, 1.30%. -é2'.-12.0.". +R3D2'3 =o. Substituting in(c)thevalues of114,25,2'6asgiven in(a),weobtain 111.1.1._Q(d) (R11-7~l-6,:-l-6-,;-l-6;)11—(J'v;¢2-C";;1a—dtr -ii.+L20’+12,0+12,0+l)i, -R5D2'3 =0,04 C4 -C-1;2',-R5D2'-2 +(L302 +12,0+R3D+ 2'3=0. Thgsystem (d)canbesolved for2'1,2'2,and2'3bythemethod ofLesson 31 4-54 Paonnamsz Srsrams. Srscnu. 2NDOanan EQUATIONS Chapter 8 Comment 33.34. Asremarked inComment 33.311, other equations inaddition tothose found in(b)canbeobtained byconsidering different circuits. These, however, will notbeindependent ofthose in(b). For example, wecould have taken thecircuit going completely around the border. This equation, however, canbeobtained from (b). Verify it. Can youﬁndother possible circuits? EXERCISE 33C 1.Atransformer consists ofaprimary coilandasecondary coil. Theprimary coilhasanemfofE(t)volts, aresistance ofR1ohms, andaninductance of L;henrys. Thesecondary coilhasaresistance ofR2ohms andaninductance ofL2henrys. Let2'1and2'2betherespective currents ineach coil. Ithas been shown that thedifferential equations forthecurrents inthecoils are d . d(33.4) L.§+12.».+Mf=Ea). d' . d'L275-+R.e+M§ =0. where Misaconstant called themutual inductance. (a)Solve thesystem (33.4) ifE(t) =0andM2<L1L2. (b)Show that theexponents aandb,asgiven intheanswer section, are negative quantities andtherefore that both 21and2'2approach zeroas t—>w. (c)Solve thesystem (33.4) ifE(t)isaconstant E0andM2<L1'L2. Show thatast—>0,thecurrent 2'1oftheprimary coilapproaches thesteady state current E0/R1 andthecurrent 22ofthesecondary coilapproaches zeroast—>w. 2.(a)Setupasystem ofdifferential equations forthecurrent intheelectric circuit shown inFig.33.41. 1 "EFF ; i i L it 1, E0; R1 R2 li1 I Figure 33.4-1 (b)Solve thesystem with initial conditions t=0,i=0when theswitch isclosed, andshow that thecircuit isequivalent tooneinwhich the resistors R1,R2arereplaced byoneresistor R,whose coefficient ofre- sistance isgiven by 1_1 1 _ R1R2 R'R.+R.’ R'R.+R. Remark. Ifmore resistors R3,---,~R,, were inserted inthecircuit of Fig.33.41, parallel toR1andR2,itfollows from (b)that thiscircuit is equivalent tooneinwhich theresistors R1,R2,---,Rnarereplaced by Lesson 33C—Exe1-cise 455 oneresistor Rwhose coefficient ofresistance isgiven by 1 1 1 1 t-a+a+m+a3.(a)Setupasystem ofdifferential equations forthecharges intheelectric circuit shown inFig.33.42. (b)Solve thesystem forqand2'asfunctions oftime with initial conditions t=0,ql=qg=0when theswitch isclosed, andshow thatthecircuit isequivalent tooneinwhich thecapacitors C1,C2arereplaced byone capacitor whose capacitance isgiven byC’=C1—|—C2. I ’\/\/\. yl i R iii l2 l E°:— Q1 C1 112 C2 I.TT 1 1 Figure 33.42 Remark. Ifmore capacitors C3,‘---,C,were inserted inthecircuit of Fig.33.42, parallel toC1andC2,itfollows from (b)that thiscircuit is equivalent tooneinwhich thecapacitors C1,C2,---,C,,arereplaced by onecapacitor whose capacitance isgiven byC=C1+C2+---+C... Intheproblems below, setupasystem ofdifferential equations forthe current ineach ofthecircuits diagrammed. Solve asmany asyouwish. 4.SeeFig.33.43. Att =0,2=0,q=q1= qg=0. ‘DIP 1 Il L=1 i i,i11 E°=5°_ <1.c,=1o" -1.c,=s-10-4 lqTT |( 1c-2-1o-4 i= Figure 33.43 5.Inproblem 4,replace E0=50byE(t) =100sin60t. 6.SeeFig.33.44. Att =0,i=0,q=qg=0. ¥ ’\/\/\. -5‘ R=2 . l2 " qzl-c,=a-10-‘ 50cos100: c=2-10-4 LE2 gnﬁs '5Figure 33.44 456 Pnonnams: Srsraus. Srncmr. 2m)Oannn EQUATIONS Chapter 8 7.SeeFig.33.45. Att =0,5=0,qg=0. ~'86!‘ 5L I E(t) L1 R i=i,+i, i ii q2 £2 C2 Figure 33.45 8.SeeFig.33.46. Initial conditions arel =0,11=0,q1=Q2=0. ’\/\/\. _1 § "CUP—|— R2 ‘2 _11 L1 J-I C2 Q2 111 C1 i-it-1-[2 l l .;E°. I-1 Figure 33.46 9.SeeFig.33.47. ii Ll i2 R2 qi >"um I'v\/\ K Ii‘ C2 E(t) mic, L, is Rs T R5 is 1 '\/\/\¢ ’\/\/\¢ 1 is is qa L3 R, 1 (|:( mm '\/v\. 8 Figure 33.47 10.SeeFig.33.48. ‘\/\/N R. if "trim 1 i=i-iC9IiEa) C L 2 l 1z 2 R2 ’\/V\ Figure 33.48 Lesson 33C—Answers 11.SeeFig.33.49.457 R1 i, G) i,’\/\/\. I I 1 Q”C“ 11E(t) @ L, C‘qeq2iC1 R2 L2R v *11 1 1 1 21 @ 24 ® 2, ® i2 Figure 33.49 ANSWERS 33C l.(a)2'1=c1e°'-l- C26“, where ab__-(I-2R1 +L1R2)=|=\/(D2131 —|—131132)” —4R1R2(L1L2 —-M2)I ' 2(L1L, -M2) —(L2R1+L1R2) =|=V(L2R1 -L1R2)2-l-4M2R1R2, = 2(L1L2 -M2) 1'2=L l(L1L2 -M2)(lw1e“ —|—11626“) +L2R1(¢1e“ +v2¢u)l-MR2 _ : b:(c)2'1-c1e° —|—026 -1-(E1)/R1);i2 isthesame asin1(a). 2.(8.) (LD-l— R1)’f1 +LD‘i2 =E0, R1'l1 ——R2'i2 =0. (b)1.=1.1+1.2 =Eo(R1 -l'R2)[1__e-R1n,¢/r.(n,+R,)] R1R2 _@ _—RtIL-R(1 e ). 3.(a)(RD+Cl1)q1+RDq2 =E,$111-$212 an ql =C1E0(1 ___ 6"‘/R(c1+c2)’ g2 =C2Eo(1 _ e"¢”z(c1+c2)' ...d d E_ E_ 1=11+‘2=dit1+%=7aoe tIR(C1+Cg)E_éQetIRC. . 1 1 . . . 4-D2+g%+%,=Eo. —é,;q2—aq1 =0,2=21+22- 1>”1".+1>”e+Ci11'.+§(1'.+e) =0. 1. 1.a;'t2'—E'l-1=0. 458 Paonnausz Srsrams. Srncmr. 2m)Oam-:11 EQUATIONS Chapter 8 5.In4replace E0by100sin60t.Therefore $5E(t) =6000 cos60t 6.(RD+L0’+2'1+(R0+1'2=-sooo sin1002 1.1022. -<R2D+ 1'2=0. [email protected]..\->»[email protected].>- 7.Ri-l-L——-|-L1-dT =E(t), _ 1 R1+L-+692 =E(t). 11' 1'—L1—dit1+F2q2 =0. Thethree equations arenotindependent. Thethird canbeobtained by subtracting thesecond from theﬁrst. Tosolve thesystem, itiseasier to usetheﬁrstandthird equations. d 1 . 1 8-L1dit'+CT1q1 =Eo. R212-l~(,T2q2 =Eo- dzqi 1 dqz 1 'Lrm-l-E01 =E0. R2?-l--6;¢12 =Eo- ql=C'1E() (1—COS 1 3)—E0R2C1L1 Sin \/C'1L1 VC'1L1 2'1=CIEO sin 1 t—E0 cos \/011.1 \/C1L1 R2 \/011., __ _ E _ q2=C2E°(1 __etIR3C3)' ,2=_R_<;e 1/R,c,_ 9.L1%-l" éQ4+Rois '=EU). _ 1 d' . 1Rm+,-,;q.+L.-§+ 12515-511. =0. d' . 1 . . L3 R313-F 5q3—Rats —-R515 =0. . . d . .Substitute 2..=T12;-'and24=21- . 2. 2. 1o.R1%+L% -L%Z+%(.'. ll.Junction point @:111-—2'5—2'2= Junction point @:112-—2'6—2'3= Junction point @:2'3—|—217—24= Junction point @:2.1+is—-2'1= Atjunction point @:25+ 2'11——2'7-—2'3=0,which satisﬁes thefour equations above. Setup,intheusual manner, thedifferential equation for1'2,is=1'2—la.is 11—- , d —12)=EEll) R111 +R212 =0 9990=i1—- =‘|.2_1.3 ='i4—- is=11—- 4 Lesson 34A Vanocrrr ANDACCELERATION FORMULAS 459 each circuit marked I,II,III,andIVindiagram. ForI,itis . 1 d' . R111 +Eqs+Lsdig‘=EU)- Although there areother choices ofcircuits, their equations willnotbein- dependent ofthechosen four. LESSON 34-. Plane Motions Giving Rise toSystems ofEquations. InLesson 16,wediscussed themotion ofaparticle constrained tomove along astraight line. Inthislesson, weconsider themotion ofaparticle freetomove inaplane. LESSON 34A. Derivation ofVelocity and Acceleration Formulas. Ifaparticle isfreetomove inaplane, then achange inthedirection of itsvelocity willbeequally asimportant asachange inthemagnitude of itsvelocity. Asmentioned inLesson 16C, quantities inwhich both mag- nitude anddirection play,a rolearecalled vectors. Since Newton’s second lawofmotion isalsoapplicable toparticles which move inaplane, wehave, by(16.1), (34.1) F=ma=m% Ema). Remark. Adotover avariable means itsderivative with respect to time; twodotsoverthevariable means itssecond derivative with respect totime. Hence :5:Edz:/dt, 5E112:1:/dtz, :3Edv/dt, etc. Themass misnotavector quantity. Wetherefore see,by(34.1), that theacceleration ofaparticle acted onbyaforce, notonlyhasmagnitude F/m, butalsohasthesame direction asF. InFig. 34.11, thevector Frepre- F sents themagnitude anddirection of aforce F.Itisconvenient tobreak Fy=F ‘in9 upthisvector force intotwocompo- nents, one,F,,torepresent thatpart ‘ oftheforce which accelerates the 0 F=F008 0 particle inthe:1:direction, theother, " F,,,torepresent thatpartoftheforce Figure 34-.11 which accelerates theparticle inthe ydirection. Iftheinclination oftheforce Fis0,weseefrom Fig. 34.11, that (34.12) F,=Fcos0. F,=Fsin0. Since F’,=mass times 12,,theacceleration ofaparticle inthe2:direction 460 Paontamsz Srsrams. Srncmr. 2NnOsman Eqtwrrons Chapter 8 andF,=mass times a,,,theacceleration ofaparticle intheydirection, weobtain from (34.12), thesystem ofequations 2 (34.13) Fcoso=F,=ma,=112%?Emzii, . dz ..Fsin0= F,,= ma,,= m$E my. Inasimilar manner wecanbreak upanyvector quantity intoits2:andy components. InFig. (34.14) wehave shown, forexample, the:0and y components ofavelocity vector v. v vy=v sin0=%y- =59 l0 dx .v,,=voos 0-=E=x Figure 34.14 Forcertain problems, itisoften convenient tousepolar coordinates instead ofrectangular coordinates. Thevector quantity isthen broken up into twocomponents: onealong theradial rdirection, theother ina direction perpendicular toit.InFig.34.15, wehave broken upthevector vinto these twocomponents v,and11,. V U0 P(r.y); P(r.9) v, r y=r sin0 .4x=rcos 0 Figure 34.15 Let(z,y) bethecoordinates ofapoint Pinarectangular system and (1,0) itscoordinates inapolar system, Fig. 34.15. Then weseefrom theﬁgure, that (34.16) :c=rcos0, y=rsin0. Lesson 34A—Exercise 461 Differentiation of(34.16) gives (34.17) iii?=cos0%—rsin 0%, dy_- Q‘ Q. E-s1n0d,+rcos6dt Asecond differentiation gives dz dzr .drdo do2 .d’o(34.18) ﬁ=cos0a§—2s1n0a?—‘E——rcos0<EZ) —rs1n0E;, .12 .oz. drd0 .do2 ozoﬁ= sinﬁa-t3+2cos0?E E—rs1n0(E) —|-rcos0F,;- These formulas, (34.17) and(34.18), arevalid forevery value of0.Hence they must hold inparticular when 0=0.But when 0=O,as=rand thedirection perpendicular toristheydirection. Hence thecomponents ofvelocity and acceleration inaradial direction and inadirection per- pendicular toitare dz dy dza: dzy(34.19) 1),-=-‘Er t)g=ﬁZr l1,=-(F: (lg-=35" Substituting theﬁrsttwoequations of(34.19) in(34.17), thesecond two in(34.18), weobtain with 0=0, (34.2) 12,=$5 1", v,=rgEr0. dz do2,,(34.21) a,=(T;-r(-di) Er—2'02, .1d0 .1’ ,a1;=2;-1%-(2-5-1-r-‘F2-E2r0—l-rli. Formulas (34.2) and (34.21) give respectively thecomponents ofthe velocity andacceleration vectors ofaparticle along theradial axisand inadirection perpendicular toit,atthepoint Pwhere thecurve crosses the2:axis. Since the2:axiscanbechosen inanydirection, these equations arevalid forevery point Poftheparticle’s path. EXERCISE 34A 1.InFig.34.22, wehave shown the2:andycomponents ofavelocity vector v aswellasitscomponents inaradial direction andinadirection perpendicular toit.With theaidofthisdiagram, prove (34.2). Hint. First show that da: d .(9.) v,=EE=vcosa, v,,=%=vsma. 462 Pnonu-ms: SYSTEMS. SPECIAL 2NDORDER EQUATIONS Chapter 8 Then show that (b) v,=vcos(a——0) =vcosacos0+vsinasin0, =%cos0—|—%sin0. In(b),replace dz/dt anddy/dt bytheir values asgiven in(34.17). Y vM =iivy= E; ‘ "0 I‘at P( I0): =2: P(z,y)ﬁn U’“' §$Hr% I y=rsin0 9 x=rcos6 (0,0) X Figure 34.22 Similarly, show that (c) v;=vsin(a—0) =usinacos0—vcosasin0 =%cos0 —gsinﬁ. In(c),replace dy/dt anddz/dt bytheir values asgiven in(34.17). "4 ll illsA”.2 “y H I P030), Q P(1,y) ar G-d_’“xd1’ y=rsin9 0 x=rcos0 (0,0) X Figure 34.23 2.InFig.34.23, wehave shown thezandycomponents ofanacceleration vec- toraaswellasitscomponents inaradial direction andinadirection per- Lesson 34-B Tun PLANE MOTION orAPno.mc'riLE 463 pendicular toit.With theaidofthisdiagram, prove (34.21). Hint. First show that dzaz d2y(a) a=—-=acosa a=—=asina. (b) a.‘dis '”dﬂ Then show that =acos (oz—0)=acosacos0+ asinasinﬂ d2a: 112;;.=‘(#7 C030-FFSIIIO. In(b),replace dzz/dtz anddzy/dtz bytheir values asgiven in(34.18). Similarly, show that (c) ag=asin(a-0) =asinacos0-—acosasin0 3.dzy dza: .=Wcos0 ——"E5100. In(c)replace dzy/dtz anddza:/dtz bytheir values asgiven in(34.18). Aparticle ofmass misattracted toaﬁxed point Obyaforce F.Theparticle moves inaplane with aconstant speed butnotinastraight line. Show that theparticle moves inacircle with center atO.Hint. Thecomponent a,of theacceleration vector ainadirection tangent tothepath oftheparticle measures thechange inthespeed oftheparticle. Since thisspeed iscon- stant, a,=0.Hence theacceleration actsonly inadirection perpendicular tothepath oftheparticle. And since Fandaareinthesame direction, F andtheconstant speed 00areperpendicular toeach other. Then show dy/dz: =——z/y. LESSON 34B. ThePlane Motion ofaProjectile. a it 1. 2. 3. 4. 5.Example 34.3. Aparticle ofmass misprojected from theearth with velocity voatanangle 0:with thehorizontal. The only force acting on isthat ofgravity. Assuming alevel terrain, ﬁnd: Theequation oftheparticle’s path. Itshorizontal range. Themaximum height itwillreach. The value ofaforwhich therange willbeamaximum. When theparticle willreach theground. Solution. Refer toFig. 34.31. Wetake the:2:andyaxes inaplane which isperpendicular totheground, and which contains thegiven velocity vector vo.The 2axisis,asusual, perpendicular toboth a:andy axes. Since byassumption theonly active force Fisthat ofgravity, the components ofFinthe2:,y,zdirections arerespectively F,=0,F,= -mg, F,=O.Hence (a) mﬁi=0, my=-mg, mé‘=0. 464 Pnonnnms: Srsrmrs. Sracmr. 21mOno!-:11 Eqtwrrons Chapter 8 Integration of(a)gives (b) v,,=a':=c1, v,,=g]=—gt+c-2, v,=2=c3. Since thevelocity vector voliesinthexyplane, itfollows thatatt=0, i.e.,atthemoment ofprojection, seeFig.34.31, (c) v,=vocosa, v,,=nosinoz, v,=0, where ozistheangle which vomakes with thehorizontal 2:axis. Sub- stituting these values in(b),weﬁnd (d) cl=vocosoz, C2=vosina, c3=0. Hence (b)becomes (e) v,=:i:=v0cosa, v,,=y=—gt+vosina, v,=é=0. Integration of(e)gives (f)2:=(vocosa)t+c4, y=—gg +(vosina)t+c5, z=O5. Ifwenowchoose ourorigin atthepoint where theparticle isprojected, then att=0,x=0,y=0,z=0.Substituting these values in(f),we y l+ V0 -ib vosina 0 vocosa x z Figure 34.31 ﬁndc4=0,c5=0,65=0.Theparametric equations ofthepath ofthe particle aretherefore2 (g) :1:=(1)0cosa)t, y=(vosina)t—25-» z=0. Since z=0,follows thataprojectile subject onlytoagravitational force moves inapla containing thevector vo. Toﬁndtheequation ofthepath inrectangular coordinates, weeliminate tbysolving theﬁrstequation in(g)fortandsubstituting thisvalue in thesecond equation. There results2 (h) y=(tana)u:-<5-Lgiig) x2,U02 Lesson 34B—Exercise 4-65 which istheequation ofaparabola through theorigin. Since thecoeﬂi- cient of1:2isnegative, thecurve isconcave downward. Thehorizontal range oftheprojectile, i.e.,thedistance from theorigin tothepoint where theparticle strikes theground, isobtained bysetting y=0in(h),forwhen y=0theparticle isatground level. Equation (h)then becomes, ifarﬁ1r/2, 2 2 (i) 0=(signa)x——g r :c=%(2 sinacos a), v.:1:=%sin2a, which gives thehorizontal range oftheparticle. Themaximum height isreached when y’=0.Therefore, differentiat- ing(h)with respect toxandsetting y’=0,weobtain 2 2. sec v.(J) 0=tana—g——-Ex, x=l-smacosa."02 9 When anhasthevalue in(j),yby(h)hasthevalue "02-2 "oz-2 "02-2(k) y=7sm a—§;sm a=§'—s1n a, which gives themaximum height reached bytheparticle. The range willbeamaximum when, in(i),aissochosen that 2:isa maximum. Hence differentiating thelastequation in(i)with respect to a,andsetting dx/da =0,wehave 2 2 (1) 0=%-cos2a, e={- Therange therefore willbeamaximum iftheprojectile isﬁred atanangle of45°. By(i),thismaximum range is1202/g. _ The particle willreach theground when y=O.Setting y=0in(g), weobtain2 (m) (vosina)t -2%=0,t=%oioa, which isthetime ittakes theparticle toreach theground. EXERCISE 34-B Inproblems 1-14, assume theairresistance isnegligible. 1.Aprojectile isﬁred from theearth with avelocity of1600 ft/sec atanangle of45°. Find theequation ofmotion, themaximum height reached andthe range oftheprojectile. 4-66 Pnosnnmsz Sxsrrms. Srncnu. 2m)Onnan Equarrons Chapter 8 2Start with theequation ofmotion (h)ofExample 34.3. (a)Show thatthecoordinates ofthevertex oftheparabola are (002 sinozcosa v02sin2oz)__._.i_ ,_____ . 9 20 (b)Show thatthedistance ofthevertex tothefocus isv02cos2oz/2g, and therefore thattheequation ofthedirectrix isy=v02/2g. Note thatthe equation hasnoozinit.Hence theparabolic orbits ofallprojectiles ﬁred with agiven velocity have thesame directrix regardless ofthe angle atwhich they areﬁred. (c)Finally show that thisconstant height ofthedirectrix above thehori- zontal isthedistance aprojectile reaches when ﬁred straight upwith an initial velocity ofvoft/sec. Ifyouhave succeeded inanswering theabove questions, youhave proved that every parabolic orbit ofaprojectile ﬁred with thesame velocity vo hasthesame constant directrix whose height isthedistance theprojectile would reach ifﬁred vertically. Weshowed inExample 34.3thatifaprojectile isﬁred atanangle of45°, itsrange willbeamaximum andwillequal v02/g. Anartillery piece, whose muzzle velocity isvoft/sec, islocated atadistance D<v02/g from anob- jectatthesame level asitself. Show thatthere aretwoangles atwhich the artillery piece canbeﬁred andhittheobject—one asmuch greater than 45°astheother isless. Find these angles. Hint. In(h)ofExample 34.3, youwant ozsuch that when y=0,2:=D.Make useoftheidentity 2sinacosoz =sin2a =cos<2a—%>. Themuzzle velocity ofanartillery piece is800ft/sec. Assuming alevel terrain, answer thefollowing questions. (a)Anobject is3.8miaway. Canitbehit? (b)Anobject is15,000 ftaway. Atwhat angles must theartillery piece be ﬁred inorder tohittheobject? (c)What isthemaximum height reached bytheshell of(b)? (d)When didtheshell reach theobject? (e)Ifamountain ofheight 6000 ftis4000 ftfrom theartillery piece, isit stillpossible tohittheobject? Aprojectile, ﬁred with avelocity of96ft/sec, reaches itsmaximum height in2sec. Assume alevel terrain. ( . .. . .dl__dy/dta)Find theangle ofprojection oftheparticle. Hint. dx-T/dt -There- re.dy/dz =0ifdy/dt=0.(b)Find themaximum height reached bytheparticle. (c)What istherange oftheprojectile from thepoint ﬁred? Aprojectile isﬁredfrom aheight ofyoftabove alevel terrain, withavelocity ofvoft/sec andatanangle ozwith thehorizontal. Find: (a)Theequation oftheparticle’s path. (b)Itshorizontal range—take the:0:axisonground level. (c)Itsmaximum height. (d)When itwillreach theground. (e)Atwhat angle andwith what velocity itwillstrike theground. (Hint. If0istheangle, tan0=(dy/dt)/(dz/dt) andIv]=\/(dz/dt)? +(dy/dt)5). (f)Thevalue ofozthatwillmake therange amaximum. Lesson 34-B—Exei-cise 4-67 Aprojectile isﬁredfrom aheight of50ftabove alevel terrain withavelocity of64ft/sec atanangle of45°. Answer thequestions asked forin6,with theexception of(f).Inregard to(f),ﬁndtheangle ofprojection thatwill make thehorizontal range amaximum andthevalue ofthismaximum range. Themaximum range ofaprojectile when ﬁred onalevel terrain is1000 ft. (a)What isitsmuzzle velocity? (b)What isthemaximum horizontal distance itcantravel iftheprojectile isﬁred from a40-ft-high platform? Hint. First ﬁndtheﬁring angle for maximum horizontal range, seeproblem 6. Themaximum distance aboycanthrow aballonlevel ground is100ft. Neglecting theheight oftheboy,ﬁnd (a)Thevelocity with which theballleaves hishand, (b)Themaximum horizontal distance hecanthrow theballifhestands on aroofwhich is40ftabove theground. Hint. First ﬁndthethrowing angle formaximum horizontal range, seeproblem 6. Twoathletes, one6}fttall,theother 5%fttall,caneach putashotwiththe same velocity of36ft/sec. Atwhat angle should theshotleave eachathlete’s hand inorder togetthemaximum horizontal range? Assume theshot leaves from heights of6ftand5ftrespectively. How much farther willthe taller athlete’s throw go? Answer thequestions inproblem 6,excepting (f),if0:=0,i.e.,ifthepro- jectile isﬁred horizontally from adistance yoftabove thehorizontal. Aprojectile isﬁred with avelocity v0atanangle awiththehorizontal. The terrain makes anangle Bwith thehorizontal. (a)Find therange oftheprojectile. Hint. CallRtherange oftheprojectile. Then theprojectile willhittheterrain when :2:=Rcosﬂ andy= Rsin/3.Substitute in(h)ofExample 34.3. (b)Find thevalue ofozwhich willmake therange amaximum. Hint. Make useofdouble angle formulas. (c)What isthemaximum range? Inproblem 3,wegave twoangles atwhich aprojectile could beﬁred in order tohitanobject located within range andonthesame level astheﬁring weapon. (a)Solve thissame problem iftheobject tobehitisonthetopofahill whose angle ofinclination is13andwhose distance Dfrom theﬁring point islessthan orequal totherange 1202/g(1 +sin|3),asgiven in(c)of problem 12.Hint. Call (X,Y)thecoordinates oftheobject. Then 0=\/X2+Y2,sinﬁ=Y/D,cosﬂ=X/D. Replace Rby1)inanswer to12(a). Solve foroz.Usethefactthat cosAsinB=ﬂsin (A+B)——sin(A——B)] and sin(2a _,e)=cos(2a-13 (b)Show, bymeans ofthesolution found in(a),thatitispossible tohitan object only ifitsXandYcoordinates satisfy theinequality \/X2+Y2+Ys1102/0- 'Hint. Usethefact—see answer to(a)—that (QX2+Yvo2)/ (1/02\/X 2+Y2) 468 Pnonuamsz Sxsrsms. Srncnu. 2NDORDER Equarrons Chapter 8 must be§1.Note thatifY=0,sothat object isonalevel terrain, X§v02/g aswesawpreviously. 14.Aman ishunting with agunwhose muzzle velocity is224ft/sec. Heaims forabirdonthetopofatree150fthigh and1500 ftaway. Isthebirdin danger ofbeing hit? Inproblems l—14, weignored airresistance. Intheproblems below, we shall assume theprojectile isﬁred from theearth andissubject notonly toagravitational force butalsotoanairresistance which isproportional totheﬁrstpower ofthevelocity. Weshall alsoassume thattheforce of theairresistance acts inadirection opposite tothat ofthevelocity, i.e., that itacts along atangent totheprojectile’s path andinadirection to oppose themotion. Call Rtheproportionality factor oftheairresistance. 15.Aprojectile isﬁred onalevel terrain atanangle ozwith thehorizontal and with avelocity ofvoft/sec. (a)Find theparametric equations oftheparticle’s path. Hint. Modify equation (a)ofExample 34.3totake intoaccount thecomponents of theforce oftheairresistance inthe:2:andydirections. (b)What isthemaximum height reached bytheparticle? Hint. Set Q_dy/dt Itdx-——dz/dt equa ozero. 16.Aprojectile isﬁred inahorizontal direction with avelocity ofvsft/sec from aheight ofyoft.Find theparametric equations ofitspath. 17.Ananti-aircraft gunﬁresashell almost vertically with aninitial velocity of voft/sec. Thehorizontal component oftheairresistance istherefore negligi- ble.Assume thegunmakes anangle ozwith thehorizontal, andthevertical component ofresistance isRdy/dt. (a)Find theparametric equations ofthepath oftheshell. (b)Assume theshell weighs 601b, themuzzle velocity is2000 ft/sec, the angle ofelevation is80°,andthevertical component oftheairresistance is1/20dy/dt. Find theparametric equations oftheshell’s path, themaxi- mum height attained byit,andthetimerequired toreach thismaximum height. ANSWERS 34B 1.y=2:-1:2/80,000, 20,000rt,s0,000rt. 1r\1 Dg3.a=ZEl:§ArccosF- 4.(a)No.Themaximum rangeis3.7879mi. (b)G:1:0.36)rad. (c)1693 ftor8307 ft,approx. (d)20.6secor45.6sec,approx. (e)Yes. When :2:=4000 ft,theheight yoftheprojectile isapproximately 6498 ft,ifthelarger angle ofelevation isused. 5.(a)oz=Arcsin(§). (b)y=64ft. (c)1=143ft. 2 6.(a):1:=(vocosa)t, y=yo+(vosina)t—-9%; gS802 (X21/=110+ (l§&D(!)Z T3 0 Lesson 34-B—Answers (b)Range isgiven bythepositive value of:1:forwhich 2 x2 ——(tana)a: —yo=0. 002sinzozo+-———- (c)1/=y 20 (d)t=1/(vocosoz),where :1:isgiven by(b), (e)tan0 =mosina _at»vocosa =V0 —vo(sin a)gt+(gt)2, where tisgiven by(d). IUI Q2 2 (f)sina=to/\/2(0o2 +gyo).2 7.(a)y=50+:2: ~i- (b)1=166411;. (c)82ft.I28 3.sec. (e)122°00’, 85.6ft/sec. (d)t 7 (f)sina =0.6,oz=37°,approx., 171ft,approx. 8.(a)v=80\/5ft/sec. (b)1039 ft,approx. (oz=44°,approx.). 9.(a)40\/2 ft/sec. (b)134ft(a=37°,approx.). 10.Usetheanswer given in6(f)toﬁndtheangle aforeach athlete. Then use therange equation asgiven in6(b). 2 11-(a)c=v6t.z/=uo~—%1z/=yo—$w2-0 (b):c (c)y(6)tI)0\/ 2110/ gft. 1/0ft. 2:/vo, where 2:isgiven by(b). (9)155119 =-95/vo, lvl=V1102 -l"(902- 12_()R R._ . (b)or 13.(a)a=€+—=|= §Arccos(a ne -ag gcoszﬂ T+_ .2 (c)Maximumrange R3- -Nib I N812:0X2+Yvo2 _ 002x/X2 +Y2 14.No. See13(b). 15.(a):6=%vccosa(1-—-e_m""), ll %(%+ vosina) (1-e-’“”") -%1, where Ristheproportionality factor oftheairresistance. (b)t 1/%log<1+%)£sina), %sina—ﬁlog 1+-Iﬂsina -R R2 mg 2 16.2:=1wR2(1 —e_E”"'), y—yo+ El (1—-e_m"") —%t. 17.(a):1: (b)w_ R2 (vocosa)t,y=asinl5(a). 347.3t, y=118,861(1 —e‘°-0267‘) —-1200t, 30,153 ft,36.4 sec469 470 Pnonu-ms: Srsrnms. Srncnu. 2NDOanna Eouxrroxs Chapter 8 LESSON 34-C. Deﬁnition ofaCentral Force. Properties ofthe Motion ofaParticle Subject toaCentral Force. Assume aparticle inmotion isattracted toaﬁxed point O,byaforce F.Insuch cases we saytheparticle moves subject toacentral force, and calltheﬁxed point Otowhich theparticle isattracted, thecenter ofattraction. InExample 28.15, wediscussed aspecial central force problem where theparticle moved onaline. Intheremainder ofthislesson weconsider themotion ofaparticle subject toacentral force where theparticle is freetomove inspace. Weprove below certain properties which arecom- mon tothemotions ofallparticles subject toacentral force. Property A.AParticle inMotion Subject toaCentral Force Moves inaPlane Which Contains theFixed Point O.Weassume a particle, moving inspace, issubject toacentral force F.Lettheﬁxed point O,toward which theforce isdirected, betheorigin ofacoordinate system. LetP(:c,y,z) betherectangular coordinates oftheparticle, letr bethedistance oftheparticle from Oattime tandletoz,)3,‘Ybethedirec- tion angles oftheforce F,Fig.34.4. F1¢:3s='Y4*‘ 1 .8ziF. B -Fcosat a _ F5 Rihtanle K 8 lay: I Y Fy FcosB X 16> <1») \ Figure 34-.4 Thecomponents ofFinthea:,y,z directions arerespectively, seeFigs. 34.4(a) and(b), (34.41) F,=Fooee=F§. F,,=Feoe6=F§. F,=Fcos'Y=FE- Since F,=mat,F,=my,F,=F2,weobtain, by(34.41), thesystem (34.42) mi:=F3. my=F’i. mz=F§-T T T Lesson 34C Panrrcna Susmcr 'roACENTRAL Foaca 471 Multiply theﬁrstequation in(34.42) by—y,thesecond by:1:andaddthe two. There results (34.43) rnxf] —myﬁ =0, xi)-—-yzii=0. Inananalogous manner, wecanobtain, respectively, from thesecond andthird equations in(34.42), andfrom itsﬁrstandthird equations, (34.44) yﬁ-zi)=0,zii-2:5=0. Integrating each oftheequations in(34.43) and(34.44) with respect to time, weobtain (34.45) 2:1]—yi:=cl, yé-—-211}=C2, z:i:—xi=03. [Verify thatthederivative ofeach equation in(34.45) gives therespective equation in(34.43) or(34.44).] Wenow multiply theﬁrstequation in (34.45) byz,thesecond byx,thethird byy,andaddallthree. There results (34.46) 022+03y—|—clz=0, which istheequation ofaplane through theorigin, i.e.,through theﬁxed point O. Property B.AParticle inMotion Subject toaCentral Force Satisﬁes theLaw oftheConservation ofAngular Momentum. We assume thataparticle inmotion issubject toacentral force F.Byprop- ertyA,theparticle moves inaplane. Bythedeﬁnition ofacentral force, Fisalways directed toward aﬁxed point O,which wetake astheorigin ofapolar coordinate system. LetP(r,0) bethecoordinates ofthepar- ticle’s position attime t.CallF,thecomponent ofFacting along the radial axisr,andF;thecomponent ofFacting inadirection perpendicular to1'.Since Falways actstoward Oalong aradius vector, itscomponent F;iszero. Therefore, by(34.21), (a) F;=ma;=m(2r0 —|—rt!)=0, 2r‘9+rd=O. Ifwemultiply thelastequation in(a)byr,itbecomes 2m)+1'25=0, which isequivalent to (1.) $020) =0. Integration of(b)gives (34.47) 1-20=h, 472 Pnonnnmsz Srsrnms. Srncmt 2NDOnona Equxrrons Chapter 8 where hisaconstant. Bydeﬁnition, theangular momentum ofa particle ofmass mrotating about anaxisperpendicular totheplane of itsmotion ismrzﬁ, where risthedistance oftheparticle from theaxis ofrotation and0isitsangular velocity about thisaxis. Wesee,therefore, by(34.47), that histheangular momentum ofaparticle perunitmass. Andsince hisaconstant, theequation tellsusthattheangular momentum oftheparticle isconserved. Wehave thusproved thataparticle inmotion subject toacentral force satisﬁes thelawoftheconservation ofangular momentum. Property C.AParticle inMotion Subject toaCentral Force Sweeps out Equal Areas inEqual Intervals ofTime. (NOTE. This property isessentially arestatement ofproperty B.)Thearea ofacircular ‘M rd0 Z "’ Figure 34-.43 sector ofradius randcentral angle 0isr20/2, Fig.34.48. Hence when a radius vector rturns through aninﬁnitesimal angle d0,itsweeps outan areaequal to dA d0(34431) 6.4=woo, W=wa- Substituting (34.47) inthesecond equation of(34.481), weobtain, with initial conditions A=O,t=0, A t (34.49) %= /A_°d.4 =ftﬂgdt, A= Inwords theﬁrstequation in(34.49) saysthattherateofchange ofthe area Aisaconstant. (Remember hisaconstant.) Thelastequation saysthatequal areas areswept outinequal times. Wehave thusproved thataparticle inmotion subject toacentral force sweeps outequal areas inequal time intervals. EXERCISE 34C 1.(a)By(34.47), r20=h,where (r,0) arethepolar coordinates ofaparticle moving subject toacentral force. Show that inrectangular coordinates h=my—-ye.Hint. tan!) =y/2:. Differentiate with respect totime andsolve for0. Lesson 34-D Foncs FIELD. POTENTIAL. Consnnvurva FIELD 473 (b)Hence show that theareal velocity dA/dt, inrectangular coordinates, is given bydA/dt =§(a:;1] —y:i:). Hint. See(34.49). LESSON 34D. Deﬁnitions ofForce Field, Potential, Conservative Field. Conservation ofEnergy inaConservative Field. Assume a force Facts onaunit mass placed ateach point (:z:,y,z) ofaregion of space. Hence ateach point oftheregion, wecanrepresent themagnitude anddirection ofFbydrawing avector Fthere. WecallFavector point- function throughout thisregion, since itisavector whose components along theX,Y,Zaxes arefunctions ofthespace coordinates x,y,z of theunit mass. Aregion ofthistype isanexample ofaforce ﬁeld. Its formal deﬁnition follows. Deﬁnition 34.5. Aregion inspace, having theproperty that atevery oneofitspoints avector point-function Fexists that gives themagnitude anddirection oftheforce acting onaparticle ofunit mass placed there, iscalled aﬁeld offorce oraforce ﬁeld. Aregion intheneighborhood ofthesolar system isaﬁeld offorce. At each point oftheregion, avector point-function exists duetothemem- bers ofthesolar system. The region intheneighborhood ofacurrent bearing wireisaforce ﬁeld. Itiscalled anelectromagnetic ﬁeld. ByDeﬁnition 9.23, F,da:+F,dy+F,dziscalled anexact differential ifthere exists afunction U(a:,y,z) such that (34.51) dU=F,dz:+F,dy+F,dz, orequivalently, such that (34.52) %=11",, ‘%=F,,, $1=F,. Deﬁnition 34.53. Aforce ﬁeld oraﬁeld offorce iscalled conserva- tiveifthere exists afunction U(:c,y,z) such thatitsdifferential dUsatis- ﬁes(34.51), orequivalently ifitspartial derivatives with respect tox,y,z respectively satisfy (34.52), where F,,,Fy,and F,arethex,y,z com- ponents ofaforce Facting intheﬁeld. The function U(:z:,y,z) itself is called aforce function andthenegative ofU(:z:,y,z) iscalled thepotential orthepotential energy oftheforce ﬁeld. Comment 34.54. Not every force ﬁeld hasapotential —U(:c,y,z). Comment 34.6. The potential (-—U) may belooked upon asafunc- tion whose partial derivatives with respect to:c,y,z give respectively the components ofaforce Finthenegative rt,y,andzdirections. 474- PnoB1.1-ms: Srsrmrs. S1>nc1.u. 2m)Onnnn Eouyrrons Chapter 8 Example 34.61. Aparticle moving inaforce ﬁeld issubject toacen- tralforce Fwhose magnitude isproportional toitsdistance rfrom aﬁxed point O.Show that theforce ﬁeld isconservative. Solution. Byhypothesis (a) F=——kr, where lc>0isaproportionality constant. Themagnitude ofthecom- ponents ofFinthea:,y,z directions aregiven in(34.41). Hence substitut- ing(a)in(34.41), weobtain (b) F,=§(-hr) =—Icx, F,=§(-la) =—log, F,=§(-tr) =-Icz. Letustake fortheforce function UofDeﬁnition 34.53, I672 _ IC 2 2 2 (0) U(m/.#)=—7+6'=—§(w +11+z)+0- Therefore U Ic U av(<1) ‘Z7=-50¢)=-—ka:, ‘Z’?=—ky, 5=—lcz. Acomparison of(d)with (b)shows that thevalues ontheright ofthe equations in(d)arerespectively F,,F,,,F,.Hence byDeﬁnition 34.53, theforce ﬁeld isconservative. Example 34.62. Aparticle moving inaforce ﬁeld issubject toacen- tralforce Fwhose magnitude isinversely proportional tothesquare ofits distance rfrom aﬁxed point O.Show that theforce ﬁeld isconservative. Solution. Byhypothesis k(8.) F=-'ﬁr where k>Oisaproportionality constant. Substituting thisvalue ofF in(34.41), weobtain :1: k kw: Icy Icz(b) F,=;<-r—2)=—F, F,,=—F, F,=—r—3- Letustake fortheforce function UofDeﬁnition 34.53, 7“ _A Lesson 34D—Exercise 4-75 Therefore 6U_ lax _ kw 6U_ Icy 8U_ kz (d) 6__ -—-“, 6__r3, 6z—_-13'” \/(w‘*’+y”+z’)3 ’ y Acomparison of(d)with (b)shows that thevalues ontheright ofthe equations in(d)arerespectively F,,,F”,F,.Hence byDeﬁnition 34.53, theforce ﬁeld isconservative. Property D. Conservation ofEnergy inaConservative Field. LetFbeaforce acting onaparticle moving inaconservative ﬁeld. There- forebyDeﬁnition 34.53, there exists afunction U(a:,y,z) such that 0U 6U 8U(34.63) dU_-556$+55-dy+56¢, where 6U aU_ aU_E--F1, W-F,,, 3?-F,. Since F,=mzi,F,=mg‘),F,=mi,wehave aU aU_ 6U__.(34.64) -6?_ms, -6?_mg, az_mt. Multiplying theﬁrst equation in(34.64) bydz,thesecond bydy,the third bydz,andadding allthree, weobtain, with thehelp of(34.63), .. .. .._6U 6U QQ __(34.65) mxdx+mydy+mzdz-—5;dx+@dy+6z dz-dU. Integration of(34.65) with respect totime gives -2 -2 -2 (34.66) -"'§-+l”é”-+%=U+c. The leftside of(34.66) isdeﬁned asthekinetic energy ofaparticle. ByDeﬁnition 34.53, -—Uisitspotential energy. Hence (34.66) tellsus thatthesumofthekinetic andpotential energies ofaparticle inacon- servative ﬁeld isaconstant. This fact, namely that thesum ofkinetic andpotential energies ofaparticle isaconstant, isknown asthelawof theconservation ofenergy. Wehave thus proved thelawofthe conservation ofenergy foraparticle moving inaconservative ﬁeld. EXERCISE 34D 1.Aparticle moving inaforce ﬁeldissubject toacentral force Fwhose magni- tude isproportional toitsdistance 12from aﬁxed point O.Show that the force ﬁeldisconservative. 476 Pnostsmsz Srsrnms. SPECIAL 2N0Onnnn Equxrrons Chapter 8 2.Aparticle moving inaforce ﬁeldissubject toacentral force Fwhose mag- nitude isproportional toitsdistance r3from aﬁxed point O.Show thatthe force ﬁeldisconservative. 3.Aparticle moving inaforce ﬁeldissubject toacentral force Fwhose mag- nitude isproportional toitsdistance r"from aﬁxed point O,where nisa positive number. Show thattheforce ﬁeldisconservative. 4.Aparticle moving inaforce ﬁeldissubject toacentral force Fwhose mag- nitude isinversely proportional toitsdistance rfrom aﬁxed point O.Show thattheforce ﬁeldisconservative. 5.Aparticle moving inaforce ﬁeldissubject toacentral force Fwhose mag- nitude isinversely proportional toitsdistance r"from aﬁxed point O,where nisapositive number greater than 1.Show thattheforce ﬁeldisconservative. 6.Canyouthink ofaforce ﬁeldwhich isnotconservative? ANSWERS 34D . ka1.Force function U=-—Er+C’. 2.Force function U=—2T4+C. 76 1.3.Force function U=——ﬁ r+1—|—C’. 4.Force function U=—2logr2. _ k5.Force function U= 6.Aﬁeldinwhich energy isbeing dissipated astheparticle moves. Forexample, aﬁeld inwhich aresisting force, proportional tovelocity, ispresent cannot beaconservative ﬁeld. LESSON 34E. Path ofaParticle inMotion Subject toaCentral Force Whose Magnitude IsProportional toItsDistance from a Fixed Point O. Weassume that aparticle inmotion ofmass missub- jecttoacentral force Fwhose magnitude isproportional toitsdistance 1' from aﬁxed point O.Wealready know many facts about theparticle. Byproperties A,B,CofLesson 34C, weknow thatitmoves inaplane, that itsatisﬁes thelawoftheconservation ofangular momentum and that itsweeps outequal areas inequal times. ByDeﬁnition 34.5, the region inwhich theparticle moves isaforce ﬁeld. ByExample 34.61, this ﬁeld isconservative. Hence byproperty Dfollowing Example 34.62, we alsoknow that theparticle satisﬁes thelawoftheconservation ofenergy. Toﬁnd theequation ofitspath, wetake the:c,yaxes intheplane of thepartic-le’s motion with theorigin attheﬁxed point Otoward which theforce acts. LetP(z,y) bethecoordinates oftheposition oftheparticle attime tinarectangular system and (r,0) itscoordinates inapolar sys- tem. Thecomponents ofFinthe:1:andydirections are,by(34.42) (remem- Lesson 34E CENTRAL Foncs Pnoronrronxt T0Drsrxncn 477 bertheforce iscentral sothat F,=O), (34.7) F,,=m:i':=-F?’ F,,=m§=%- Byhypothesis (34.71) F=—k2mr, where forconvenience wehave used kzmfortheproportionality constant. Theminus signisnecessary because theforce isacting toward Oandthe positive direction isoutward from O.Substituting (34.71) in(34.7), we obtain thelinear system ofequations (34.72) 3=-16%, 17=-—k2y. Their respective solutions, obtained byanyofthemethods previously dis- cussed, are (34.73) :0=clcoskt—|—C2sinkt, y=c3coskt—|—c4sinkt. These aretheparametric equations ofthepath. You canverify byrefer- ring toTheorem 31.33 that thepair offunctions in(34.73) contains the correct number offour arbitrary constants. Hence inaspeciﬁc problem fourinitial conditions willbeneeded, 2(0), :c'(0), y(0), y’(0). Theperiod ofthemotion oftheparticle, byDeﬁnition 28.34, is21r/k. Itisthetime ittakes theparticle toreturn toitsinitial starting position, headed inthe same starting direction. Toﬁnd theequation ofthepath inrectangular coordinates, wemust eliminate theparameter tbetween thetwo equations in(34.73). The easiest Way todothisistoﬁrst solve them simultaneously forsinktand cosktinterms of:7:andy.The result is Sln kt=- : COS kl= 10104 —C263 75 62037-6164 61647-0263 Squaring both equations in(34.74) andthen adding them, weobtain for thepath oftheparticle inrectangular coordinates, _ 2 _ 2 (34.76) 1= 3,6.-6,63¢0, which canbewritten as (34-76) (632-l"642)“? "2(¢1¢3 +6264):‘?! +(612+022)!/2 _(C164 7-¢2¢a)2 =0,0164 —6263 5*0- From analytic geometry weknow thatiftheconstants in (34.77) A3”+2Bxy+Cy’+D=0 478 Pnonmms: Srsrsms. Srncrxn 2m:Onnnn Equxrrons Chapter 8 aresuch that B2-—AC<0,and Case 1.D96O,AD <0,then theequation represents anellipse with center attheorigin. Case 2.D;-50,AD >0,theequation hasnolocus. Case 3.D=0,theequation represents asingle point. Acomparison of(34.76) with (34.77) shows that (5) B2—AC=(6163 -l"62¢-Q2 “(632+642)(°12 +622) ="((3164 '—62¢a)2> and (bi AD ="(C32 +¢42)(¢1¢4 '“¢2¢3)2- Both oftheabove expressions areless than zero ifc104 -—c2c3 960. Hence if,in(34.76), c1c4 —C263, which corresponds toDof(34.77), is notzero, Case 1above applies and (34.76) istheequation ofanellipse with center attheorigin. Wehave thus proved that theorbit ofaparticle, attracted toanorigin Obyacentral force Fwhose magnitude isproportional toitsdistance from O,isanellipse with center attheﬁxed point O.Hence wehave also proved forthiscase that theforce isdirected toward thecenter of theellipse. Comment 34.78. Iftheparticle isconstrained tomove toward the ﬁxed point Oalong aradius vector sothat 0isconstant, then d0/dt =0, F=F,,andby(34.21) and(34.71), F,=ma,=mi‘=—k2mr, F=—k2r. This lastequation, asweshowed inExample 28.15, is,asitshould be, thedifferential equation ofmotion ofaparticle executing simple harmonic motion. Example 34.79. Aparticle weighing 16pounds is10feetfrom aﬁxed point Oandisgiven aninitial velocity of15ft/sec inadirection perpen- dicular tothea:axis. Ifacentral force Factsontheparticle with amag- nitude which isone-eighth ofthedistance oftheparticle from theﬁxed point O,ﬁndtheequation ofitspath andtheperiod ofthemotion. Solution. Wetake theorigin attheﬁxed point O,and the:z:,yaxes intheplane oftheparticle’s motion. Byhypothesis w F=—ﬂ andby(34.42), (b) F,=m,5i=%» F,,=mg=@. Lesson 34E CENTRAL Foncr-1 Pnoronrronu. roDrsrxncn 479 where risthedistance oftheparticle from O.Substituting (a)in(b),we obtain (c) ma':':=—g» mg']=--%- Byhypothesis m=§-§=inHence (c)becomes (o 3+§=0, t+§=c whose solutions, byanymethod youwish tochoose, are (e) z=c1cos%+c2sin%1 y=c;;cos%+c4sin%- Therefore dw_ C1-2 ¢_22<11/__a-2 242. (f) E-—§s1n2+2cos2» E_ 2s1n2+2cos2 The initial conditions aret=0,x=10,y=0,dz:/dt =0,dy/dt =15. Substituting these values in(e)and(f),weobtain (g) c1=10, C2=0, c3=0, c4=30. Hence (e)becomes t .t(h) x=10cos§, y=30s1n§- Byeliminating theparameter t,weobtain _ $2 g2 <° W+W=L which istheequation ofanellipse with center attheorigin orﬁxed point O. Itsgraph isshown inFig. 34.791. The period ofthemotion, obtained (0.30) vo=15'/sec (10,0) Figure 34.791 from (h),is41rseconds. Itisthetime itwilltaketheparticle tomake a complete circuit oftheellipse. 480 Pnonu-ms: Svsrrms. Srncuu. 2N0Onnnn Equxrrous Chapter 8 EXERCISE 34E Verify theaccuracy ofthesolution of(34.72) asgiven in(34.73). Verify theaccuracy of(34.74), (34.75), and(34.76). Aparticle inmotion ofmass missubject toacentral force Fwhose magnitude isproportional toitsdistance from aﬁxed point O.Initially itisxoftfrom theorigin andisgiven avelocity of00ft/sec inadirection perpendicular to thezaxis. (a)Find theparametric equations ofitspath; alsotheequation ofitspath inrectangular coordinates. Take kfortheproportionality constant in- stead ofkzmaswedidinthetext. (b)Forwhat relative values of2:0,vowillthepath beacircle? Abody weighing 16lbisattracted toaﬁxed point Obyaforce whose magni- tude isone-eighth thedistance oftheparticle from 0.Initially itis12ftfrom Oandgiven avelocity ofvoft/sec inadirection perpendicular tothe2:axis. (a)Find theparametric equations ofmotion; alsotheequation ofmotion in rectangular coordinates. (b)What initial velocity vowillmake theeccentricity oftheorbit {.7 Solve problem 3(a), ifinitially theparticle is:60ftfrom theorigin andisgiven avelocity ofvoft/sec inadirection making anangle 0with the:4:axis. Abody weighing 16lbis12ftfrom aﬁxed point O.Itisgiven aninitial velocity of20ft/sec inadirection making anangle of45°with the2:axis. A central force actsontheparticle with amagnitude equal toone-eighteenth ofthedistance ofthebody from O.Find theequation ofitspath andthe period ofitsmotion. Verify that theequation satisﬁes Case 1after (34.77) andistherefore theequation ofanellipse. Aparticle inmotion ofmass misrepelled from aﬁxed point Owith aforce proportional toitsdistance from O.Initially itis2:0ftfrom theorigin andis given avelocity ofvoft/sec inadirection perpendicular tothe2:axis. (a)Find theequation ofmotion. (b)What type conic isit?(c)Show thatproperties A,B,C,DofLessons 34CandDarealsovalid when thecentral force is repelling instead ofattracting. “ Setupthesystem ofdifferential equations ofmotion fortheparticle ofprob- lem3ifinaddition there isaforce ofresistance proportional tothevelocity. ANSWERS 34E 2 2 (a)2:=:cocos\/k/mt, y=vovm/ksinvk/mt;;£5+ki2 =1.2 2 0 "W0 (b)Z0 =mvo 2 2 _ _ - L L= (a)2: 12cqs<}t,y 2120sinQt,144+4002 1. (b)00=3\/3 ft/sec, or4\/3 ft/sec. 2:=socosmt—|— mm cos0sin\/ﬁt, y=v0msin0sin\/Wt; 2 (sinz 0)a:2 —(2sin0cos0)xy+(%§ -l—cosz0)yz-—2:02sin20=0. 2:=12cosﬁt+30\/2 sinQt,y=30x/2 sinQt;61rsec; $2-—-2:cy+ 1.081/2 —144=0. Lesson 34F FORCE INVERSELY Pnororrrromu. roSQUARE orDrsrxnca 481 7.(a)2:=socosh\/k/m t,y=v°\/m/ksinhvk/mt, 2:2 kg/2 _ "-1-(b)Hyperbola. 8.mi=-kz ——rat,my’=-—ky —rg,where kandraretheproportionality constants respectively fortheforce andtheresistance. LESSON 34F. Path ofaParticle inMotion Subject toaCentral Force Whose Magnitude IsInversely Proportional tothe Square ofItsDistance from aFixed Point O. Weassume that aparticle in motion, ofmass m,issubject toacentral force Fwhose magnitude is inversely proportional tothesquare ofitsdistance rfrom aﬁxed point O. Byproperties A,B,CofLesson 34C, weknow that: 1.Theparticle moves inaplane. 2.Itsatisﬁes thelawoftheconservation ofangular momentum. 3.Itsweeps outequal areas inequal times. ByDeﬁnition 34.5, theregion inwhich theparticle moves isaforce ﬁeld. Hence, byExample 34.62 andproperty Dfollowing it,wealsoknow that: 4.This ﬁeld isconservative andtheparticle therefore satisﬁes thelawof theconservation ofenergy. Toﬁndtheequation ofthepath oftheparticle, wetake thex,yaxes in theplane oftheparticle’s motion and theorigin attheﬁxed point O toward which theforce Fisdirected. LetP(z,y) bethecoordinates ofthe position oftheparticle attime tinarectangular system and (7,0) its coordinates inapolar system. By(34.42), (34.8) F,=m£=€_E» F,=mi]=FTy- Byhypothesis Km where forconvenience wehave taken Km,K>0,fortheproportionality constant. The minus sign isnecessary because theforce acts toward 0 and thepositive direction isoutward from 0.Substituting (34.81) in (34.8), weobtain thesystem ofequations .. K ,, K(34.82) at=—--ﬁx, y=-ﬁg. Ifin(34.82), wesubstitute forritsequal \/:02 —|—yz,theresulting equations form anonlinear system which isdifﬁcult tosolve. Itturns outthat thepath oftheparticle canbefound more easily byusing polar coordinates. Call F,thecomponent ofFintheradial rdirection andF; 482 Pnostnms: Srsrrzns. Srrzcnu. 2NDOnnnn Eouxrrons Chapter 8 itscomponent inadirection perpendicular toF,.Then by(34.21), (34.83) F,=ma, =m(i‘—r02), F,=ma,=m(21‘d +rd). Since Fisacentral force, F,=0.Setting thesecond equation in(34.83) equal tozero, andthen multiplying itbyr/m, weﬁnd 2rr0 —|—rzd=0, which isequivalent to (34.34) $620) =o,Ho=1.,6=h/T2, where histhesame constant weintroduced in(34.47), i.e.,histheangu- larmomentum oftheparticle perunit mass. Intheﬁrst equation of (34.83), substitute forF,itsvalue asgiven in(34.81) (remember here F,EFsince theforce actsonlyalong r)andfor0itsvalue h/1'2asgiven in(34.84). Wethus obtain Km ,_h’ ,,1.’ K Although methods ofsolving thenonlinear equation (34.85) aregiven in both Lessons 35A and35C which follow———see alsoExercise 35,11—use of either ofthese methods willgive asolution oftasafunction ofr.Itturns outtobeeasier toanalyze thepath oftheparticle ifwesolve thesecond equation in(34.85) forrasafunction of0.Toaccomplish thisend, we usethesubstitution 1 1(34.86) M-—-';I T-—E‘ [Note that uasdeﬁned in(34.86) istheforce function UofExample 34.62.] Substituting (34.86) inthelastequation of(34.84), weobtain (34.87) 9=huz. Two differentiations ofthesecond equation in(34.86) give, with thehelp of(34.87), ____1du___1dud0____1du 2____d__n (3488) 1- u2dt_ u2d0dt_ a2dﬁhu — hd0’ __ .1’ 6%. 6%7‘= = = —'h2'll.2a—03' Substituting thelastvalue of5‘of(34.88) andthevalue ofrof(34.86) in thesecond equation of(34.85), weobtain 22dz" 23____ 2(34.89) -hu To-2-—hu - Ku, which simpliﬁes tothelinear equation .121. KW +U»=Fr h760. Lesson 34F Foncr: Irwnnsnnr Pnoron'rroNAL 'roSQUARE orDISTANCE 483 Itssolution, byanymethod youwish tochoose, is (34.892) 1.=g+6663(0-0.),h.40, where cand 00arearbitrary constants. Since K,h,andcareconstants, wecanwrite (34.892) inamore useful form byreplacing cbyanewcon- stant Ke/h2. Wethusobtain (34.893) u=%[1 +ecos(0—00)], h360. By(34.86), u=1/r. Ifwenow make thissubstitution in(34.893) and choose ouraxes sothat 00=O,theequation simpliﬁes to h2 which istheequation, inpolar coordinates, ofthepath ofaparticle moving subject toacentral force whose magnitude varies inversely asthe square ofitsdistance from aﬁxed point 0. P1 d, Semi-focal2 "6i.=i'"‘ K P(r,0) .1 P0181‘ ﬂxiﬂ 0 rcos0 V(perigee) O(focus) DirectrixConic section Figure 34-.895 Wedigress momentarily toreview foryou theproof that (34.894) is theequation ofaconic section whose eccentricity ise,whose semifocal width ish2/K andwhich hasonefocus attheorigin. InFig.34.895, we have illustrated such aconic. Thepoint onaparticle’s path that isnearest thepoint Otowhich theparticle isattracted iscalled theperigee ofthe path. Bydeﬁnition, theratio r/dforevery point Ponaconic isequal to itseccentricity e.Hence, forthetwopoints Pand P1ontheconic we have respectively, 2 2 (a) e=§ and e=KLdl; d=£and d1=g(»h#0. 484 Pnonu-znsz Srsrnms. SPECIAL 2NDOnnsn EQUATIONS Chapter 8 From theﬁgure, weseethat (b) d1= d+rcos0. In(b),replace dandd1bytheir values asgiven inthelasttwoequations of(a). There results h2 r hz(C) ‘i=5-i"1'COS0, T= : which isthesame as(34.894). Additionally, weknow from analytic geometry that if (d) e<1,theconic isanellipse, e=1,theconic isaparabola, e>1,theconic isanhyperbola. Ife=0,(34.894) becomes r=ha/K, which, inpolar coordinates, isthe equation ofacircle whose radius ishz/K. Comment 34.896. Wehave thus proved that, ifh9-40,theorbit of aparticle, attracted toaﬁxed point Obyacentral force which satisﬁes theinverse square law(34.81), isaconic section with onefocus atthe ﬁxed point O.Ife<1,theconic section isanellipse. Since onefocus is attheﬁxed point O,which wealsotook tobetheorigin ofourcoordinate system, Ocannot bethecenter oftheellipse. Iftherefore aparticle, sub- jecttoaforce which satisﬁes theinverse square law(34.81), moves inan elliptical orbit, thecentral force isdirected toward afocus oftheellipse andnottoward itscenter. Contrast thisresult with that obtained in Lesson 34E. Wefound there that ifFvaries directly asthedistance r from O,then theforce isdirected toward thecenter oftheellipse. Comment 34.8961. Ifh=0,then by(34.84), 0=0.Therefore 0isaconstant. Hence theparticle must move onaline. By(34.85), with h=0,thedifferential equation ofmotion simpliﬁes to Kl‘——r—2- For possible methods ofsolving it,seeLesson 35,also Exercise 35,7 and22. Comment 34.897. Determining theConstants ofIntegration h, e,0,;of(34.893). The path oftheparticle, by(34.893), with ureplaced byitsequal 1/r,is <4) §=gt+ms<0—0.)]. Lesson 34F Foncs Invnnsanv PROPORTIONAL TOSQUARE orDISTANCE 485 where h,e,and00areconstants which were introduced byintegrations. Assume thatwhen theparticle isatthepoint P0ofitspath, itsdistance from Oisroandthatitismoving with avelocity voinadirection making V0 vosinA=rodo P°(ro,0) vocosA=F0 To 96 O Figure 34-.893 anangle Awith thelinejoining OtoPo,Fig.34.898. Forconvenience, wemeasure subsequent values oftheangle 0from thislineOP0. The initial conditions are,therefore, [for12,,v,values seeFig.34.898 and(34.2)] (b) r=ro, 0=0, v=vo, A=angle shown inFig.34.898, v,=720cosA=to, v;=vosinA=redo. Thesubstitution of(b)in(a)gives 1K h’(c) R=F[1+ecos00], ecos00=ﬁ—1. Let0:betheangle measured from theradius vector toatangent toa curve atP(r,0), Fig.34.899. Letvbethevelocity oftheparticle atP. v _ .v0=v sinor.-r0 P030) v,.=vcosa-1" 0 O Figure 34-.899 Therefore thecomponents ofvintheradial direction andinadirection perpendicular toitarerespectively (d) v,=vcosa, v;=vsina. 486 Pnonmms: Srsrsms. SPECIAL 21mORDER EQUATIONS Chapter 8 By(34.2), v,=7"andvs=rd.Hence (d)becomes (e) 1"=vcosoz, r0=vsinoz. By(34.84), r0=h/r. Therefore thesecond equation in(e)canbewritten as (f) h=rvsinoz. Hence when r=ro,v=voandoz=A,weobtain, by(f)and(b), (g) h=rovosinA=r0200. Differentiation of(a)with respect to0gives 1d K.(11) -FI;=-F,€s1n(0—— 0,). Also _lﬁ___1_£1£§lr2d0_ r'*’dtd0 1dr2=-r-27:7[by(34.34)) ___l£i£_ hdt =—%vcosa[by(e)above]. Hence, byequating thislastexpression with theright sideof(h),we obtain (i) %sin(0—00)=vcosoz. Inserting in(i)theinitial conditions 0=0,v=vo,or=A,vocosA= fo,there results (j) %sin (-00) =vocosA, ——esin00='11-'29cosA=I-'%r,,. Squaring thesecond equations in(c)and(j)andadding them, wehave h2 2 h.2 h4 2h2 h2 2 (k) 62=($—1)+(Er°)=w"$+1+%Cos2A. Wecansimplify (k)somewhat, bynoting from (g)that (1) h2=r2v2sin2A —"2-=1-6662.4 00 r rozvog 1 coszA=1—T021102 Lesson 34F Foncm Invsnssu PROPORTIONAL '1-oSQUARE orDrsrmcn 487 Substituting thelastequation of(1)in(k),weﬁnd 2_h‘_g1f h’v.,’_ h‘(ml 6_r02K3 r0K+1+K2 1‘02K22 22 =1_E+_hL.TQK K2 By(g),wecandetermine h.By(k)or(m), wecanthen determine e; remember Kisaproportionality constant andnotaconstant ofintegra- tion. With eandhknown, wecandetermine 00bythesecond equations in(c)and (j).After 00isknown, wecanthen choose ouraxes tomake 00=0,andthus obtain (34.894). Comment 34.9. Energy Considerations Related totheInverse Square Law. Inproperty Dfollowing Example 34.62, weshowed that thesum ofthekinetic and potential energies ofaparticle moving ina conservative ﬁeld isaconstant, see(34.66). Therefore, by(34.66), (a) Qmvz ——U=E, where wehave replaced theconstant cbytheenergy constant E’,andthe velocity components, ¢,y,z, byv.InExample 34.62, weshowed that if F=-k/r2, then theforce ﬁeld isconservative andthepotential energy function ——U=——Ic/r. Intheexample ofthisLesson 34F, F=—-Km/r2, see(34.81). Hence thepotential energy function -—U=——Km/r. Sub- stituting thisvalue ofUin(a),itbecomes (b) §mv2-Q=E. Inserting in(b), theinitial values v=voand r=ro,weobtain- remember Eisconstant forallvandr— (°) §"Wo2 '-M =E-To By(m)ofComment 34.897, _==2_""'_ "”"_<>’. (d) 1 e MK K2 Multiply (c)by—2h2/K2m. There results- 2h” _2h”(2Km)_hw 2h’ <°)—mE- -M *""’<>"7""7<T+r—°1'<" Since theright sides of(d)and(e)arethesame, wecanequate their left sides. Wehave thusshown that _=_,2h’__2(i)” (f) 1 e- K2mE- mK E. 488 Pnonmsmsz Srsrams. Srncuu. 2m)Oam-:11 EQUATIONS Chapter 8 Themass misapositive quantity; soalsois(h/K) 2.Hence weconclude from (f),thatifE<0, (S) 1—e2>0, e2<1, e<1, andtheparticle, therefore, must move inanelliptic orbit. IfE=0, then 1——e2=O,e=1,andtheparticle must move inaparabolic orbit; ifE>0,then e>1andtheparticle must move inahyperbolic orbit. Weinfer from allthisthataparticle inmotion inaconservative ﬁeld, whose orbit iselliptic, must initially have hadnegative energy. Conversely, ifaparticle with negative energy isprojected into aconserva- tiveﬁeld, itwillmove inanelliptic orbit. Analogous remarks canbe made fortheother twotypes oforbits. Remark. Aparticle willhave negative energy initially, ifitskinetic energy, which istheﬁrstterm of(a),islessthan theforce function U= Km/r. Comment 34.91. Equation (34.894) gives theposition rofaparticle asafunction of0,hsé0.Itwould bedesirable toexpress rasafunction oftsothat wecanknow where theparticle isatanymoment. By(34.85) 5‘=ha/1'3 —K/r2. Asmentioned previously, two methods ofsolving thisequation willbegiven inLesson 35.(Also seeExercise 35,11.) Un- fortunately, useofeither method gives tasafunction ofr.Theproblem ofsolving theresulting equation forrasafunction oftturns outtobe exceedingly difﬁcult. EXERCISE 34F 1.Verify theaccuracy ofthesolution‘ of(34.891) asgiven in(34.892). 2.Abody weighing 16lbis12ftfrom aﬁxed point O.Itisgiven aninitial velocity of6ft/sec inadirection perpendicular tothe:1:axis. Find itsequation ofmotion ifitissubject toacentral force whose magnitude isequal to120/r2, where risthedistance oftheparticle from O.Hint. Follow themethod ofthe text. Initial conditions inrectangular coordinates aret=0,2:=2:0=12, 1/=yo=0,£0=0,170=vo=6.Inpolar coordinates initial conditions al‘et=0,T=T0=12,0:0Q=0,1‘Q=0,0=o()=U0/T()=j§§=in What istheeccentricity ofthepath? 3.Abody weighing 16lbis10ftfrom aﬁxed point 0.Itissubject toacentral force Fwhose magnitude is100/r2, where risthedistance oftheparticle from 0.What initial velocity should begiven theparticle, inadirection perpendicular tothe2:axis, inorder that theparticle may (a)move inan elliptic orbit ofeccentricity 1},(b)move inacircular orbit (hint, orbit is circular ife=0),(c)move inaparabolic orbit, (d)move inahyperbolic orbit ofeccentricity 2? Intheproblems below, weshall consider themotion ofasatellite of theearth, where thesatellite hasbeen setinmotion bybeing ejected from arocket. These problems arecentral force problems, obeying theinverse Lesson 34F—Exe1-cise 439 square law(34.81). Thesatellite willbeattracted toward thecenter of theearth with aforce inversely proportional tothesquare ofthedistance rofthesatellite from thiscenter. CallRtheradius oftheearth. 4.In(34.81), replace Fbyma,where aistheacceleration oftheparticle. (a)Show that (34.911) K==gR2. Hint. Whenr =R,a=—g. (b)Show thattheequation ofmotion (34.892) becomes 2 (34.912) u=%+Ccos(0—00), andthat (34.894) becomesh2 (34.913) 7'= 1 where (34.914) h=1'29. 5.Assume in(34.913) that att=0,0=0,that thelastrocket isﬁred ata distance r=rofrom thecenter oftheearth andthatithasejected thesatel- litewith avelocity voinadirection making anangle Awith theradius vector joining therocket tothecenter oftheearth. SeeFig.34.898. Show thatthe constants h,e,and00in(3-1.913) aregiven respectively by (34915) h=T0260, 32 2 2 20 6(34916) 3=<5g7,% -1)+(';’T" #0). 1ro3do2 . 1ro2001‘°005 00=Z -"17 S111 00=—Z TE-2-’ 1 where toandr000aretheinitial velocity components ofvointheradial direc- tionandinadirection perpendicular totheradial axis. Hint. With K=gR2 asgiven in(34.911), equation (3-1.913) isthesame as(a)ofComment 34.897. Make useof(g),(k),(c),and(j)ofthiscomment. 6.Show thattheorbit ofasatellite oftheearth willbeacircle if (34.918) 1113902 =gR2 and to=0. Hint. By(3-1.913), theorbit iscircular ife=0;by(3-1.916) e=0if(3-1.918) holds. 7.Asatellite isejected byarocket intoacircular orbit 300miabove theearth’s surface. Find itsperiod ofrotation. Hint. Theperiod ofthesatellite, by Deﬁnition 28.34, isT=21r/w, where w=0isitsangular velocity. There- foreby(3-1.918), T=(21rro/R)\/H75. Here R=4000 mi,ro=4300 mi. 8.Ifthesatellite isvery close tothesurface oftheearth sothatr0isvery close toR,then (34.918) canbewritten as (s4.919) 0'0=\/E/R. 490 Pnosnnmsz Srsrams. Srncuu. 2m)Onnnn Eqtwrrons Chapter 8 Show thattheperiod inthiscase, foracircular orbit, isapproximately 85min. Seehintinproblem 7forperiod formula. [Compare with answer toExercise 28AandB,2(c).] 9.Atadistance 1'0from thecenter oftheearth, arocket propels asatellite ina direction perpendicular totheradius vector joining therocket tothecenter. Thevelocity ofthesatellite isvo.Hence 1‘0=0,90=vo/ro andtheangle A inFig.34.898 is90°. (a)Show that if602<gR2/1'03, then 00=1r.Hint. In(34.916), to=0. Solve foreand,since theeccentricity isalways positive, choose theproper signtomake e>0.Substitute thisvalue ofein(3-1.917). (b)Show thattheorbit ofthesatellite isthen r—i—T040O2 —where e—1—ﬁ or_gR2(1 —ecos0)' _ gR2 i _ 11,4002 TgR2 —(gR2 ——r0360?) cos0i Hint. Use(34.913), (34.915), and(34.916). (c)Show that thesatellite isfarthest from thecenter oftheearth, called theapogee oftheorbit, when r=ro,0=0,i.e.,theapogee isatthe point where thesatellite isreleased. Hint. Thedistance rislargest when thedenominator inequation (b)above issmallest. Thedenominator is smallest when thenegative term initislargest. This negative term is largest when 0=0,cos0=1.Solve forrwith cos0=1. (d)Show that thesatellite’s perigee, i.e.,thepoint ofthesatellite’s orbit nearest thecenter oftheearth, occurs when 0=1r,andthatitsdistance from thecenter oftheearth isthen r0490’/(2gR2 —r036o2). Seehint in(c)above. (e)Show that thesatellite willmake acomplete orbit without hitting the earth if902>2gR3/[ro3(ro —|—R)]. Hint. Theperigee oftheorbit, as given in(d)above, must begreater than theradius Roftheearth. (f)Show thatif902>9R2/1'03, then00=0.Seehintin(a)ofthisproblem. Show thattheorbit ofthesatellite isthen _ T04002 h _r0302 T— rWer€e—?§*l,OT T= T041002 gR2+(11,390? —gR2) cos0i Seehintin(b)above. Show that theperigee oftheorbit occurs when r=ro,0=0,i.e.,atthepoint where thesatellite isreleased; thatthe apogee oftheorbit occurs when 0=1randthat itsdistance from the center oftheearth isthen r0400’/(2gR? —rosdoz), provided 2gR2 > 1113602. Seehints in(c)and(d)above. Finally show that if2gR” § r030o2, theorbit willnothave anapogee. Hint. Thedenominator ofthe apogee’s distance formula above willthen benegative orzero. If2gR2 =r0390”, then rozdoz =2gR2/ro. Ifthesatellite isejected atornearthesurface oftheearth sothat1'0=R,then thelastequation becomes 1-090 =\/2gR. Butvo=r090 sothat vo=x/2gR which is theescape velocity ofabody ﬁred from thesurface oftheearth, see(i) ofExample (16.36). Lesson 34-G PLANETARY Morton 491 ANSWERS 34-F =rovo =72,m=Q,mk=120, k=240, e=0.8,r=21.6/(1+ 0.8cos0). 3.(a)5.48ft/sec. (b)4.47ft/sec. (c)6.32ft/sec. (d)7.75ft/sec. 7.5700 sec,approximately, or95min. Theﬁrstsatellite putintoorbit in1957 bytheU.S.S.R., known astheSputnik, hadanearly circular orbit of300miles above theearth’ ssurface andaperiod of96min.P=- LESSON 34-G. Planetary Motion. Newton’s law ofuniversal gravitation states that every twobodies intheuniverse attract each other with aforce proportional totheproduct oftheir masses andin- versely proportional tothesquare ofthedistance separating them. Let Mbethemass ofthesunandmthemass ofaplanet. Itcanbeproved thatwedonotcommit aserious error ifweconsider thesunasﬁxed, its mass Masconcentrated atitscenter, theplanet asaparticle, andsun andplanet asisolated bodies. Then byNewton’s lawofuniversal gravita- tion, GM(34.92) F=- where risthedistance ofaplanet from thesun's center, andGisapro- portionality constant called thegravitational constant. Replacing the constant GMin(34.92) byanewconstant K,itbecomes F=—Km/1'2, which isthesame equation as(34.81) ofLesson 34F. Since thisforce Fis directed. toward aﬁxed point O,namely thesun's center, itisacentral force. Hence planetary motion isexactly thesame asthemotion ofthe particle discussed inLesson 34F. Wecantherefore assert that: 1.Aplanet moves inaplane. 2.Theorbit ofaplanet isaconic section whose equation, by(34.894), is 2 T= 1h760, where histheangular momentum oftheplanet perunit mass, eisthe eccentricity ofitsorbit andKistheproduct ofthegravitational con- stant Gandthemass Mofthesun. 3.The planets satisfy thelawoftheconservation ofangular momentum. 4.The planets sweep outequal areas inequal intervals oftime. 5.Theforce ﬁeldinwhich theplanets move isconservative; hence the planets satisfy thelawoftheconservation ofenergy. 6.Thesunisatonefocus oftheplanet's orbit. [SeeComment 34.896.] Comment 34.93. TheOrbits oftheEarth andtheOther Planets ofOur Solar System AreEllipses with theSunatOneFocus. Hence fortheplanets ofoursolar system e<1.This means, asweshowed inComment 34.9, that each planet, atthebeginning ofitsexistence, had negative energy. 4-92 Pnontnusz Srsrnns. SPECIAL 2NDOnnnn EQUATIONS Chapter 8 The orbits ofcomets* which appear after long intervals oftime areex- tremely elongated ellipses whose eccentricity isnear 1,almost close to parabolas. Those bodies forwhich eg1have parabolic orhyperbolic orbits. They leave thesolar system andnever return. EXERCISE 34-G 1.Find theequation ofmotion ofaplanet ofmass mifitsdistance atperigee, i.e.,itsdistance nearest thesun,isroanditsvelocity there isvo.Hint. Take theaxisoftheellipse through theperigee. Then att=0,r=ro,0=00=0, 1‘=0,6=vo/ro. SeeFig.34.898. 2.Find theapproximate equation ofHalley’s comet. Hint. Seefootnote atthe bottom ofpage: e=0.967, a—c=0.587, where aisthesemimajor axis andcisthedistance ofthefocus from thecenter oftheelliptic orbit. 3.Acomet atrestataninﬁnite distance away from thesunisattracted toward thesuninaccordance with theinverse square law. Ifitsdistance atperigee isro,ﬁndtheequation ofitspath andshow thatitsorbit isparabolic. Hint. Take axissothat00=0in(34.892). Att=0,0=1r,u =1/r=0.When 0=0,u=1/ro. ANSWERS 34-G 22 1,=__i__...' K+(10002 —K)cos0 2 2 2.ﬁ +Egg =1.Figures inastronomical units. 210 3.r=ﬁé-5- Orbit isparabolic since e,thecoefficient ofcos0,isone. LESSON 34H. Kepler’s (1571-1630) Laws ofPlanetary Motion. Proof ofNewton’s Inverse Square Law. Kepler’s three laws ofplane- tary motion are: 1.Each planet moves inanelliptical orbit with thesunatonefocus. 2.The radius vector connecting sunandplanet sweeps outequal areas in equal times. 3.The square oftheperiod ofaplanet isproportional tothecube ofthe semimajor axisofitsorbit. Wehave already proved 1and2:seenumbers 6and4ofLesson 34G. Weshall now prove 3. Proof of3.By(34.894), theorbit ofaplanet isgiven by (a) ,-__L “K(1+ecos0)' ‘The famous Halley’s comet hasanelliptical orbit whose eccentricity is0.967. Its period is76years. Itsperigee is0.587 astronomical units (anastronomical unit isthe distance oftheearth tothesun, approximately 92,900,000 mi). Since itlastvisited us in1910, itwillbeagain visible in1986. Lesson 34H Km>Lan’s Laws. PROOF orInvansa SQUARE LAW 493 where onefocus isattheorigin ofacoordinate system, called (0,0) in Fig. 34.94, andthesemifocal width ish2/K, called Lintheﬁgure. Let (0,0) bethecenter oftheellipse. With respect tothecenter oftheellipse, (0.6) (c,L) L=L’ F(¢»°) a0 66K (I) (») Figure 34.94 let(0,0) bethecoordinates ofthefocus, (a,0), (0,b), bethecoordinates of theends ofthesemimaj orandsemiminor axes respectively. Theequation oftheellipse with respect toitscenter asanorigin is,therefore, 2 2 i(:§+ll:§=1i where c2=a2—b2.If$2=c2=a2-—b2,then g2=L2,andthere- fore, by(b), a2 __b2 L2 L2 b2 b2 @ -?“+w=L F=F L=7' Thesemifocal width Lalsoequals h2/K. Substituting thisvalue ofLin thelastequation of(c),weobtain h2__b2 2__ah2 @ r"7' b-Y" Theﬁrstequation in(34.49) holds forevery particle subject toacen- tralforce. Ittherefore holds fortheplanets. Thelastequation in(34.49) resulted when wetook forourinitial conditions A=0,t=0.Hence if A=0,t=O, (8) A=iht gives thearea Aswept outbyaplanet intime t.CallTtheperiod ofa planet’s orbit, i.e.,thetime ittakes aplanet tomake acomplete circuit ofitsorbit. When theplanet hasmade acomplete circuit, ithasswept outthearea oftheellipse, namely 1rab. Hence when A=1rabandt=T, 494 PROBLEMS: SYSTEMS. SPECIAL 2m)ORDER EQUATIONS Chapter 8 weobtain by(e) __21rab 2__41r2a2b2 In(f),replace b2byitsvalue asgiven in(d). There results 2__41r2a2 £__41r2a3 _(34.95) T-———h2 K———K Since K(=GM)isaconstant, andaisthesemimajor axis oftheellipse, (34.95) saysthatthesquare oftheperiod ofaplanet isproportional tothe cube ofthesemimajor axisofitsorbit. Proof ofNewton’s Inverse Square Law from Kepler’s Laws. We have proved Kepler’s three laws from Newton’s universal lawofgravita- tion. Historically, however, Kepler preceded Newton and hence the former knew nothing ofthelawofgravitation. Itisindeed remarkable thatKepler wasabletodeduce histhree lawsfrom anintensive study of therecordings ofthepositions oftheplanets made bydirect observations. ItwasNewton who used Kepler’s laws asahypothesis todevelop hisown universal lawofgravitation. Part ofhisproblem wasthus theinverse of theonewesolved. Heassumed that aplanet moves inanelliptical orbit with thesunatonefocus, thatitsweeps outequal areas inequal times, andthen setouttoprove thattheplanet must therefore besubject toa central force directed toward afocus, whose magnitude varies inversely asthesquare ofthedistance oftheplanet from thesun. Theproof follows. Proof. ByKepler’s second law, dA/dt isaconstant. Therefore by thesecond equation in(34.481), (a) =}r20=-5» 1'26=c, where cisaconstant. Differentiation ofthesecond equation in(a)and multiplying theresult byaconstant mass m,gives (b) m(2r1‘9 +r20) =0, m(2r0 +rd)=0. Thesecond equation of(b),by(34.21), isma;=0.Butma,isthecom- ponent offorce acting onaparticle inadirection perpendicular tothe radius vector. Since thiscomponent iszero, theforce acting onaplanet must beacentral one; i.e.,theforce always acts along aradius vector toward oraway from thesun. ByKepler’s ﬁrstlaw,aplanet moves inanelliptical orbit with thesun atonefocus. Weshowed inLesson 34F, thattheequation ofanellipse inpolar coordinates with onefocus attheorigin is A <°> '-aw’ Lesson 34H Kai>Lan’s Laws. Paoor orInvnnsa Sooann Law 495 where e<1isitseccentricity andAisitssemifocal width. Differentia- tionof(c)gives Aesin0 A2 esin0 (d) 1i=(1+ecos0)20=(1+ecos0)2 A0' Making useof(c)and (a),wecanwrite (d)as _ esin6 c ce.(8) 1'=7'2——Z—;2-=-A-S1116. Differentiating (e)andthen using (a),weobtain 2 (r) r=%(ecosa)c=ﬁ(ecos0). Solving (c)forecos0,there results A_ (g) ecos0=—;——r - Substituting thisvalue in(f),weobtain __ 02 A—r __c2 02 <1‘) '—A—.=(*.—)—.—3*z§' Thecomponent ofaforce inaradial direction is,bytheﬁrstequation in(34.21), (i) F,=ma, =m(i‘—r62). In(i),replace Fbyitsvalue in(h)and0byitsvalue in(a).There results . c2 c2 c2 mcz <1) F""”[r?_F_r_3j__Tﬁ' Weshowed above that theforce acting onaplanet istoward oraway from thesun. Since m,c2,andAarepositive constants, (j)tells usthat theforce acting onaplanet isdirected toward thesun, anditsmagnitude isinversely proportional tothesquare ofitsdistance from thesun. Comment 34.951. Theinverse square lawjustproved isonly part of theuniversal lawofgravitation. Newton’s studies ofthegravitational force oftheearth plus hisobservations ofthemoon's orbit about the earth, plus hisown genius, enabled him toformulate hisfamous lawof universal gravitation asstated atthebeginning ofLesson 34G. Comment 34.96. InLesson 34C, weproved that every particle sub- jecttoacentral force obeys Kepler’s second law, i.e.,itsweeps outequal areas inequal times. Hence Newton’s inverse square lawisonly asu_ﬂi- cient condition forKepler’s equal area law, notanecessary one. However, theinverse square lawisanecessary andsuﬂicient condition forKepler’s 496 Pnonnnus: SYSTEMS. SPECIAL 2m)Ononn EQUATIONS Chapter 8 ﬁrstlaw: theorbit ofaparticle subject toacentral force isanellipse with theforce directed toward afocus. Note thatifFisproportional to rasinLesson 34E, theorbit isalsoelliptical, buttheforce isdirected toward thecenter oftheellipse. EXERCISE 34H 1.(a)Take 240,000 miles asthesemimajor axisofthemoon’s orbit andits period as27.3days. UseKepler’s third lawtoﬁndthevalue ofthe proportionality constant lofortheearth, where Ioreplaces 412/K in (34.95). Useforunits 1000 miandhour. (b)InExercise 34F,7, wefound T=95minfortheperiod ofthecircular orbit ofasatellite oftheearth 300miabove itssurface. Take 4300 mi forthesemimajor axisofthesatellite’s orbit andcalculate lo.Usefor units 1000 miandhour. (c)InExercise 34F,8, wefound T=85minfortheperiod oftheorbit ofa satellite close totheearth’s surface. Take 4000 miforthesemimajor axis oftheorbit andcalculate k.Useforunits 1000 miandhour. Compare results in(a),(b),and(c). Ana. It=0.031. Remark. Next time youread ofanearth satellite which hasbeen success- fully orbited, anditsperigee andapogee aregiven, usethisvalue ofkand Kepler’s third lawtocalculate theperiod oftheorbit andseeifitagrees with theobserved period. 2.(a)UseKepler’s third lawtocalculate thevalue oftheproportionality con- stant kforthesun. Useforunits 1,000,000 miandday. Take theperiod oftheearth as365days andthesemimajor axisofitsorbit as93,000,000 mi. Ans. lc=0.165. (b)Usethisvalue ofktocalculate theperiod ofoneoftheother planets from itsknown distance from thesun,orcalculate itsmean distance from the sunfrom itsknown period. By(34.s91), dz K(8) ﬁi2£+u=i,,;» hiéo. By(34.81) and(34.86), K Frz F <'°> F="§' K="w=-anHence (a)becomes dzu Fr2 FE5§+u=—W=— s which isthedifferential equation, inpolar coordinates, oftheorbit ofapar- ticlesubject toacentral force. Fordifferent values oftheforce F,there willbedifferent orbits. Conversely, iftheequation r=r(9)inpolar coordinates oftheorbit ofaparticle isknown, (34.97) willgivethecen- tralforce Fwhich causes theparticle tomove inthisorbit. Alloneneed Exercise 34M MISCELLANEOUS PROBLEMS Lmnmo roSYSTEMS 497 dotoﬁnd Fistosubstitute in(34.97) thevalues ofuanddzu/d02 and solve forF.(Remember u=1/r.) Usetheabove facts and(34.97) tosolve thefollowing problems. Assume inallcases that Fisacentral force and that theﬁxed point Otoward which theforce actsisattheorigin. 3.Theorbit ofaparticle isanellipse withonefocus attheorigin. Show thatthe force Fobeys theinverse square law. (Nora. Thisassertion hasalready been proved inthislesson; seeproof ofNewton’s inverse square lawfrom Kepler’s laws.) Hint. Theequation oftheorbit inpolar coordinates is ,=__A___.1—|—ecos0 Therefore 2 u= i %=-—-f{cos0. Substitute thelasttwovalues in(34.97). Solve forF.Remember thatm,h, andAareconstants. Ans. F=-mhz/Ar”. 4.Theorbit ofaparticle isacircle, with theorigin apoint ofthecircumference. Show that theforce Fisinversely proportional totheﬁfth power ofthe distance oftheparticle from theorigin. Hint. Theequation oftheorbit in polar coordinates isr=2acos0,where aistheradius ofthecircle. There- foresec0=2a/r =2au. Alsomake useofthefactthatsec20=1+tan?0. Ans. F=—8a2h2m/r5. 5.Theorbit ofaparticle isanellipse with theorigin atthecenter oftheellipse. Show thattheforce Fisproportional tothedistance oftheparticle from the center, seeLesson 34E. Hint. Theequation oftheorbit inpolar coordinates is2 r2= ﬁ .Follow suggestions inproblem 3. Ans. F=-mh2(l -e2)r/A4. 6.Theorbit ofaparticle isthespiral r=e'.Find theforce F. Ans. F=-—2mh2/r3. 7.Theorbit ofaparticle isthelemniscate r2=a2cos20.Find theforce F. Ans. F=—3mh2a4/r7. 8.Theorbit ofaparticle isthecardioid r=a(1+cos0).Find theforce F. Ans. F=—3mh2a/r4. 9.Theorbit ofaparticle isacircle with center attheorigin. Find theforce F. Hint. Theequation oftheorbit isr=-=a.Ans. F=—mh2/a3. EXERCISE 34M MISCELLANEOUS TYPES OF PROBLEMS LEADING TO SYSTEMS OF EQUATIONS 1.Aparticle moves inaplane. Ifthe2:andycomponents ofitsvelocity are equal respectively totheyand:2:coordinates ofitsposition, ﬁndtheequation ofitspath. 2.(a)Solve problem 1,iftheword velocity ischanged toacceleration. (b)Find theequation ofitspath ifinitially theparticle isattheorigin andhasa velocity of15ft/sec inadirection whose slope is2. 498 Pitonm-ms: Srsrnms. SPECIAL 2NDORDER EQUATIONS Chapter 8 3.Solve thefollowing system ofdifferential equations. They areused incertain problems ofelectron motion. (122: at .121; da:WIW-j-6.HE-—6E, mw-——eHE—0, where m=themass oftheelectron, e=thecharge oftheelectron, H=theintensity ofthemagnetic ﬁeld, E=theintensity oftheelectric ﬁeld. Assume thatinitially theelectron isattheorigin anditsvelocity iszero. InLesson 30M-C, wediscussed theproblem ofawire twisted byro- tating abobatoneend, where theresulting torque ormoment offorce was proportional totheangle oftwist, see(30.63), (30.64), and (30.67). Forconvenience werecopy (30.67). 2 ($4.971) 1%=-to, where kisaproportionality constant, called thetorsional stiffness constant, and 0istheangle through which thewire hastwisted from anequilibrium position. Make useof(34.971) tosolve problems 4-6. 4.Three disks areconnected byshafts. Themoment ofinertia ofthetwoend disks isI,ofthemiddle disk2I,Fig.34.98. Thetorsional stiffness constant ofeach ofthetwoshafts connecting thethree disks isk.Ifatorque 2T0sinwt 1. 1. 21 0, 0, 0, Figure 34.98 isapplied tothecenter disk, ﬁndtheangular motion ofthedisks. Assume no resistance andthat initially thedisks areatrestandtheshafts areintheir untwisted equilibrium position. Hint. Call01theangular displacement from equilibrium attime tofanenddiskand02theangular displacement from equilibrium ofthemiddle disk. Iftheenddisks areconsidered ﬁxed atthat instant, then attime t,theshafts connecting them tothemiddle diskhave twisted through anangle, 02-01.Hence therestoring torque acting onthe middle diskis2k(02 -01).Ifthemiddle diskisconsidered asﬁxed, then the shaft connecting ittoanenddiskhastwisted through anangle, —(02 —01). Hence therestoring torque acting onanenddiskisk(01 -02).Now make useof(34.971) taking intoaccount theapplied torque acting onthecenter disk. Thedifferential equations canbefound intheanswer section. 5.Three disks andadriving wheel areconnected byshafts, Fig.34.99. Theright enddiskisfree. Each diskhasthesame moment ofinertia I,andthethree shafts have thesame torsional stiffness constant, k.Setupthesystem of Exercise 34-M MISCELLANEOUS PROBLEMS LEADING 'roSYSTEMS 499 6 Mi-1 3 4 5.differential equations fortheangular motion 01(t), 02(t), 03(t) ofthethree disks from their equilibrium positions duetoanangular motion 0ofthedriv- ingwheel. Assume that initially thedisks areatrestandtheshafts arein their untwisted equilibrium position. (Seehintgiven inproblem 4.) ,k( 1. 5 k_(- O 1 1 1 9 0, 0, \o, Figure34-.99 Solve problem 5ifthere isalsoaresisting torque operating onthedisks proportional totheﬁrstpower oftheir angular velocities. Thisproportionality constant iscalled thetorsional resistance constant. Assume thetorsional resistance constant foreach diskisR. ANSWERS 34M =c1e‘+c2e"‘, y=cie‘-cze"'; 2:2—yz=c. =c1e‘+cge"‘ +03cost+c4sint, =c1e‘+cge-‘ ——03cost -c4sint. (b)x=341(e' —e")+-gsin t,y=341(e‘ —e")-§-sin t. :2:=E—m(1—cosH§t). y=Et~@sine—IiteH2 m H eH2 m' They aretheparametric equations ofacycloid. Forthedeﬁnition ofacycloid, seeExercise 28C,34. 2{£8\_/ ‘£8 I9% =—k(01 —02),foreach enddisk, 2 2Iid}? =—2k(92 —01)+2T0sinwt,forthemiddle disk. Solutions are, 01=C1+at+63sinx/2k/I z+6.cosx/2k/I t+ 2. 02=C1+at-c3.-.5.“/21¢/I: -6.cos\/21¢/1 t+ Initial conditions aret =0,01=0,02=0,61=0,92=0. a’oI-,7‘=—tan—0)—kw.~0.). .120IY?=-M02 -0,)-mo,-03), .120I-,-5,3=—tan-92); 500 Pnosmms: SYSTEMS. SPECIAL 2NDOnnnn EQUATIONS Chapter 8 or (ID2 +2k)01 —k0-2=k0(t), -—k01 +(ID2 +2lc)02 -I003=0, -k0; —|—(ID2 —|—lc)03 =0. Initial conditions att=0are0;=02=03=61=62=63=0. 6.Addtheterm —Rd0;/dt totheright sideoftheﬁrstequation of5;—Rd0g/dt totheright sideofthesecond equation; —Rd03/dt totheright sideofthe third equation. LESSON 35. Special Types ofSecond Order Linear and Nonlinear Differential Equations Solvable byReduction toa System ofTwo First Order Equations. Inprevious lessons, weoutlined standard methods bywhich solutions interms ofelementary functions could beobtained forcertain types ofﬁrst order differential equations andforlinear differential equations with con- stant coeﬁicients oforder n>1.Forthenonlinear differential equation oforder n>1and forlinear equations with nonconstant coeﬁicients of order n>1,such standard methods areavailable only iftheequation belongs tooneofseveral special kinds. InLesson 23andExercise 23,18 and22,wediscussed such special kinds oflinear equations with noncon- stant coefficients. Inthislesson wediscuss three special types ofnonlinear equations oforder twoforwhich astandard method ofsolution isavail- able. (See also Exercise 35,23foranadditional type.) These same methods can, ofcourse, alsobeused iftheequation islinear. LESSON 35A. Solution ofaSecond Order Nonlinear Differential Equation inWhich y’and theIndependent Variable xAreAbsent. Equations ofthistype that weshall consider willbethose which canbe written intheform (35-1) y”=f(t), where f(y)isdeﬁned onaninterval I:a§y§b.Note that y’and:1: aremissing. Thesubstitution u=y’,u’=y”willchange (35.1) intothe equivalent ﬁrst order system <35-11> jg=1»,‘$3=re). The second equation in(35.11) canbesolved foruasfollows. Multiply itby2utoobtain 2uu’ =2uf(y). Replace 2uu’ byitsequal (d/d:e)(u2), andubydy/dx. Hence (35.12) %cu”)=2j%r<y>. Lesson 35A y’ANDINDEPENDENT VARIABLE :2:ABSENT 501 Therefore (35.13) d(u2) =2f(y) dy. Integration of(35.13) gives (35.14) u2=2/f(y) dy=F(y) +cl. Substituting (35.14) intheﬁrstequation of(35.11), wehave (35.15) %=:h\/F(?/) +cl. If\/F(y) +c1aé0,weobtain from (35.15) 1 Remark. Wedonotwish toimply that (35.16) willalways bein- tegrable interms ofelementary functions. Intheexample below, f(y) has been chosen carefully sothat itwillbe.Ingeneral, itwillnotbe. Example 35.17. Solve thenonlinear equation (a) 1/’=4u‘°, y¢0- Solution. Following themethod outlined above, wesubstitute u=y’, u’=y"in(a)toobtain theequivalent ﬁrst order system. d d _ 51=1».i=4y3. Multiplying thesecond equation in(b)by2u,there results d _ d _d(c) Zuﬁ =8y3u, (E012) =8y3d—Z., d(u’)=81/‘adv, 112=/'8y‘3dy =—4y" +C1,61>0, V011/2 ——4 2 ——2u==|=\/c —4y"2=:|=—————» y>—ory<—- ‘ 1/ \/T ~/T c c Substituting thislastvalue ofuintheﬁrst equation of(b),wehave dy(.1) 1”-- =dz.:l=\/clg,/2 —4 Integration of(d)now gives 1(9) =|=;;\/611/2-4=$+62, =l=\/611/2—4=¢1Z+61¢2> 2 —2cy2=(cx+cc)2+4, y>-—— y0. 1 1 12 \/5;, ‘\/E71 502 PROBLEMS! SYSTEMS. SPECIAL 2NDORDER EQUATIONS Chapter 8 LESSON 35B. Solution ofaSecond Order Nonlinear Differential Equation inWhich the Dependent Variable yIsAbsent. Equa- tions ofthistype that weshall consider willbethose which canbewritten intheform (35-2) y"=f(w.y')- Note that yismissing. The substitution u=y’,u’=y”will change (35.2) into theequivalent ﬁrst order system dy_ du_(35.21) E-u, E5-f(x,u). The second equation isnow aﬁrst order equation inuandhence may be solvable bythemethods ofChapter 2.Ifitisand itssolution isu=- u(x) +cl,then bytheﬁrstequation in(35.21), (35.22) y=/[u(:z:) +c1]dx +C2. Example 35.23. Solve thenonlinear equation (a) y”=w(y')’- Solution. Following themethod outlined above, wesubstitute u=y’, u’=y”in(a)toobtain theequivalent ﬁrst order system Q_ E_2(b) dx_u’ dx—W' Ifu#50,wecanwrite thesecond equation in(b)as d(0) 7’;=xdx. Itssolution isu=-2/(x2 +c),which wewrite as -2‘M—ﬁg :|:C12; with theunderstanding that theplus sign istobeused ifc>0;the minus sign ifc<0.Substituting (d)intheﬁrst equation of(b),wehave dy__ 2 __—-2dx (6) (it— :z:3=!=c12' dz/_:c2:hc12' Byintegrating (e),weobtain 1 1 —y=—-2 Arctan +C2]and y=—51-log +c2. Lesson 35C INDEPENDENT VARIABLE 1ABSENT 503 Ifc1=0,then by(e), (f) dy=—%da:, 1/=2-l-C’, ac;-$0. LESSON 35C. Solution ofaSecond Order Nonlinear Equation in Which the Independent Variable xIsAbsent. Equations ofthis type that weshall consider willbethose which canbewritten intheform (35.3) y”=f(y,y')- Note that2:ismissing. Ifweletu=y’,then d dd d(35.31) y"=I:=Ti;‘T;=ﬁn. Substituting these values in(35.3) willchange itinto theequivalent ﬁrst order system d a(35.32) $1:=u,ta;=j(y,u). Thesecond equation in(35.32) isnow aﬁrstorder equation inuand hence may besolvable bythemethods ofChapter 2.Ifitisanditssolu- tion isu=u(y) +0|,then bytheﬁrstequation in(35.32), (35.33) =[dz =x+Cg. Example 35.34. Solve thenonlinear equation (11) yy”=ya—|—(y’)’- Solution. Following themethod outlined above, wesubstitute u=y’, y"=ujig:in(a)toobtain theﬁrstorder system (b) ;‘§§j=u. uy%;=1/“+1.”. The second equation in(b)canbewritten as d 1 _(c) %;—§u=y2u 1,u¢0,y;-60, which isaBernoulli equation. Hence, following themethod outlined in Lesson 11D, wemultiply (c)by2utoobtain (d) 2ug5 —--E142 =2;/2, 504- Pnonm-ms: Srsraus. SPECIAL 2NDORDER EQUATIONS Chapter 8 which canbewritten as d 2(6) @012) __50-42) =23/2: u#0)y750: anequation linear inu’.Itssolution bythemethod ofLesson 11Bis (theintegrating factor iseI'“""’ =F2‘°"'=1/y’) 2 u=:l:yV2y+cl1 u9£0;y#0r2l/+c1>0- Substituting these values ofuintheﬁrstequation of(b),weobtain §2_ .i__ (g)dz-:|:y\/2y+c1, yx/gym-—:bda:, y¢0,2y+c1>0. The solution ofthesecond equation in(g),ifyaé0,23/+c1>0,is (h) %log i~— ==|=a:+c2,c1>0, 1 1/ c iArctan—-i= =b:c+c2, c1<0. V—c1 V——c1 Ifcl=0,then (f)becomes u==l=\/2 ya/2. Therefore by(b) (1)%=:l:\/2ya/2, iy-may =\/2d:c, =F2y'1'2 =\/5.5+ ¢, (\/ix —62)23/=4.2/¢0- No'rE. Thefunction y=Owhich wehadtodiscard inorder toobtain (h)and(i)alsosatisﬁes (a).Itisaparticular solution of(a)notobtain- able from thefamilies (h)or(i). EXERCISE 35 Solve each ofthefollowing differential equations. =211115 6-(1/+1)y”=3(u')"’-=lc. 7.P=—k/1'2. =(y')2 —1. (SeeComment 34.8961.) y”—|— my’=1. 8.y"=§-lcyz. (Note thattheequation islinear.) 9.y”=2ky3. 5.xy”—y’=2:2. 10.yy”+ (y')2 —y’=0. (Note thattheequation islinear.) 1.’kLi‘9°5°1"aw:<e“<e\=‘§‘\ ll.F=175—E. (SeeComment 34.91.) 12-yy”+(y’)3—(1/)2=0-13.yy”—3(y')2 =0. 14-(1+r’)y" +(1/)2+1=0-15.(1+:c2)y” +2:c(y’ —|—1)=0.(Note thattheequation islinear.) Lesson 35—Exercise 505 Find aparticular solution ofeach ofthefollowing differential equations satisfying thegiven initial conditions. 16-(1'),-l-1)y" =3(1/)2, 1/(1)=0.1/(1) =—t-17-y”=1/'6", 1/(3) =0.1!/(3) =1- 18-yI=2:1/y’. 21(0)=1,1/(0)=2-19.2y’=e",y(0)=0,y’(0)=1.20.xzy” —|—ray’=1,y(1) =1,y'(1) =2.(Note thattheequation islinear.) 21.ray”——y’=2:2, y(1) =0,1/(1) =—1. (Note thattheequation islinear.) 22.r=-$.40) =1,no)=0,r(0)=-5.Seeproblem 1above. Hm». After ﬁmging tasafunction ofrinintegral form, substitute either r=coszu orr=u. 23.Adifferential equation issaid tobehomogeneous inx,ifbyarbitrarily assigning degree ntox"anddegree —ntoy<"),each nonzero term ofthe equation isofthesame degree. Incomputing thedegree ofaterm, thede- greeofeach ofitsmembers isadded. Forexample, 2:2isofdegree 2;yisof degree zero; y”isofdegree -2;:c2yy” isofdegree zero. Equations homoge- neous in:2:may besolvable bythesubstitutions, seealsoExercise 23,18, :c=e", u=logx, a-1 Q_1it_ad:c_a:du’ a¢=‘¢2 aw <11.’ etc.,andthen making useofoneofthemethods ofLesson 35.Verify that each ofthefollowing nonlinear differential equations ishomogeneous in2: andsolve. (a)12/y”—2w(z/)2 +z/y’=0-(b)I1/:1/"+1(z/)2 —111/’=0-(c)11/y"——2¢(y')’ +(1/+1):/’=0- ANSWERS 35 -1/=61ta11[¢1(r¢+ 62)]- 1 2.011/2 =(cm:+c1c2)2 +k. 3.cly=sinh(c1a: —|—C2). 4__(l0gw)2.y-gd-c1loga:+cg. 3 5.y=%—+c1a:2+cg. 6.(y+1)"2 =61St—l— 02. = lr . =27.tI if+CITdr+C2.Toevaluate theintegral, letru.See22. 8.=fi~+. dy ~/It/*+¢. C2 506 PROBLEMS: SYSTEMS. SPECIAL 2NDORDER Equxrrons Chapter 8 10-1=z/+c1l0s(z/ —c1)+c2;y =@- 1 lc 2 —11.t=cg=1:c—1\/c1r2+2kr ——hz:1;qTmlog[r+£+‘i ]- 12.y=a:+c1log(cgy). 13.y‘2=c1:2:+ cg. 14-.clzy =(012+1)log(c1x—|—1)-c1a:+ cg. 15.y=c1Arctana: —-1+cg. 16.(y—|— 1)'2 =2:. 17.y=—log (4——2:). 1r18.y=tan<a:—|—Z)- v/2_ 2_19.e -——-——2__x 1 2 20.y=(l2i)+2logx+ 1. I3 2 22l.y=-§——:c +§- 22.k=Qc1=—1,t=[1=!=‘l%dr ==F(Arccos\/;+\/r(1—r)). l No'rE. Since thelimits ofintegration arefrom 1torandr<1,useofthe plussignintheintegrand willgiveanegative time; useoftheminus sign willgiveapositive time. 23.(a)y'1=c1loga:+ cg. (b)y2=c1z2—|— cg. (c)2Arctan(cly) =c1loga:+cg. LESSON 36. Problems Giving Rise toSpecial Types of Second Order Nonlinear Equations. LESSON 36A. The Suspension Cable. Acable, chain, string, or similar object supported attwoends iscalled asuspension cable. It may support aload attached toitasinthecase ofabridge, oritmay hang under itsown weight. Weconsider thelatter possibility ﬁrst. We shall determine forittheshape ofthecurve that thecable assumes. InFig.36.1, wehave drawn acable, assumed tobeperfectly ﬂexible and inextensible, supported attwo ends, A,B,and hanging under its own weight. Whether thecable willhave theappearance shown inFig. 36.1(a) orinFig. 36.1(b) willdepend onthelength softhecable relative tothelength ABbetween thepoints ofsupport. Weconsider themore general case shown inFig. 36.1(a), where thecable atitslowest point does notnecessarily have ahorizontal tangent asitdoes inFig. 36.1(b). LetP(z,y) andP1(:z: —|—Arc,y—|—Ay)betwoneighboring points onthe curve AB, and callAsthelength ofthearcbetween them. The forces acting onthispiece ofcable, considered asisolated from therestofthe system are[refer toFig.36.1(a)]: Lesson 36A THE SUSPENSION CABLE 507 TY B 1 Y 1+ ,,P1(x+Ax, L Ay+L)‘ T,sin0, A AQX Tlcos9| P 1 \ (xy) wAs A \H \.(x..5) ATs“” (*0,Ii)\wI’ T T-*' 0=0 Tl’T Tcos 0 l}? (x010) X (x010) X (<1) (5) Figure 36.1 1.Atension TatPdirected along thetangent tothecurve. 2.Atension T1atP1alsodirected along thetangent tothecurve butin adirection opposite tothat ofT. 3.Aforce duetotheweight ofthecable oflength As,directed downward. Ifweassume ahomogeneous cable whose weight isuniformly dis- tributed andiswpounds perfoot, thentheweight ofthecable oflength AsiswAs. Byhypothesis, thecable isperfectly ﬂexible and inextensible. And since itisinequilibrium, theportion PP1 ofthecable will retain its shape under theaction ofthethree forces 1,2,3above, even ifitwere cutatPand P1,just asifitwere arigid body. But thecondition for static equilibrium ofthepiece ofcable PP; isthat thealgebraic sum of thecomponents ofthesystem offorces acting along anarbitrary direction bezero. Hence equating tozero, thealgebraic sum ofthehorizontal com- ponents ofthethree forces 1,2,3above, weobtain [keep referring toFig. 36.1(a)]. (36.11) T1cos01-—Tcos 0=0, where 0and01aretheangles shown inFig.36.1(a). Equating tozero thevertical components ofthese forces, weobtain (36.12) T1sin01—Tsin 0—wAs =O. The change inthehorizontal tension (Tcos0),is (36.13) A(Tcos0)=T1cos01—Tcos0; 508 PROBLEMSI Srsrnms. SPECIAL 2NDORDER Equxrrons Chapter 8 thechange inthevertical tension Tsin0,is (36.14) A(Tsin0)=T1sin01—Tsin0. Substituting (36.13) in(36.11) and (36.14) in(36.12), weobtain respec- tively (36.15) A(Tcos0)=0, A(Tsin0)=wAs. Dividing each oftheequations in(36.15) byAxandletting Ax—>0, there results thesystem ofequations d d . ds(36.16) a(Tcos0)-0, ‘E(Tsin0)-wa- From theﬁrst equation in(36.16), weﬁndthat (36.17) Tcos 0=H, where Hisaconstant. Since Tcos0isaconstant andequal toH,we may write thesecond equation in(36.16) as dTsin0 wds <36-18> a -17a’ which simpliﬁes to (36.19) %(tano)=1.gs. where (36.2) k= Buttan0istheslope y’ofthecurve ofthecable atP.Hence (36.21) tano=$- Substituting (36.21) in(36.19) andrecalling thatthedifferential dsofthe arcPP1 ofthecable isgiven bytheformula ds=\/dz” +dy2sothat 3=1+(dy/dw)’. weobtain dzy_(36.22) 53-Icx/1 +(dy/d:z:)2. This isanonlinear second order equation ofthetype discussed inLesson 35Bwith ymissing. Following themethod outlined there, weletu=y’, Lesson 36A THE SUSPENSION CABLE 509 u’=y"andthus obtain from (36.22), thesystem dy__ du_(36.23) E-u, 5-k\/1 +11.2, Bythemethod ofLesson 6C,andwith theconstant ofintegration taken tobekxo,thesolution ofthesecond equation in(36.23) is (36.24) log(u+\/1-1-142)=kw—lcxo, u+\/U-12? =6""-‘°>. The second equation in(36.24) canbesolved forubytheusual algebraic means. Asimpler method, however, isthefollowing. Take thereciprocal ofeach side ofthesecond equation in(36.24) and rationalize thede- nominator oftheresulting leftside. There results (36.25) u-\/TE =-6-'=<'-'2’. Adding (36.25) andthesecond equation in(36.24), weobtain (36.26) u=§[e'°(’_‘°) -e-'"<‘-">1. By(18.9), wecanwrite (36.26) as (36.261) u=sinh[lc(x-—160)]. Replacing uin(36.26l) byitsvalue asgiven in(36.23) andlcbyitsvalue asgiven in(36.2), weobtain (36.27) y’=sinh (6-1.0)]. Integration of(36.27) gives (36.28) y=gcosh (:1:—-a:o)]+c, which istheequation ofthecurve ofahanging cable supported attwo points PandP1.Since thesame form ofequation (36.28) results ifthe points PandP1aretaken anywhere onAB, (36.28) istheequation ofa hanging cable supported attwoends A,B. Keep inmind that asPand P1shift, thevalues ofHandxin(36.28) change. There arethree constants inoursolution (36.28), H,2:0,andc.From (36.27), weseethat y’=Owhen a:=1:0,(remember sinh0=0).This means that 2:0isthe:1:coordinate ofthat point onthecurve (36.28) where itsslope iszero. This point is,therefore, aminimum point ofthecurve. Ifwhen :0=:60,wenow choose our:2:axis sothat yhasthevalue H/w, then by(36.28), c=0,(remember cosh0 =1).InFig. 36.1(a), we have shown thepoint (:30,H/w). Ifwenow choose ouryaxis sothat 510 PRoBLEMs: Srsrnns. SPECIAL 2NDORDER Eqtwrrons Chapter 8 Y YB B \~~ A A x \~ 0 + H H=T + (o,U) (0,3;) __. (0,0) X (0,0) X (11) (bl Figure 36.29 itgoes through thepoint (xo,H/w), then 2:0=0.With reference to these new axes, i.e.,where theyaxis goes through thelowest point of thecurve (36.28), and the2:axis isH/wunits below this lowest point, seeFig.36.29(a), (36.28) simpliﬁes to H(36.3) y=-5cosh 2:), and(36.27) simpliﬁes to (36.31) y’=sinh 6). Acurve whose equation hastheform (36.3) iscalled acatenary. Finally toﬁnd Hwemake useofthefact that thelength ABofthe cable iss.Letabethe2:coordinate ofthepoint A;bthe:1:coordinate of thepoint B.Then from thecalculus, weknow 6 (36.32) s=/V1+(1/)2 dz.G By(36.31), wecanwrite (36.32) as b (36321) 8=/,/1+sinh’ 6)dz 6 =/L coshggdx —€sinh-‘P-sub_w H Lesson 36A THE SUSPENSION CABLE 511 Therefore (36.33) 8=%[sinh 6)-sinh In(36.33), s,w,a,baregiven constants. This equation therefore deter- mines H. Comment 36.34. Letusassume thatsissuﬁiciently longsothatthe cable hangs asinFig. 36.1(b). Hence thepoint (:c0,H/w) atwhich the curve hasahorizontal tangent isapoint ofthecable. Thetension Tof thecable atthat point istherefore alsohorizontal. Butwhen thetension Tishorizontal, theangle 0inFig. 36.1(a) iszero. And when 0=0,we seefrom (36.17) thatT=H.Hence forthisspecial case,Histhetension ofthecable atitslowest point, seeFig. 36.29(b). And if2:0=0,thelength sofacatemrry from itslowest point (0,H/w) toanypoint P(z,y) onit,for thisspecial case, istherefore [in(36.33) take a=0,b=2:], H.w(36.35) s=5sinhif2:. Example 36.36. Acable 50feet long, weighing 5lb/ft, hangs under itsown weight between twosupports 30feetapart. Find: 1.Theequation ofthecurve. 2.The tension atitslowest point. 3.The sag. 15 <15.» 25 T"+-—> H }y=_g (0.0) Figure 36.37 Solution (Fig. 36.37). Ifwechoose theorigin sothat theyaxisgoes through thelowest point (0,H/w) ofthecable, then by(36.3) theequation ofthecurve formed bythecable is H(a) y=5cosh%2:; andby(36.35), thelength ofthecable from 2:=0to:1:=:1:is (b) s=gsinh g2:. 512 PROBLEMS1 Sxsr-Ens. SPECIAL 2NDORDER EQUATIONS Chapter 8 Inserting in(b)thegiven conditions, w=5,2:=15,s=25,weﬁnd H.5 125 .75(C) 25=‘-E-81Ill1ﬁl.5, Using atable ofvalues ofsinh :z:,theapproximate value ofHwhich will satisfy (c)is (6) H=46.316. This isthetension ofthecable atitslowest point. When H=40.8and w=5,weobtain from (a) (e) y=8.16cosh . which istheequation ofthecurve ofthecable. From itweﬁnd that when 2;=15, (f) y=8.16663.11% =s.16(3.22) =26.23ft, andthat when :1:=0,y=8.16. Hence thesagofthecable is26.28 — 8.16 =18.12 feet. Comment 36.38. Ifthespeciﬁc weight ofabody isnotuniform butis afunction of2:,then itstotal weight Wfrom 2:=0to1:=2:isgiven by I (36.4) W=g/L)f(z)dx, where f(z) isthespeciﬁc weight ofthebody, i.e.,itsweight perunit length. Forexample iff(z) =50+:0pounds, then when 2:=10feet, thespeciﬁc weight ofthebody atthat point is60lb/ft; when as=10.1 feetthespeciﬁc weight atthat point is60.1lb/ft. By(36.4), itstotal weight Wfrom :1:=0tosayx=20feetis 2° $220 W=/6 (50+:11)dx=[5032 + =1200 lb. o o Assume thataperfectly ﬂexible cable ofnegligible weight supports, by means ofvertical rods, ahorizontal load such asabridge, whose speciﬁc weight isf(z). The rods areequally spaced and close enough toform a continuous system. The weights oftherods arealsonegligible. Wewish toﬁndtheequation ofthecurve formed bythecable. Following thesteps used toarrive attheequation ofthehanging cable with noload attached, weﬁndthat nochanges occur until wereach (36.12). Inthisequation, wAsmust bereplaced, seeFig.36.41, byf(2:)A:c, where f(z) isthespeciﬁc weight ofthehorizontal load measured from the origin, which istaken tobethelowest point ofthecurve. Continuing from there on,andreplacing wAsbyf(:z:)Ax, weobtain eventually, in Lesson 36A THE SUSPENSION CABLE 513 As H' ‘(0,0) T4g1i ‘Ax T+i> f(1) Figure 36.41 place of(36.18), dTsin0 f(z) d(tan 0)=ﬁg.)dx, d(g)=@1936. Integrating thelastequation andmaking useofthefactthat when x=0, dy/dx =0,wehave3 (36.43) %=Ii,lo/(6)dz. Asecond initial condition is2:=0,y=O.ByComment 36.34, theH inthisequation isthetension atthelowest point ofthecable. Example 36.44. Aperfectly ﬂexible suspension cable, attached totwo towers atthesame level, 100feetapart, supports abridge (assumed rigid) bymeans ofvertical rods connecting cable and bridge. The rods are equally spaced and close enough toform anapproximately continuous system. The weights ofcable androds arenegligible incomparison with theweight ofthebridge whose speciﬁc weight isgiven byf(z) =10+ :02/50 lb/ft. Ifthelength ofthecable issuch that itssagis25feetwhen thebridge ishorizontal, ﬁndtheequation ofthecurve inwhich thecable hangs andthetension atitslowest point. Assume thattheorigin istaken atthelowest point ofthecable. Solution. SeeFig. 36.441. By(36.43) andthegiven speciﬁc weight f(z) =10—|—2:2/50, weobtain Q_f £2)_ L3. (a) Hdx- 010+50 dx-10a:+150 Integration of(a)gives 4 (b) Hy=52:2—|—%—|—cl. 514- PRoELEms: Srsrnns. SPECIAL 2NDORDER EQUATIONS Chapter 8 Thesubstitution in(b)oftheinitial condition 2:=0,y=0gives c1=0; thesubstitution oftheinitial condition :0=50,y=25,gives 4 _ 2EL _(0) 25H —5(50) +600, H—917, (076) (50.0)(50,25) Figure 36.441 Which isthetension ofthecable atitslowest point. Hence (b)becomes I 2 $4 (‘D 1/=tn5“+as' which istheequation ofthecu1've ofthecable. EXERCISE 361 1.Acable oflength 23anduniform weight wlb/ft hangs from twosupports onthesame level, Fig.36.45. Thesupports are2Lftapart with L<s. Thetension atthelowest point ofthecable 1SH. Y Lii d H1/ax’+dyz T dy dx 1++—-> Ew (0.0)X Figure 36.45 (a)Show thatthesagdisgiven by H wL(36.46) d=:0‘(0OSl1 TI" '-1)' Hint. Use(36.3). Thesagisthedifference invalues when 2:=Land :v=0. Lesson 36A—Exercise 515 (b)Show thatthelength softhecable from 2:=0to2:=Lis (3647) s=gsinhLI?- Hint. Use(36.35) with x=L. (c)Show thatthetension Tatanypoint ofthecable isgiven by (3643) T=Hcosh%x=wy. \/2 2 Hint.By(36.17) (BeealsoFig.36.46) T=Hsec0-H%'- = H\/1+ (dy/dz)”. Replace 1/’byitsvalue asgiven in(36.31). Remark. Since cosh 2:isanincreasing function of:r,thetension isamaximum when :2:islargest, i.e.,atapoint ofsupport. (d)Show thatthetotal weight ofthecable isgiven by (3649) W=2Hsinh(wL/H). Hint. Use(36.47) which gives thelength ofcable from 2:=0,tox=L. (e)Show thatthehorizontal tension Hisgiven by (36491) e-L/H=(4+d)/(8-a), wL H' lb'Hint. In(36.46) and (36.47), change thehyperbolic functions totheir exponential forms, see(18.9) and(18.91), andthen show thattheright side oftheﬁrstequation in(36.491) simpliﬁes toe"'L”'. (f)Show thatthehorizontal tension Hisalsogiven by (36492) H=w(s2 —dz)/2d. Hint. Use(36.46), (36.47), andthefactthatcosh’ x—-sinh” 2:=1. Atelephone wire, weighing 0.04 lb/ft, isattached topoles atintervals of 300ft.Thesagatthecenter is7.5ft. (a)Find thetension atthelowest point ofthecurve. Hint. Use(36.46). Remember inthisformula, thedistance between supports is2L. (b)What isthelength ofthewirebetween poles? Hint. Use(36.47). Re- member inthisformula thelength ofwireis2s.Use(36.491) or(36.492) tocheck your result. (c)What isthetension atthesupports? Hint. Use(36.48) with 2:=150. (d)Find theequation ofthecurve. Hint. Use(36.3). Acable 100ftlong, weighing 41b/ft, hangs under itsownweight between twosupports onthesame level and80ftapart. Find: (a)Theequation ofthecurve. (b)Thetension ofthecable atitslowest point; atapoint ofthecable mid- waybetween thelowest point andapoint ofsupport; atapoint midway horizontally between thelowest point andapoint ofsupport; atapoint ofsupport. (c)Thesag. (d)Theslope ofthecurve atapoint ofsupport. Forhints, seeanswer section. 516 PRoELEMs: Srsrans. SPECIAL 2NDORDER Equxrrons Chapter 8 4. 5 6 7 8. 9. 10.Acable hangs under itsownweight between twosupports onthesame level. Theslope ofthecurve formed bythecable atapoint ofsupport is0.2013. Itssagatthecenter ofthecable is12ft.Find: (a)Thedistance between supports. (b)Thelength ofthecable. (c)Theequation ofthecurve. Hint. Use(36.31) toﬁndwL/H andsolve forH/winterms ofL.Then use (36.46), (36.47), (36.3) inthatorder. Achain 117.5 ftlonghangs under itsownweight between twosupports on thesame level 100ftapart. (a)Find thesagofthechain. Hint. Use(36.47) toﬁndH/w. Then use (36.46). (b)Find theequation ofthecurve. (c)Find themaximum tension ifthechain weighs 2lb/ft. Hint. Seere- mark after (36.48). Use(36.48). Achain hangs under itsownweight between twosupports onthesame level 100ftapart. Itssagis7.55ft. (a)Find thelength ofthechain. Hint. Solve (36.46) forH/w.Then use (36.47). (b)Find theequation ofthecurve. Acable hangs under itsownweight between twosupports onthesame level 50ftapart. Theslope ofthecurve formed bythecable atapoint ofsupport is0.5211. Find: (a)Thelength ofthecable. (b)Thesag. (c)Theequation ofthecurve. (d)Thetension ofthecable atitslowest point ifitweighs 0.5lb/ft, i.e., ﬁndH. Hint. Use(36.31) toﬁndw/H. Then use(36.47), (36.46), (36.3) inthat order. Acable, 100ftlong, hangs under itsownweight between twosupports on thesame level. Itssagatthecenter is10ft. (a)Find thedistance between thesupports. (b)Find theequation ofthecurve. Hint. Use(36.491) toﬁndwL/H. Then use(36.46) or(36.47). Acable 200ftlonghangs under itsownweight between twosupports onthe same level. Itssagatthecenter is20ft.Itsmaximum tension, seeremark after (36.48), is60lb.Find itsweight wperfoot. Forhints, seeanswer section. Start with thegeneral equation ofthehanging cable asgiven in(36.28). Itcontains three constants, 2:0,c,H.Bychoosing ourorigin inaspecial way, wewere abletomake :00=0,c=0.Hwasthen determined byusing thegiven length softhecable. This timeletustaketheorigin atthelower point Aofthehanging cable andcall(x1,y1) thecoordinates ofthehigher point B,seeFig.36.493. Show thatthevalues ofthethree constants canbe Lesson 36A—Exercis6 517 B6”1..71) y=-licosh -"l(x—x) +c .4(o,o) “’ [H 0] 1+,H ——> Figure 36.493 obtained from theequations H(A) 0=;0-oosh%o.,+o, (36494) y-1=£5cosh (:01—xo)]—|—c, :1 , H.w .w(B) s=/ \/1+(y)2d:c=—[sinhﬁ(a:1—xo)+smhﬁx0]0 w _2H .wan w(:r1 —22:0)—w[sinh 2Hcosh 2H Bysubtracting (A)from (36494), show that H _ (36495) 1/1=I;[cosh9% -cosh =if[sinh sinh“’(“‘2;,2”]- Finally show by(B)and(36.495)—square both equations andthen sub- tract thesecond from theﬁrst-— s2—- 2-5-Iifsinhz 91.M 102 2H (36496) sinh =%\/.2 -@112. Following thehintgiven after (36.48), show that thetension atanypoint ofthecable isgiven by (36.49?) T=Hcosh%(2:-2:0)=6(1,-c), where cisaconstant ofintegration whose value isgiven by(A)and(36494). 11.Using theequations found in10,solve thefollowing problem. SeeFig. 36.493. Acable oflength 100ftanduniform weight of2lb/ft hangs from twosupports AandB,whose horizontal distance is50ftapart. Support Bis20fthigher than A.Find: (a)Theequation ofthehanging cable. [Hint. Use(36.496), (36.495), and (36.494) inthat order.] (b)Thecoordinates ofitslowest point, i.e.,thepoint where y’=0. (c)Thetension atboth supports andatitslowest point. 518 PROBLEMS! Srsrsns. SPECIAL 2NDORDER Eouxrrons Chapter 8 Inproblems 12-16, assume theweights ofthecable androds areneg- ligible incomparison with theweight ofahorizontal supported load, that thecable isperfectly ﬂexible, that therods areequally spaced andclose enough toform acontinuous system, andthat theorigin isatthelowest point ofthecable. 12.Asuspension bridge is22:ftlong andissupported byvertical rods. The weight ofthebridge isuniformly distributed andiswlbperunithorizontal distance. Find theequation ofthecurve formed bythecable. Hint. In (36.43),f(z) =w. 13.Show thatthetension Tatanypoint ofthecable ofproblem 12isgiven by (36.498) T=\/H2 +w2:c2. See(36.48) andhintfollowing; herey’isobtained from problem 12. 14.Asuspension bridge is200ftlong. Thesupporting ends ofthecable are50ft above thebridge andthecenter ofthebridge is10ftbelow thecenter ofthe cable. Theweight ofthebridge isuniformly distributed andiswlbperunit horizontal distance. (a)Find thetension atthelowest point ofthecable, i.e.,ﬁndH.Hint. Inthesolution to12,usethefactthat2:=100,y=40. (b)Find theequation ofthecurve formed bythecable. (c)What isthetension atthesupports? Hint. See(36.498). (d)What istheslope ofthecable atthesupports? 15.Asuspension bridge is2Lftlong. Theweight ofthebridge isuniformly distributed andiswlb/ft. Thesupports ofthecable holding thebridge are aftabove theorigin. (a)Find thetension atitslowest point, i.e.,ﬁndH.Hint. Inthesolution toproblem 12,usethefactthatwhen x=L,y=a. (b)Find theequation ofthecurve. (c)What isthetension atthesupports? Hint. See(36.498). 16.Asuspension bridge is200ftlong. Thesagatitscenter is50ft.Thespeciﬁc weight ofthebridge isgiven byf(z) =100+2:2.Find: (a)Thetension atthelowest point, i.e.,ﬁndH. (b)Theequation ofthecurve formed bythecable. 17.Auniform, ﬂexible cable weighing wllb/ft supports ahorizontal bridge. Theweight ofthebridge isuniformly distributed andiswglb/ft. Theweight ofthesupporting rodsisnegligible. Show that thedifferential equation of thecurve ofthecable isgiven by 2 (36.499) Hg=w1\/1 +(dy/dz)? +wg. Hint. Combine equations (36.22) and(36.42) withla=w1/H andf(z) =wg. 18.Iftherods inproblem 17arenotofnegligible weight andweigh w3lb/ft, show thatthedifferential equation ofthecurve ofthecable isgiven by 2 <36.499i> H3-,}=w1\/1+(dy/dc)” +1112+way. Lesson 36A-Exercise 519 where yisthelength ofarod. Assume thattherodsaresoclose together as toform anapproximate continuous system. Hint. Theweight duetoarod ofwidth A2iswayA2. 19.Holes arebored intheends ofslender uniform rodsofvarying lengths and therodsarethen strung onacordofnegligible weight. Assume thattherods justtouch each other, thateach weighs wlb/sq ftandthattheir other ends Y ($1.7) 1+,0,a) —-> Ax (0-0) X wyAx Figure 66.4992 lieonahorizontal line,Fig.36.4992. Find theequation ofthecurve formed bythecord. Take theorigin ata.distance afeetbelow thelowest point of thecurve. Hint. In(36.15), replace wAs bywyA2.Initial conditions are 2=0,11=a,dy/d2 =0. 20.Foranonhomogeneous hanging cable whose weight isafunction ofits distance from itslowest point, show that thedifferential equation ofthe curve formed bythishanging cable is 2 (36.4993) Hg=p(s)\/1 +(dy/d2)2, where p(s)isthespeciﬁc weight ofthecable, i.e.,itsweight perunitlength atadistance sunits from thelowest point ofthecable. Hint. In(36.15), replace wbyp(s). 21.Ifin(36.4993), p(s) =aH/(a2 -—-22),andtheorigin istaken atthelowest point ofthecable sothat 2=0,y=0,dy/d2 =0,show that thecurve formed bythehanging cable willbeanareofthecircle. 12+(11-4)’=(12- 22.Anarched bridge istobebuilt ofstone ofuniform density. Theweight of thestone iswlbforeach square footoffacing. Thebridge issoconstructed thatateach point ofthearch theresultant tension duetotheweight ofthe stones above itactsinadirection tangent tothearch, seeFig.36.4994. Find theequation ofthearch. Take the2axisatthetopofthebridge, theyaxis through thehighest point ofthearch, theorigin asindicated intheﬁgure andthepositive direction downward. Hint. Since thetension ateach point actsinadirection tangent tothecurve, equations (36.15) apply with wAs replaced bywyAx. Initial conditions are2=0,y=h,dy/d2 =0. 520 Pnonmms: Svsuzms. Srncuu. 2NDOman EQUATIONS Chapter 8 23 2 3 4 5 6 7 8 9 11 12. 14. 15. 16. 19. 22. 23.Ax .xans l__I— III]- II-I— ll--Il————l———_ +IKIII |———_I-2]- *III‘_— wIIIIII IIIIIIIIIIIIIIIIII‘IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII18IIII8IIIIIIIIIIIIIIIIIII_—l——I -:=l_—__—___———l P----!==:::: --I " (OJ!) *1-|III ‘I =:.I' my)7Il yms wyAx Figure 36.4-994 Assume aroad isbuilt ontopofthemasonry ofproblem 22.Theweight oftheroad isdistributed uniformly andweighs kwlbperlinear foot. Find theequation ofthecurve. Hint. In(36.15) replace wAsbywg/Ax +kwAx. ANSWERS 36A (a)601b. (b)300.5 ft. (c)60.3lb. (d)y=1500 cosh (x/1500). (a)Use(36.47) toﬁndH.Then use(36.3). (b)Use(36.35) toﬁnd2: when s=25.Then use(36.48) with xequal successively to0;itsvalue when s=25;20;40. (0)Use(36.46). (d)Use(36.31) with 2:=40. (a)239.3 ft. (b)240.9 ft. (c)1/=598cosh (1:/598). (a)27.2ft. (b)y=50cosh (2:/50). (c)154.3 lb. (a)101.5 ft. (b)y=(500/3) cosh (32:/500). (a)52.11 ft. (b)6.38ft. (c)y=50cosh (1/50). (d)25lb. (a)97.3. (b)y=1120cosh 0.00832. Use(36492) toﬁndH/w.Use(36.47) toﬁndL.Use(36.48) with 2:=L (write H=w(H/w) andsolve forw);w=3/13 lb/ft. (a)H=23.35 approx., :00=22.6approx., c=—41.3 approx., 2:—22.61]—11.7 COSl'l —'41.3. (b)(22.4, —29.7). (c)82.7lbatA,122.7 lbatB,23.35 lbatitslowest point. 2 W8Nuy= ,aparabola. (a)H=125w. (b)y=12/250. (6)T=25w\/4-1.(d)Slope ==l=0.8. (a)H=wL2/2a. (b)y=G222/L2. (0)(wL/2a)\/L2 +4..¢.4 =(3/530,000) (so?+ - acoshVw/H x. Same asin19with areplaced byh.Since thetension ateach point ofthe bridge istangent tothecurve, itfollows that nomortar willbeneeded if thebridge hastheshape ofacatenary. y=(h-l-k)coshVw/Ha: —k.(E)H=530,000/s. (b)y y=g(,¢.7z.+e-./zm.) = Lesson 36B ASracnu. CENTRAL Foncn Pnonmm 521 LESSON 36B. ASpecial Central Force Problem. Example 36.5. Aparticle weighing 16pounds and10feetfrom a ﬁxed point isgiven avelocity of15ft/sec inadirection perpendicular to thexaxis. The particle isattracted totheﬁxed point byaforce Fwhose magnitude isinversely proportional tothecube ofitsdistance from the point. Iftheproportionality constant is8,ﬁndthedistance oftheparticle from theﬁxed point asafunction ofthetime. Solution. Letustake theorigin ofacoordinate system attheﬁxed point. Wehave already proved inLesson 34C that aparticle subject toa central force moves inaplane. Since theforce Facts only inadirection along theradius vector, thecomponents ofFare,with theproportionality constant equal to8, (a) F,= -—%=m.a,-, F,=0=ma,. Here themass m=Q= Hence by(34.21), (a)becomes (b) %(¢=_ra*)=-7%. %(21"0+r5)=0. After multiplication by2r,thesecond equation in(b)isequivalent to d (0) 3(T29)=0, from which weobtain T20 =C1. The initial conditions aret =0,r=10,v,=15.Therefore, bythesec- ondequation of(34.2), rd0/dt =15.Substituting these values in(d),we ﬁndcl=150. Hence (d)becomes 150 Intheﬁrstequation of(b)replace 0byitsvalue asgiven in(e).There results 1 150” 8 ..150”-1622,484 <9§(*"7?)= -F’ ’='".T'"=“r=»—' anequation which isofthetype discussed inLesson 35A. Assuggested there, weletu=dr/dt andobtain thesystem . . 22484<g> 1=u,u=;—.- Following theprocedure outlined inLesson 35A, wemultiply thesecond 522 Pnonmamsz Srsrnms. Sracmn 2m)Oannn Eqnurons Chapter 8 equation of(g)by2u. The remaining steps aregiven without further comment. . 22484 22484 .(h) 2uu=2%u=2—-aq-r, 22,4s4 22,484d(u2) =2'-;:?'dT, ‘I42=—T +C2. Att=0,r=10,u=1‘=0.Substituting these values inthelast equation of(h),weﬁndC2=22,484/100. Hence by(h)and(g), . . 1 1 1’-100(1) M2=T2=-22,484 - =22,4841 (iii 2_. 1mr2 dr"22,484(r2 -100)’ 101at=---—-a.149.9\/1'2 _100T Integration of(i)gives . 1(1) t='T4-®\/1'2-100-I-C3. Att=0,r=10.Therefore c3=0.Hence (j)becomes (k) (l4.99t)’ =r2—100 1'2=100+224.84? r= . Comment 36.51. Note thatforthisspecial central force problem, we were able toﬁndrasafunction oftrelatively easily. InComment 34.91, weremarked onthediﬁiculty inﬁnding rasafunction oftinthecase of aparticle moving subject totheinverse square law. EXERCISE 36B 1.Aparticle ofmass misattracted toaﬁxed point 0byaforce Fthatvaries inversely asthecube ofthedistance roftheparticle from O.Initially the particle isaunits from Oandisgiven avelocity 00inadirection perpendicular tothe:4:axis. Find: (a)tasafunction ofr. (b)rasafunction oft. (c)rasafunction of0,where 0isthepolar angle. Take mkfortheproportionality constant. Hint. In(c),after youobtain anequation forP2divide theequation by02=azvoz/r4 andnotethat1‘/9= dr/d0. Initial conditions aret =0,r=a,0=0,1‘=0,vo=ado. 2.Solve problem 1iftheparticle isrepelled from instead ofbeing attracted to0. Lesson 36C APnnsurr Pnonnsm 523 3.Aparticle ofmass misattracted toaﬁxed point Obyaforce Fthat varies inversely astheﬁfth power ofthedistance roftheparticle from O.Take kzmfortheproportionality constant. Initially theparticle isaunits from 0 andisgiven avelocity vo=k/(a2\/2) inadirection perpendicular tothe xaxis. Find rasafunction 0.SeealsoExercise 34H,4. Hint. You should obtain #2=—k2/(2a2r2) +I02/(2r4) and02=k2/(2a2r‘*). Divide 1‘2by92 andnotethat1‘/6=dr/d0. Initial conditions arel =0,r=a,0=0,1‘=0, vo=ado=k/(a2\/2). ANSWERS 36B 2 1.(a)la=E£—_-I; (T2'"“2), b)2_2+a2vo2-—kt2( T — a a2 7 \/a2v 2k 0_ (c)r-asec W0 0. Note that there arethree diﬁerent possible orbits depending onwhether k islessthan, equal toorgreater than a2v02. 2t2__ 112 (2__ 2) 2_ 2+¢l2vo2-l-ktz.-—J2vo2+k r a, r-a ——————-G2 , \/aﬁvoz +k0 ave ' 3.r=acos0,which istheequation ofacircle through theorigin.T=G580 LESSON 36C. APursuit Problem Leading toaSecond Order Nonlinear Differential Equation. InLesson 17,wesolved pursuit problems leading toaﬁrstorder differential equation. Weshall nowsolve apursuit problem which leads toasecond order nonlinear differential equation. Example 36.6. Aﬁghter pilot sights anenemy plane atadistance andstarts inpursuit, always keeping thenoseofhisplane inthedirection oftheenemy plane. Assume theenemy plane isﬂying inastraight line atV5miles/hour, andtheﬁghter plane isﬂying atVpmiles/hour. Find theequation ofthepath oftheﬁghter plane asseenbyanobserver onthe ground. Nora. This isthesame type ofproblem astherelative pursuit problem ofLesson 17B. Here, however, theorigin isﬁxed inspace and does notmove with theenemy plane. Solution (Fig. 36.61). Lettheorigin betheposition oftheenemy plane, andPo(:z:o,y0) betheposition oftheﬁghter plane, attime t=0. Letthe2:axisrepresent thelineofﬂight oftheenemy plane. Then atthe endoftseconds, theenemy plane hasgone adistance equal toVgtmiles along the1:axis, andtheﬁghter plane, which isnowatP(z,y), adistance 524 Pnonnamsz Svsrnus. Srncnu. 2NDOsman Eouurons Chapter 8 Vptmiles along thearcofhispath. Hence theﬁghter plane’s distance in time tisgiven by ll II (=1) Vrt=—L \/1+(dw/dz/)’dz/, l=—ViF£ \/1+(d$/dy)’dy- (The minus signisnecessary since yo>y.)AtPtheﬁghter’s direction istoward thebomber’s position (V;;t,0) andistangent tothecurve ofpur- Y Po(xc, yo) V1,‘t P(x,y) yl y 017 V5;l’-—x I (0,0) (VEt,0) X Figure 36.61 suitthere. Hence theslope ofthecurve atPisy’=tan0=—y/(Vgt —:0). Therefore atP dy__ y _dx__:c—VEt_ _i( __ (b)d:|:_x—VEt’ dy_ y'‘-1/E" ydy Equating thevalues oftin(a)and(b),weobtain 1 dc: 1 V_ __=___ _(e) VE<2 ydy) VF/LoV1+(dw/dy)2 dy Differentiation of(c)with respect toygives, by(9.15), (<1) 1("” dz” -,}lP~/i1+(aw/am TrTy_yﬁ—____ which simpliﬁes to,with VE/Vy replaced byanew constant k, 2 <e> yj—y’§=k\/___1+(dz/dy)", k= This equation isofthetype discussed inLesson 35B, with a:andyinter- changed. Following theprocedure outlined there, welet dz: du dzx “l “er2’at-as‘ Lesson 36C APnnsurr PROBLEM 525 Substituting these values in(e),weobtain thesystem ofﬁrstorder equa- tions dz du(8) 31-/-14. yjy-—k\/1+“? Thesolution ofthesecond equation in(g)is (1.) log(1.+WW) =klogy+log0,, which wecanwrite as (i) WW =cw"—u- Squaring (i)andsimplifying theresult, weobtain (i) 1=C322" —2m/"w Replacing ubyitsvalue in(f),wehave 2_.~.._1_1[1._£2].(k) dy— 2611/,‘ —2cly 61 Iflc=V5/Vp, isnotequal toone,i.e.,ifthevelocities ofﬁghter and bomber arenotthesame, then integration of(k)gives _1 01 n+1 1/Pk] __Vl (1) ’”"‘2ll.+1y +¢,(l@-1)+°” '°‘VF"‘1' Att=0,1:=xo,y=yo,dy/dz: =yo/xo, which implies, by(f),that u=dz/dy =xo/yo. Substituting these values in(1)and(k),weobtain respectively (m) z=li1_y'°+1_|_._ﬁ__+c °2k+1 ° c1(k—1) 2' E_C123/02» _1. (n) I/0 2611/0" With theaidofequations (m)and(n),wecanobtain, inaparticular problem, thevalues ofc1andC2.Thecurve ofpursuit isgiven by(l). Iflc=1,then by(k) dy_2 lg cly ' 1 21x= §(c12/5 —-alogg/+02). 526 Paonnamsz Srsraus. Srscuu. 2NDOsman EQUATIONS Chapter 8 EXERCISE 36C Adogseesacatandstarts inpursuit always running inthedirection ofthe cat. Thedog’s speed isvdft/sec, thecat's v,ft/sec. Assume that initially thedogisat(0,a), thecatat(0,0) andthat thecatruns straight foratree located onthe:2:axis. (a)Letk=v,/v4. Determine theequation ofthedog’s path ifIoaé1and ifk=1.Hint. Follow themethod ofExample 36.6. Initial conditions aret =0,2:=0,y=a,u=dz/dy =0. (b)Show that ifk<1,i.e.,ifthecat’s speed islessthan thedog’s speed, thecatwillreach thetreeinsafety provided thedistance ofthetreefrom the origin is<alc/ (1-I02). Hint. Show that when y=0, 2:=ak/(1 —k2)andtherefore that dogandcatwould meet atthis distance from theorigin. (c)When will thedog reach thecat, assuming thetree’s distance is >ak/(1 —k2)? (d)Showthatifk >1,x—><=° asy—>0andifk =1,z—><=0 asy—>0. Inboth cases, itisevident thatthedogcannot reach thecat. Aﬁghter plane, located at(0,0), sights abomber at(a,0) andstarts inpursuit always keeping thenose ofhisplane inthedirection ofthebomber. Assume thebomber isﬂying parallel totheyaxisatU3mi/hr, andtheﬁghter plane’s speed isatupmi/hr. (a)Letlc=03/Up. Find theequation oftheﬁghter’s path ifk751,andif k=1.Hint. Follow method ofExample 36.6. Initial conditions are t=0,2:=0,y =0,dy/dx=0. (b)Show thatifk=U3/Up <1,i.e.,ifthebomber’s speed islessthan the ﬁghter’s speed, theﬁghter will reach thebomber atx=a,1/= ak/(1-—I02). Hence show thattheﬁghter willreach thebomber intime t_ ak _ av; _ vB(1 "'52) W2—UB2 (c)Assume a=2mi,Up=300mi/hr, U3=200mi/hr. When willthe ﬁghter reach thebomber? Y y=f(x) P(X,Y) Y—y Q(r.y)F(X,Y)=0 (0,0) X Figure 36.7 Assume that thepursued object does notmove along oneoftheaxes, but along agiven curve F(X,Y)=0,andthatattime t,itsposition isP(X,Y), Fig.36.7. Thepursuer, asusual, always moves toward theposition Pofthe pursued object. Ifthepursuer's position attime tisQ(:r,y) andtheequation Lesson 36C—Exercise 527 ofhispath isy=f(z), wehave, fortheequation ofthelinePQ, (36.71) (X—a:)y' =Y—y. With Q(z,y) considered asﬁxed, thedifferential of(36.71) is (36.72) (X—2:)dy’—|—y’dX=dl’. Thedifferential of (36.73) F(X,Y) =0 is OF 8F(36.74) 5‘;dX—|—5?dY—0. Ifthespeed ofQisktimes thespeed ofP,wehave [speed =|v|=|ds/dt| = \/(dz/dt)? +(dy/dt)”, |ds|=\/.112+<32] (36.75) \/dz? —|—dy2=kvdX2+dY2. With thehelp ofequations (36.71) to(36.75), solve thefollowing problem. Apursued plane Pﬂiesinastraight linemaking anangle of45°with the:1: axis. Find theequation ofthepath ofthepursuing plane Qifitsspeed is twice thatofthepursued plane P.Choose theorigin onP’spath. Hint.Here (36.73) isX-—Y=0,(36.74) isdX—dY=0,(36.75) is (36.76) \/1+ (1/)2d:t =2\/1+ (dY/dX)2dX =2\/1+ 1dX =2\/2dX, (36.71) is(X—z)y' =X-y,(36.72) is(X——2:)dy'=(1—y’)dX. -_— I i i NowshowthatdX= andy'—1 =§%—1 =ft-%, __ __ I 2:——X= Hence dX= Substitute thisvalue ofdX in(36.76) toobtain 2 (36.71) <1/'—1>”~/1+<1/>2=2\/2<11—x)$7?- Tosolve (36.77), make thesubstitution atdy dy at a’y ah.u=y—:r, E=a'*1, ‘d-z'=1+E) WE-' There should result a2/ 2 <2’(36.79) 1+(1+%)=2\/211$. Since (36.79) does notcontain theindependent variable 1,themethod of Lesson 35Cisapplicable. ANSWERS 36C _a 1 yH1 1 y1-)‘ ak 1-‘*2’-§lm(;) +711 2‘K’ “*1- 528 Pnosnnms: Sxsrsus. Sracmn 2m)Oman EQUATIONS Chapter 8 12__ 2 :c=§[1-97-l-l——-alogﬂi k=1. (C)¢=.m =_i_sec_v¢(1—k)".12—v.’ _ 1-l-k It _ 1-1: 2'(“)”=il(:*(1-l)k)"a(‘i-air) l+1ikk2’ '°"1' 1/'il“‘°*..—i"..+Hl' (c)t=43.2sec. a/2 3.2\/2:0: =2c1\/1/—a:—\/2(y—:c) — +c2. LESSON 36D. Geometric Problems. InLesson 13,Wesolved geo- metric problems which gave risetoaﬁrstorder diﬁerential equation. In thislesson, weshall solve ageometric problem giving risetooneofthe speciﬁc types ofsecond order equations discussed inLesson 35. Inrectangular coordinates theradius ofcurvature Rofacurve y=f(z)atapoint P(x,y) onitisgiven bytheformula (36.8) R= Y R Y y>0;also y">0since thecurve is concave up P(xr y) atP(x' y) '3 y<0;also E P(x,y) y"<0since thecurve is concave down 0 X atP(x,y) (11) (5) Figure 36.81NoQ N1VOphqlN Ifanormal tothecurve isdrawn from thepoint Ptothe:1:axis, then it andtheradius ofcm'vature have: 1.Opposite directions when yandy"have thesame signs, seeFig.36.81 (a) and(b). 2.Thesame directions when yandy”have opposite signs, seeFig.36.82(a) and(b). Lesson 36D GEOMETRIC Pnonmms 529 Y O X 'P(x|y) Q 5 y>0;also o _,, . Q y<0,alsoR y<0since y”>0since thecurve is .g R thecurve 18concave down concave up lei atP(x» atP(x) y) P(1.y) 0 X Y (fl) (5) Figure 36.82 Example 36.83. Find thefamily ofcurves whose radius ofcurvature istwice thelength ofthenormal segment from apoint onthecurve to thea:axis, (a)ifnormal andradius ofcurvature have opposite directions, (b)ifthetwohave thesame directions. Solution. Thelength ofthenormal segment from apoint onthe curve tothe2:axisis[see(i)ofExercise 13,1] |y\/1 +(y’)2|.Hence by (36.8) andthehypothesis oftheproblem, <.> =-:21/\/FEE, where theplussignistobeused when normal andradius ofcurvature have opposite directions, theminus signwhen they have thesame direc- tion. Division of(a)by[1+(1/)2] 1/2gives (b) 1+(1/)2==I=2yz/". anequation which isofthetype described inLesson 35C. In(b),wethere- foremake thesubstitutions d(C) 1/'=u. y"=udi;» thusobtaining du 2udu dyI-l-'tl2=:l:21I/ll.-‘-ii-1 The solution of(d)is (e) log(1+u2)==l;log y+logcl. From (e)and(c),weobtain theequation d d (fl 1+"2 =611/. 5%==!=\/C19 —1, '# ==i=d$../“S _ 530 Pnonums: Srsrnus. Sracufn 21mOnnan EQUATIONS Chapter 8 when normal and radius ofcurvature have opposite directions, and the equation — \/zldy 12=a, @=i 1 _____= d,‘E’+“1/drVy’~/I. *” when they have thesame directions. The solution ofthelastequation in(f)is (h)éx/my —1==I=(w+cg). 4(¢1y—1)=c1”(w+C2)’. which isafamily ofparabolas with vertices at(—c¢,1/cl) andaxes parallel totheyaxis. Thesolution ofthelastequation in(g)is(substitute u=ﬂ) (i) v1Arc sin\/1//C1 —\/y(¢1 —2/)=ix+62, which isafamily ofcycloids. (For deﬁnition ofacycloid, seeExercise 28c,34.) EXERCISE 36D 1.Find thefamily ofcurves whose radius ofcurvature isequal tothelength of thenormal segment from apoint onthecurve tothe:2:axisandhasthesame direction asthenormal. Identify thefamily. 2.Solve problem 1,iftheradius ofcurvature andnormal have opposite direc- tions. Identify thefamily. 3.Find thefamily ofcurves whose radius ofcurvature hasaconstant value lc. ANSWERS 36D 1.(2:—cl)’—|—y2=62.Afamily ofcircles with centers at(c1,0) andradius \/E.2.cly=cosh (=|=c1:c —l-C2).Afamily ofcatenaries. 3-(I—¢1)2+(1!—¢2)2=k2- Chapter 9 Series Methods Introductory Remarks. Aswehave repeatedly emphasized, few differential equations have solutions which canbeexpressed explicitly or implicitly interms ofelementary functions. When asolution cannot be expressed inthisway, theproblem ofﬁnding asolution ofadifferential equation isnotentirely hopeless. There areavailable graphical methods, some ofwhich were described inearlier lessons, numerical methods which willbediscussed inthenext chapter and series methods which weshall consider inthis chapter. Furthermore, itisfrequently true that anim- plicit solution interms ofelementary functions islessuseful than aseries solution oranumerical one. Implicit solutions areusually such compli- cated expressions that itisextremely difﬁcult toﬁndvalues ofthede- pendent variable forgiven values oftheindependent variable. Foraclearer understanding ofthesubject matter ofthis chapter, it willbenecessary tohave aknowledge ofTaylor series andtoknow certain deﬁnitions andtheorems from analysis. Hence, before beginning adiscus- sion ofseries methods forsolving differential equations, weshall ﬁrst re- view thisneeded material foryou. LESSON 37. Power Series Solutions ofLinear Differential Equations. LESSON 37A. Review ofTaylor Series and Related Matters. A series oftheform (371) do+'11($ —1¢o)+ ¢l2(f¢ "-$o)2+¢1a($ —ivola'|'''', where ao,al,a2,---,2:0areconstants andxisavariable iscalled apower series. Apower series may: 1.Converge only forthesingle value a:=mo. 2.Converge absolutely forvalues ofxinaneighborhood of2:0,i.e.,con- verge forIx—xol<h;diverge forIx—x0|>h.Attheendpoints xo:l:h,itmay either converge ordiverge. 3.Converge absolutely forallvalues ofx,i.e.,for—-oo <:1:<oo. 531 532 Seams Mrrmons Chapter 9 Incases 2and 3,thesetofvalues of:1:forwhich thepower series con- verges iscalled theinterval ofconvergence-of theseries. Incase 2,for example, ifaseries alsoconverges forx=mo:|=h,then itsinterval ofcon- vergence isxo—h§x§1:0+h;ifitconverges only for2:=to+h,but notfor:1:=zco—h,then theinterval ofconvergence is:00-—h<as§ xo+h;etc.Incase3,theinterval ofconvergence istheentire realaxis. Comment 37.11. Interval ofCmvergence. Inthecalculus youwere taught certain tests bywhich youcould determine aninterval ofcon- vergence ofapower series. Asimple oneandonewhich isfrequently used isknown asthe“ratio test.”Itstates that theseries u1+u¢+ua+--- +u..+--- converges absolutely if (37.12) lim =k<1.ﬂ—')@ Wegivebelow examples ofeach ofthethree types ofseries. Forcon- venience wehave taken xo=0. Example 37.13. Determine theinterval ofconvergence ofthepower series (a) 1+1l:z:+2l:c2+3!:c3+---+n!x"+---. Solution. Here u,,=nlx”, u,,+1 =(n+1)!:c"+1. Therefore <1») I“?= =|<n+1>x|- Foreach x¢0,|(n+1)x|—-> ooasn—> oo.Since thislimit 96Ic<1, theseries (a),byComment 37.11, converges only for2:=0. Example 37.14. Determine theinterval ofconvergence ofthepower series (a) 1+x+§x’+§x“+---+},x"+--- Solution. Here u,,=as”,u,,+1 =x"+1. Therefore .u,, .(n+1)x"+1 ‘bl l‘i‘3.ﬁ=.l‘_'.“..—.r";F=|“'- Hence, byComment 37.11, theseries (a)converges absolutely foreach ac whose absolute value islessthan one. Theinterval ofconvergence, how- ever, is—1§an<1,since theseries converges forx=-1,butdiverges forac=1. Lesson 37A Rsvmw orTntoa Seams 533 Example 37.15. Determine theinterval ofconvergence ofthepower series Z2 Z4 we (__1)»-1x2»-2 (*1 1-§i+Ii—a+"'+"@%W+"“_ x2(n—1) 1:21» S9lutlon- HBIB =(Wm I|u,,+1| =6;’-)-i‘ Tl18l‘6f0l‘6, .'u.,,+1 _.1:2” (2n—2)!__. 2:2 = (blP33.T.—13*}.‘co!F_=»—='P33.T<2n"'- 1)°*foreach 2:.Hence, byComment 37.11,theseries (a)converges absolutely forall2:.Itsinterval ofconvergence istherefore theentire realaxis. Wenow state anumber ofrelevant theorems inconnection with power series. Theorem 37.16. Ifapower series (37.1) converges onaninterval I: I2:——zcol<R,where Risapositive constant, thenthepower series deﬁnes afunction f(z)which iscontinuous foreachatinI. Comment 37.17. Ifonewrites apower series, say (8) 1+w+w’+r§+-~-. which isconvergent onI:—1<x<1,then byTheorem 37.16, the series (a)deﬁnes afunction which iscontinuous onthisinterval. The question naturally arises: which function? This question is,ingeneral, noteasytoanswer although itisfor(a),because theseries isageometric one. This series, foreach :1:forwhich Ix]<1,converges to1/(1 -—x). Hence, <1»r<x>=;-ﬁ=1+s+@”+w’+---. lwl<1-1 Forexample, when 1:==},then f(§)=Q =2andthegeometric series ontheright converges to2.Butif1:=2,thenf(z)=1/(1—-2)= —1,andtheseries ontheright of(b)certainly doesnotconverge to—l. Now consider thepower series a 5 1 2»-1 <c>Z-%+%-97+---+<-1)"-‘%+---. whose interval ofconvergence isalso—1<x<1.Hence byTheorem 37.16, itdeﬁnes afunction f(z)which iscontinuous onthisinterval. In thiscaseonly onefamiliar with series might recognize thattheseries (c) deﬁnes thefimction Arctanac.Itisafact, however, thatmany conver- gent power series cannot belinked toelementary functions forthevery good reason thatmany convergent power series donotdeﬁne elementary f\1I1015l0I1B. 534 Seams Mnrnons Chapter 9 Theconverse ofthequestion raised above issomewhat easier toanswer, i.e.,given acontinuous function onaninterval I,isthere apower series which deﬁnes it?Weshall state certain theorems below which notonly willgive ustheanswer tothisquestion, butalsowillshow usatthesame time how toﬁndthedeﬁning power series, ifthefunction hasone. Theorem 37.2. Iff(:z:) isdeﬁned byapower series, i.e.,if (37-21) f(t)=(lo+a1(I—$0)+a2(rv-f'30)2 +<1s(¢-$o)3+"', I-'|$—f'Io| <R, then (37-22) f'(w)=<11+2w2(=v —$0)+3a3(rv —we)’+---. I:Ix—xol<R, i.e.,thepower series obtained bydiﬁerentiating each term of(37.21) deﬁnes (orconverges to)thederivative off(z)onthesame interval I. Theorem 37.23. Iff(z) andg(2:) aredeﬁned bypower series, i.e.,if (37.231) f(z) =an+a1(:c —xo)+a2(:c —:00)” —l----,Ix—2:0]<R and (37-332) 9(1)=bo+b1(w-$0)+b2(@>—Ivo)2+''',I2-wol<R, thenf(z) =g(a:) ifandonly if (37.233) ao=bo, a1=bl, a2=b2, ---. Theorem 37.24. Iff(z)isdeﬁned byapower series, i.e.,if (37-25) f(z)=(10+<11(w-we)+¢12(w-—f'3o)2+--- +an(x_’1:0)n+"'s lx_x0l<R1 then (37-26) (10=f($o), 21=f'(5"o), Proof. ByTheorem 37.2, thesuccessive derivatives of(37.25) are (2)f'(¢)=a1+2¢l2($ —1o)+ 3aa(1 —$o)2+-~- +Mn($ “‘$o)"_1-lr '''1 f”(f'3) =222+3l¢la($ —270)-l-'''-l-n(n“‘1)2n($ -'$o)"_2 +''‘, f”’(x) ==3!a3 +---—|—n(n-1)(n -—-2)a,,(x -:z:o)"_3 —|—---, f‘"’(e) =n!m.+ (n+1)!a..+1(w —$0)+---- Lesson 37A Ravn-:w orTarpon Snnms 535 In(37.25) and(a),set:1:=1:0ineach equation. There results (b) f(¢o) =20, f’(wo) =<11, f"(¢o) =2312, f"’(a:o) =3!a3, ---, f"(:c0) =n!a,,. Hence by(b), (0) (10=f($o)» (11=f'($o)» f”(¢) f”’( ) f(”)( ) _ 0 _ Io _ i'30_ “"‘ 2!’ as‘ 3!’ '°"‘“ nl Iff(z)isdeﬁned byapower series, then by(37.25) and(37.26), (31.21) re)=fa.)+/'<w.><x —to+ o—wo)2 +~53,(f‘—")(x—w0>3+--- + (x—xo)"+---, |:c—-:co| <R. If2:0=0,then (37.27) becomes (37.28) f(z)=r<0>+r'(0>w+ i’+ x“+--- + x”+---, <R. Deﬁnition 37.3. Theseries ontheright of(37.27) iscalled theTaylor series expansion off(z)inpowers of(a:—xo),orinaneighborhood ofxo. Deﬁnition 37.31. The series ontheright of(37.28) iscalled the Maclaurin series expansion off(:0)inpowers of:c,orinaneighborhood ofzero. Example 37.32. Assume thefunction f(z) =sinx isdeﬁned bya Maclaurin series (37.28) onsome interval. Find theseries anditsinterval ofconvergence. Solution. Taking successive derivatives off(z) =sin:0andevaluat- ingthem at:1:=0,weobtain (2) f(='=)=Sinw, f(0)=0, f'($)=0081. f'(0)=1- f”(w) =-Siniv..f”(0) =0. f'”(I) =-008$. f"'(0) =—1, f(4)(:v) =sinx, f(‘)(0) =O, f(5)(x) =cos2:, f(5)(0) =1. 536 Smuss Mamoos Chapter 9 Substituting theright-hand values of(a)in(37.28), there results __ xa $5 n_1 $2»-1 Sln$—Z_§T+'5!"""+(-I) +"' Here |u,,|=x2"'1/(2n —1)i.Therefore .u,,.,.1 _. x2”+1 (2n—1)!’__ .I 2:2 i__ (°),l‘fi.u,.‘lit. (2n+1)! $2»--1 ",l‘I..’?., (2n)(2n+1) ‘O’ foreach x.Hence byComment 37.11, theseries (b)converges forallav. Comment 37.33. Theorem 37.24 says that iff(z) isdeﬁned bya power series, then thecoeﬁicients intheseries aregiven by(37.26). But itdoesnottelluswhether f(z)canbedeﬁned byapower series. Perhaps itcannot. Here isanexample ofafunction which cannot bethusdeﬁned. Let (a) re)=e-1"’.w¢0=0,:c=0. Thefunction f(z)iscontinuous atas=0.Itsderivatives atx=0are [usethedeﬁnition ofthederivative f’(0) = (b) f’(0) =0,f"(0) =0,f”'(0) =0. Substituting these values in(37.28), weobtain (c)f(x)=o+o¢+%x’+_%+---=o+o+o+---, —-oo <:t< co. Foranx¢O,theright sideof(c)certainly does notconverge tothe function f(z)deﬁned by(a).Forexample, ifrc=0.1,thenby(a),f(0.1) = e_1/°'°1. Theseries ontheright sideof(c),however, converges tozero foralla:. Iftherefore westart with acontinuous function f(z) andobtain a power series whose coeﬂicients aregiven by(37.26), westillneed atheorem which willtelluswhether thepower series thusobtained actually deﬁnes orconverges tothegiven function f(z). Forthispurpose weintroduce the following theorem. Theorem 37.34. Taylor's Theorem. Ifafunction f(z)hasderivatives ofallorders rmaninterval I:|x—xol<h,then <31-35> re)=/<-».>+r'<x.>o-we+ o—23o)2+--- + o—230)"+ac). Lesson 37B Seams SOLUTION orALINEAR EQUATION 537 where R,.(x), thereminder term orthesumofallterms after the(n—|—1) term, isgiven by __r‘"+"(X><x —Z>"+‘ ———ii(1T r andXisbetween 2:0andx. IfR,,(x) —>0asn—>oo,thenandonlythenis (swanf(t)=re.)+1'<-:.>o—we+ o—xv’+---. la:_x0l <hr i.e.,theinﬁnite series ontheright actually converges toanddeﬁnes thefunc- tirmf(:c) onI:la:-—2:0]<h. Remark. The series ontheright of(37.35) iscalled aTaylor series with remainder. Deﬁnition 37.4. Afunction f(z)issaid tobeanalytic atapoint at=:00ifithasaTaylor series expansion inpowers of(av—xo)valid foreach :2:inaneighborhood of2:0. Deﬁnition 37.41. Afunction f(z)issaidtobeanalytic onanin- terval ifitisanalytic ateach point oftheinterval. Examples ofAnalytic Funetions. Ithasbeen proved thateach ofthe functions listed below isanalytic ontheinterval indicated. ItsTaylor series expansion isgiven bytheseries ontheright. 2 s (37.42) e"=1+a:+g-i+%+---, —-oo<a:<oo; 2 4 o cos:c=1—%,+%—%,+---, —oo <x<oo; 3 5 7 8iI11==v-%+%-%+---, —oo <:c< oo; 3 6 7 Amianz=@--’§T+%~-3;-+---, —1§:c§ 1. Comment 37.43. Because of(37.26), ananalytic function hasone and only oneTaylor series expansion. This fact implies that thesame series results nomatter what method isused toobtain it.Verify, for example, that theseries 1/(1 —:c)=1+re+:02+---, <1,can beobtained byTheorem 37.24 aswell asbyordinary division. LESSON 37B. Solution ofLinear Differential Equations bySeries Methods. The series methods weshall describe inthislesson areespe- cially well suited forﬁnding asolution ofthelinear differential equation 538 $111111-:s Marnons Chapter 9 with nonconstant coeﬂicients, (37-5) v‘"’+f.._1(w)y‘""’ +---+f1(r)y' +fo(w)v =Q(x)- Weshall therefore consider thisclass ofequations ﬁrst, andthen proceed toadiscussion ofother types ofdifferential equations andtosystems of differential equations. InTheorem 65.2, westate and prove asufficient condition forthe existence anduniqueness ofasolution of(37.5) satisfying ninitial condi- tions. Here wemerely state, without proof, asufficient condition forthe existence ofapower series solution of(37.5). Theorem 37.51. Ifeach function f0(1:), f1(1:), ---,f,,_1(1:), Q(1:) in (37.5)isanalytic at1:=1:0,i.e.,ifeaehfunction hasaTaylor series expan- sion inpowers of(1:—1:0)valid* forI1:——1:o|<r,thenthere isaunique solution y(:c) of(37.5) which isalso analytic at1:=1:0,satisfying then initial conditions l/($0) =a0; if/($0) =alt '''1if/(n_D(x0) =an-11 i.e.,thesolution hasaTaylor series expansion inpowers of(1:—-1:0)also valid forI1:——1:0]<r. Comment 37.53. Apolynomial isaﬁnite series. Hence theseries is valid forall1:.Iftherefore thefunctions fo(:c), f1(a:), ---,f,,_1(1:), Q(1:) of (37.5) areeach polynomials, then, byTheorem 37.51, every solution of (37.5) hasaTaylor series expansion valid forall1:. Comment 37.54. Anexistence theorem tells youonly whether asolu- tionofadifferential equation exists. Itdoes nottellyouhow toﬁndthe solution. First Series Method. BySuccessive Differentiations. Weshall illustrate byexamples theﬁrst method ofﬁnding apower series solution ofalinear differential equation. Example 37.541. Find byseries methods, aparticular solution ofthe linear equation (a) v”—(w+1)v’+w’y=w forwhich y(0) =1,y’(0) =1. Solution. Comparing (a)with (37.5), weseethatf°(1:) =1:2,f1(1:) = -1;——1,Q(x) =1:.Since allthese functions arepolynomials, theseries solution weshall obtain, byComment 37.53, isvalid forall1:.Because ‘We shall usetheword “valid” tomean that foreach 1:inaneighborhood of1:0the series expansion ofthefunction converges tothevalue ofthefunction. Lesson 37B Snares SOLUTION orALINEAR EQUATION 539 wehave been given values ofthesolution anditsderivative when 1:=0, weseek asolution intheform oftheMaclaurin series (37.28). With f(z) replaced byy(z)itbecomes (w uo=y@+iwn+l§@#+2%QH _*_y(‘:§0)x4_*__._‘ Bytheinitial conditions, 1:=0,y=1,y’=1.Substituting these values in(a)gives, (0) 1/"(0) —1=0, y”(0) =1- In(b),wenow know, bytheinitial conditions and (c),thevalues of y(0), y'(0), y"(0). Toﬁnd thevalues ofsucceeding coefficients, wetake successive derivatives of(a)andevaluate them at(0,1). The next two derivatives are (d) 2/”’—(rv+1)v”—2/’+wzv’+2wv=1.y(4) __(x+1)yII! __2:,/n +x2yu +42,111 +2y=_0. Hence when 1:=0,y=1,y’=1,y"=1,weobtain from (d) c) ww=& yWm=a Substituting in(b),theinitial conditions, (c)and(e),wehave $2 xv Z4 (f) y(1:)=1+x+§~+§-+§+'-', -oo<1:<oo, which gives theﬁrst ﬁveterms ofaseries solution of(a)satisfying the given initial conditions. Example 37.55. Find bypower series methods, aparticular solution ofthelinear equation (2) 1/"+ff_xj,§1/'-%y=0,|$|?‘1, forwhich y(0) =1,y’(0) =1. Solution. Since theinitial conditions have been given interms of 1:=0,weseek aseries solution inpowers of1:.Comparing (a)with (37.5),we seethatf0(1:) =—1/(1 -—-1:2),f1(1:) =1:/(1 —:c2),Q(1:) =0. The Maclaurin series expansion of—1/(1—-1:2)is (b) —,?1g,,=-(1+x’+¢‘+¢°+---),|e|<1. 540 Snan-:s Mnrnoos Chapter 9 The Maclaurin series expansion of1:/(1 —1:2)isalso valid for <1. Hence byTheorem 37.51, each solution of(a)hasapower series expansion which isvalid for <1.Bytheinitial conditions, 1:=O,y=1, y’=1.Substituting these values in(a)gives (<1) y”(0) =1- IntheMaclaurin series (37.28), namely o>m=wwwm#%H%@#%%wm. wenow know, by(c)andtheinitial conditions, thevalues ofy(0), y’(0), y"(0). Toﬁnd thevalues ofsucceeding coefﬁcients, wemultiply (a)by (1-—1:2), then take itssuccessive derivatives and evaluate them at (0,1). The next twoderivatives are (e)(1-w’)y”' ——wv”=0. (1—w’);/“" —3wy”’ —v”=0- Hence when 1:=0,y=1,y’=1,y”=1,weﬁndfrom (e) w w@=o W@=1 Substituting in(d)theinitial conditions, (c)and(f),weobtain 2 (o M@=Hw+%+§+~,%<w<L which gives theﬁrst ﬁveterms ofaseries solution of(a)satisfying the given initial conditions. Example 37.56. Find bypower series methods aparticular solution ofthelinear equation (a) v'”+%v'-$v=0. $960. forwhich y(1) =1,y'(l) =0,y"(l) =1. Solution. Since theinitial conditions have been given interms of x=1,weseek aseries solution inpowers of1:—1.Comparing (a)with (37.5), weseethat f0(1:) =—1/1:2, f1(:c) =1/1:, Q(x) =0.The series representation of1/1:2 inpowers of1:—1is1-—-2(1:-—-1)+3(1:—-1)2-- 4(1:-—-1)3-1----,which isvalid for0 <1:<2.Theseries representation of1/xinpowersof (1:— 1)is1—-(1:— 1)—l-(1:— 1)2— (1:—- 1)“-l----, which isalsovalid forO<1:<2.Hence, byTheorem 37.51, each solu- tion of(a)hasaTaylor series expansion inpowers of1:——1,valid for O<1:<2.Bytheinitial conditions, 1:=1,y=1,y’=0,y”=1. Lesson 37B SERIES SoLu'r1o1~; orALINEAR EQUATION 541 Substituting these values in(a)gives (b) 1/”'(1) +0—1=0, I/”(1) =1- IntheTaylor series (37.27), namely <<=>ye)=1(1)+y'<1><1—1)+ <1—1)’+ <1—1)“ +%l(x_1)4+%l(,;__1)5+..., wenow know thevalues ofy(1), y'(1), y”(1), y"'(1). Toﬁnd thevalues ofsucceeding coefﬁcients, wemultiply (a)by1:2,then take itssuccessive derivatives, andevaluate them at(1,1). The next twoderivatives are Z21,/(4) +2xyn/ +xyn +y/_yl=0, xyﬂ) +2y/u +yn =0’ xi/(5) +3:,/(4) +y/u = When 1:=1,y=1,y’=0,y”=1,y”'=1,weﬁndfrom (d) (e) v“’(1) =-3. 2/“’(1) =8- Substituting in(c),theinitial conditions, (b)and(e),weobtain _(x—1)’(1—1>3_ o—1)‘(1-—1>‘____ 0'<1:<2, which gives theﬁrst sixterms ofaseries solution of(a)satisfying the given initial conditions. Second Series Method. Undetermined Coefficients. Weoutline asecond method ofobtaining aseries solution, onewhich does notdepend ontaking derivatives. This method willtherefore bemore useful than the preceding one whenever itbecomes toodifficult toobtain successive derivatives. Example 37.6. Find bypower series methods aparticular solution of thelinear equation (a) y”—(w+1):/’+$211=w forwhich y(0)=1,y’(0) =1. Solution. This example isthesame as37.541. Hence weknow that (a)hasaseries solution inpowers of1:valid forall1:.Aseries solution in powers of1:hastheform (b) 1/fr)=(lo+aw+(12372+11313+arr‘+---- 54-2 Ssnnas METHODS Chapter 9 ByTheorem 37.2, itstwosuccessive derivatives, alsovalid forallan,are (c) y’(:c) =a1+2a2:z: +3a;;:c2 +4a4x3 +---, y”(:c) =2:12+61131: +12a4x2 +---. Substituting (b)and(0)in(a),weseethat y(z) willbeasolution of(a)if (d) 2:12—|—6:131: +12a4a:2 +--- —(vv+1)(¢11+2a2w+30312 +'4a4w3 +---) +f¢2(¢lo+air+@2312+aw“+aw‘+---)=w- Carrying outtheindicated operations in(d)andsimplifying theresult, we obtain (e) (2:12 —a1)—|—(6a;,~ —2112—a1—1):: +(12a4-3113-2a2+am?+---=0. Because ofTheorem 37.23 [take y(z)=0+Ox+Ox”+---],equation (e)willbeanidentity inxifandonly ifeach coefficient iszero. Hence we must have (f) 2112-<11=0, 412=%' 603'-'2(l2'—'(l1—1=0, a3= - 12a4—3a3—2a-2-]-a0=0, a4= - By(37.26) andtheinitial conditions y(0) =1,y'(0) =1,weknow, with (I39 =0,that (2) "0=1/(0)=1,<11=1/(0)==1- Hence by(f), (11) a2=%1 ¢la=§» a4= =%' Substituting (g)and(h)in(b),weobtain $2 ms $4 (i) Z/($)=1+w+-5+5-+§+"', —°°<$<°°, which agrees with (f)ofExample 37.541. Example 37.61. Find bypower series methods aparticular solution ofthelinear equation II 1 <a> y+1—_f—x§y'—;—_—1;y=0, forwhich 1/(0) =1,y’(0) =1. Lesson 37B Smmas SOLUTION orALINEAR EQUATION 543 Solution. This example isthesame as37.55. Hence weknow that (a)hasaseries solution inpowers ofxvalid for <1.Aseries solution inpowers ofxhastheform (b) y=ao+a1x+a2x’+aar’+a4w‘+---- ByTheorem 37.2, itstwosuccessive derivatives, also valid for <1, are (c) y’(:z:) =a1+2a2:z: +3a3x2 +4a4:c3 +5:15:11‘ +---, y”(:z:) =2112+61132: +l2a4:v2 +20a5a:3 +---. The function y(z) of(b)willbeasolution of(a)ifthesubstitution of(b) and (c)in(a)yields anidentity inx.Multiplying,(a) by(1—-:02),then making these substitutions andsimplifying theresult, weobtain (d) (2a2 -—-(lo)+611311 +(12:14 ——a2)x2 +(20a5 —4a3)x3 +---=0. Because ofTheorem 37.23, equation (d)willbeanidentity inx,ifand only ifeach ofitscoefﬁcients iszero. Hence wemust have 1(e) a2=§a0, a3=0, a4=%, a5=%»---. By(37.26) andtheinitial conditions y(0)=1,y'(0) =1,weknow, with $0 =0,iihﬂt (f) do=y(0)=1, <11=1/(0) =1- Hence by(e)and (f) (E) ¢l2=§, as=0, a4=§14, ‘la=0,“‘ Substituting (f)and(g)in(b),wehave 21 <h> y=1+x+%+ﬁ@‘+---. lwl<1, which agrees with (g)ofExample 37.55. Example 37.62. Find bypower series methods aparticular solution of 1 1 (*1) 11"’-l‘;?/"'-E?/=0, forwhich y(1) =1,y’(1) =0,y"(1) =1. Solution. This example isthesame as37.56. Hence weknow that (a)hasaseries solution inpowers of(2:—1)valid for0<:0<2.A series solution inpowers of(:1:—1)hastheform (b)1/=ao+a1(1=—1)+<12(rv—1)2+aa(r—1)3+a4(1=-1)“+---- 544 Sanms Mmuons Chapter 9 ByTheorem 37.2, itsthree successive derivatives, alsovalid for0<:1:<2, are (c)y’=a1+2az(w —1)+3a3(w -—1)’ +4a4(rv —1)“+5a5(w -1)‘ +---, y"=2ag+3!a3(x —-1)—|—3-4a4(a: -—1)2 +4'5a5(x ‘-1-)3'i""v y"'=3!a3 +4!a4(:z: —-1)+60a5(:c —1)2+---. Substituting (b)and(c)in(a),weobtain, after multiplication by1:2, (d) :z:2[3!a3 +4!a4(x —1)+60a5(:c —1)2+---] +1?l¢l1+2¢12($ —1)+3¢ls($ '"1)2 +411-1(=v —1)3+5¢1s($ -1)‘-l" '‘l ""lao +¢l1($ —1)+¢12(1l= —1)2+''‘l=0; Itwillbeeasier toequate coeﬂicients oflikepowers of:ctozero, ifwe express 1:2anda:inpowers of(2:—1).Their respective series are,see (38.25), (6) x2=1+2(:z:—1)+(x—1)2, :0=1+(x—1). Substituting (e)in(d),andsimplifying theresulting expression, wehave (fl (6113 +G1"-110)+(24114 +1211:; +2¢l2)($ '"1) +(60115 +48a4 +9a3+a2)(a: ——1)2+---=0. Equating each coeﬁicient in(f)tozero, weobtain (8) as=1%’ I14=i(-1203 -'202), a5=;15(—-48a4 -—9a3—a2). By(37.26) andtheinitial conditions, weknow, with 2:0=1,that (11) do=y(1)=1,'11=1/(1)=0. "2=%y"(1) =it Hence by(g)and(h) (i)¢1a=‘l‘» "4=a§r(-2-1)= -t, ¢1s=a‘o(6—%-%)=1‘s- Substituting (h)and(i)in(b),wehave _ 2 __a _4 __5G)y=1+(:c21) +(a: 61) _(x 81) +(:c151) +___’ 0<x<2, which agrees with (f)ofExample 37.56. Lesson 37B Seams SOLUTION orALINEAR EQUATION 54-5 Example 37.621. Find bypower series methods, ageneral solution of thelinear equation (B) y”+(sinw)y’+e”y=0- Solution. Allcoeﬂicients in(a)areanalytic at2:=0.We shall therefore seek aseries solution oftheform (b) y=a0+a1x+a2:z:2+a3x3+a4x‘+a5:z:5+---. Since, by(37.42), a 5 (c) sina:=:z:-—%+%—---, —oo<z<oo, $2 $3 8z=1-|-(I3-|-5-|-i-§i'-,l""", —(D <£l3< w, each series solution of(a),byTheorem 37.51, isvalid forallzc.ByTheo- rem37.2, thetwosuccessive derivatives of(b)are (d) y’=a1+2a22: +3a3:v2 +4a4x3 +5a5:v‘ +---, y”=2a;+61132: +121142;’ +20a5:c3 +---. Making thesubstitutions (b),(c),and(d)in(a),weobtain (e) (2a2 +6a3x +12a4a:2 +20a5a:3 +---) s 5 +(w—i%+%§-— (111+2a2w+3aaw2+4¢wv3+5a5w"+---) $2 $3 > -i*(1-irili-ti-i"i+"' X(ao+alx+(Z2122 +a3x3 —|—a4:v4 +a5x5 +---)=0. Performing theindicated operations in(e)andsimplifying theresult, we have (f)(2%+at)+(6113+21».+am+(12a. +sag+a1+9;)$2 +(20a5+4a3+a2+%+%)x’+--- =0. Equating each coefficient in(f)tozero, weobtain 2 a=_§g__¢l1__ao=ao__a1_ao=ao_¢l1, ‘41224812241212 546 Ssnrss Mm-nons Chapter 9 ___B_B__ﬂ__&"5' 52060120 =_1_a1_a0 __1 _a0 __a1_ao 5 3 6 20 2 60 120 _ﬂ Q._20+20 Substituting these values ofthea'sin(b),there results (11)1/=¢lo(1-'§$2"%x3+‘I5x4+i1ox5+"') +a1(x—§x3—T12-x‘+-215x5+---), —oo <x< oo, which gives terms toorder ﬁveofageneral solution of(a). Here aoand a1arearbitrary constants. Comment 37.63. Inﬁnding aseries solution ofadifferential equa- tion, itwillusually notbeeasy towrite thegeneral term oftheseries. Infact, itwill, inmost cases, bevery difficult ifnotimpossible. However, ineach oftheabove examples where westopped with aﬁnite number of terms, itshould beevident toyouthat foreach :1:intheinterval ofcon- vergence oftheseries, y(z)canbecomputed toadesired degree ofaccuracy byusing suﬂicient terms, justase’orsinascanbecomputed toadesired degree ofaccuracy from their respective series. EXERCISE 37 1.Find theinterval ofconvergence ofeach ofthefollowing series. 4 6 Zn <a>x2+%+§+---+3‘m+---. (b)1+s¢+s%2+---+s""‘¢""‘+---. +3<+a>’ +""‘ (c)1+%+%+...+L£__+..._ 2.Obtain theﬁrstthree nonzero terms oftheMaclaurin series foreach ofthe following functions. (a)cos2:.(b)e‘.(c)tanx. Obtain terms toorder koftheparticular solution ofeach ofthefollow- inglinear equations, where kisthenumber shown alongside each equation. Usebothmethods ofthislesson. Also ﬁndaninterval ofconvergence of each series solution. 3.y'—:cy+:c2 =0,1/(0) =2,k=5. 4-.xzy” =21+ 1,y(1) =1,1/(1) =0,k =4. 5.:cy"+ a:2y'—2y=0,y(1) =0,y’(1) = =4. 6.y”—|—3:ty'+ e’y=22:, y(0) =1,y'(0) =—- =4. 7-$211”-—2=y'+ (101;1)?!=0,1/(1)=0,:1/(1)=itk=5-8-(1—w)y'”-—211/+ 31/=0,y(0)=1.1/(0) =—1,:/"(0) =2,k=6-r .'-PrPr‘ Lesson 37-Exercise 547 Obtain terms toorder k,inpowers of(:1:—1:0)ofthegeneral solution ofeach ofthefollowing equations, where lcand 2:0areshown -alongside each equation. Also ﬁndaninterval ofconvergence ofeach series solution 9. 10. 11. 12.y"—a:y'+2y=0, k=7,a:0=0. 2(w’+8)1/”+21y'+(1+2)z/=0. k=4,wo=0. xy”+:c2y’—2y=0, k=4,:ro=1. y"—xy'—y=sin2:, k=5,a:o=0. ANSWERS 37 l.(a)—w<x<<>°. (b)|a:|<§. (c)—4<2:<—2. 2 4 2 2.(a)1—%+%—--~-. (b)1+1+%+---. 3 5 <o1+%+§i5+---. 3.1/= 4-.y= 5.y= 6.y= 7.y= 8.1/= 9.1/= 10.1/= 11.y=3 4 5 2+a:2~—%+%—%+---, ——w <:v< <=<>. 3 _ 4 .+<._.,=_e+»+@..,a_...,:c—1-—}(:t—1)2+§(x-1)3—i(:c-—1)4+---, 0<a:< 2 1_1_£2+L3+£4._... _w<x<w_ 2 6 3 ' 3(1-1)+%(1v—1)2+%(1—1)3—:*<r(I—1)‘+z1s(1—1)5—--n0<:a:<2 3 4 5 6 l—a:+:2:2--%—%§—-£6-—%+---, —1<a:<1. 3 5 1 1680 2: 2: 5:: :0: :0:“<1_.ﬁ__§E+Ei5.+...)+a1(,;._.ﬁ_E§_|_...), --2\/2<:c<2\/23 _ 4 a°(1+(,,_1)2_ +("%)+...)ao(1—a:2)+a1(z—%—%—-—£-—+--~)> ——°<> <x<w. 2 3 4 3 4 a:—12 :1:-13 2:-14-1-a1((a:—-1)-(2)+(3)——(4)—l----),0<a:<2 l2.y=(H><~> 1101+?-i"'§+"' +a1w+—§+i3+-~- $3 225_q°<3<¢Q 548 Smurzs Mrrrnons Chapter 9 LESSON 38. Series Solution ofy’=_f(x,y). Thetwomethods outlined intheprevious lesson forﬁnding aseries solution ofalinear equation carry over without change totheﬁnding of aseries solution ofageneral ﬁrst order equation (38-1) 3’=f(w,y)- However, thedetermination oftheinterval ofconvergence forwhich the series solution isvalid isamore diﬁicult task. Weshall ﬁrstgiveadeﬁni~ tion andthen state without proof therelevant theorem weshall need in thisconnection. Deﬁnition 38.11. Afunction f(z,y) isanalytic atapoint (a:o,y0) ifit hasaTaylor series expansion inpowers of(ac—mo)and(y—yo),valid insome rectangle la:—xol<b,Iy—yo]<c,i.e.,f(:c,y) isanalytic at ($0,?/0) if (38-12) f(x7y) =‘loo+l<l1o(1 ""$0)+¢lo1(1/ "-1/0)] +la2o(1F —$o)2'1"a11(f¢ —1¢o)(!/ "‘1/0)-l"l1o2(1/ '"3/o)2l '1''''1 isvalid foreach (z,y) intherectangle Ia:—-xol<b,Iy—yo]<cwhich has(:co,yo) atitscenter. Comment 38.121. The method forﬁnding the coeﬁicients a,-,-in (38.12) isessentially thesame asthatused inTheorem 37.24 toﬁndthe coefficients a,-in(37.25). These values ofa,-,~are (38-13) (loo=f($o,1/0); a __af(x0>f/0) _af(x01f/0) , 10-"lax ' 1101——'—'—'-ay 1 a=L92f(i'30,!lo) , a=Zazffilioillo) , 2° 2! 8:02 11 2! 6:2:6y 132f($0»?/0) _1102=Q ayg » a__1'33f(1¢o,Z/0) a__Q33f($o,!/0) “"3! 6:03’“*3! 8:c28y ’ __3_33f($o,?/0) __133f(930,!lo) , an_3! 8x83,/2 ' G03_ 83/3 ’ The symbol 0"f(:vo,yo)/01:" means evaluate 6"f(:c,y)/8a:" when :0=3:0, y=yo.Similarly, thesymbol 8"f(:c0,y0)/6y" means evaluate 8"f(:z:,y)/Oy” when :0=2:0,y=yo. InTheorem 58.5, westate andprove asufficient condition fortheexist- ence anduniqueness ofaparticular solution of(38.1) satisfying aninitial Lesson 38 Snares Sourrron ory’-f(z,y) 549 condition. Here wemerely state without proof asufﬁcient condition for theexistence ofapower series solution of(38.1). Theorem 38.14. Ifthefunction f(z,y) of(38.1) isanalytic at(a:0,y0), i.e.,iff(z,y) hasaTaylor series expansion inpowers of(x—1:0)and (y—yo),validff"Ix-$o|<F.Iv-vol<T,andiffvf Wm! (M/)in this2rX2rrectangle which has(xo,yo) atitscenter, (38-15) If(av)! §M, where Misapositive number, then there isaunique particular solution y(z)of(88.1), analytic at2:=xo,satisfying theinitial condition y(:z:°) =yo, i.e.,thesolution hasaTaylor series expansion (38.16) yo)=rev+i'c°>c-—30>+ <3—30>’III +3-;;_<»><.~..>=~»+._..valid inaninterval about 2:0.Thisinterval isatleastequal to (38.17) I:Ix—-zol<min(552%), where risgiven above andMisgiven in(38.15). Remark. This theorem isaspecial caseofthemore general theorem onsystems stated inLesson 39.SeeTheorem 39.12. Comment 38.18. Formula (38.17) gives aminimum interval onwhich theseries in(38.16) converges tothesolution y(z)satisfying y(x0) =yo. Theactual interval ofconvergence maybelarger. Example 38.2. Find byseries methods aparticular solution ofthe nonlinear ﬁrstorder equation (3) y’=3’+3’, forwhich (b) 3(0)=1- Solution. Here f(z,y) ofTheorem 38.14 isac’+y’and so=0, yo=1.InComment 38.21 below, weshow how toobtain theTaylor series expansion ofac”+yzinpowers ofxand(y—1).This series is (0) f(=/av) =3’+v’=1+2(v-—1)+3’+(11—1)’- Since theseries isﬁnite, itisvalid forallxandy.Wemaytherefore choose forrofTheorem 38.14 anyvalue weplease. Foranarbitrary randwith 550 Smurzs METHODS Chapter 9 |:c|<r,Iy—1|<r,(c)becomes (<1) |a:2+y2|<1+2r+r2+r2=2r2+2r+l, which isthemaximum value of:02—|—yzforall(z,y) ina2rX2rrectangle which has(0,1) atitscenter. Itistherefore theMof(38.15). Hence, by thetheorem there isaunique particular solution y(z) of(a),analytic at x=xo=0,satisfying (b),i.e.,thesolution hasaseries representation oftheform (38.16), namely <e> ye)=y<o>+y'<0>x+Pl§?x*+i$x“+---, valid inaninterval about 2:=0.By(38.17), this interval isatleast equal to[here risarbitrary, Misgiven by(d)] I.’ <IIlil'l(T, . Tomaximize I,weletu=r/[3(2r2 +21'—|—1)]. Taking itsderivative with respect tor,andsetting theresult equal tozero, weobtain 1(g)0=2r2+2r+l—r(4r—|-2), -212+ 1=0, r=-—--\/5 Forthisvalue of1',theinterval (f)becomes (11) 1.-|@|<3‘/50 +1‘/5+1)=6(1J:\/5)<0.069. ByTheorem 38.14, wenow know that theseries solution weshall obtain isvalid foratleast <0.069. Weshall ﬁndthisseries solution bythe twomethods ofLesson 37B. First Method. BySuccessive Diﬂerentiations. By(b),theinitial condi- tions arezc=O,y=1.Equation (a)anditsnext three successive de- rivatives, evaluated at(0,1), are (i) y’=w’+y’, 1/(0)=0+1=1; y"=2x+2yy’, y”(0)=0—|-2-1-1=2; =2+2w"+2(y')’, y"'(0>=2+2-1-2+2-1=8;=-2yy'”—|—6y'y”, y4(O)=2-1-8+6-1-2=28. ‘§A:§e5 Substituting (i)in(e),weobtain (i) y<x)=1+x+w’+%w“+%w‘+---, Lesson 38 Seams SOLUTION ory’=f(z,y) 551 which gives theﬁrstﬁveterms oftheseries solution of(a)satisfying (b), valid atleast intheinterval (h). Second Method. Umietermirwd Coeﬂicients. Since xo=0,weseek a power series solution oftheform (k) y=ao+a1:c+a2:c2+a3x3+a4:c4+---. Itsderivative, byTheorem 37.2, is (1) y’=a1+21122: +3a3:c2 +4a4x3 +5a5x‘ +--- Substituting (k)and(1)in(a),itbecomes (m) a1+2a2x +3a3:z:2 +4a4a:3 +5a5a:‘ +--' =iv’+(<10+air»+02132+@3113+---)2 =112+I102+211011191 +(2a0¢l2 +¢l12)932 -l-(2a0¢1a +2l1i¢l2)$3 +'''- ByTheorem 37.23, (m)willbeanidentity in:0,ifthecoeﬂicient ofeach likepower of:2:iszero. Hence wemust have (I1) 111—110’=0, 2a;-—2a0a1 =0, 3a3—1—2aoa2 —alz=0, 4114—-2a°a3 —2a1a2 =0. Theinitial condition is3/(O) =1.Therefore by(37.26), with :00=0, (0) '10=1/(0)=1- By(n)and(0) (P) <11-'=1, <12=1, ¢~=%(§+2)=%-Substituting (o)and(p)in(k),there results (q) y=1+w+x’+§w”+%w‘+---, which isthesame as(j)above. Comment 38.21. Weshall obtain theseries representation off(z,y) = 1:2+yz,asgiven in(c)above, intwoways. First weshall obtain itby useof(38.13). Thisisastandard method. Starting withf(z,y) =:02+ya, 552 Smms Mmnons Chapter 9 wetakethesuccessive partial derivatives called forin(38.13). These are (33-22) f($,y) =11’+1/2 Q_ Q_Bx_2”’ ay"2”’ 2 2 2 Q=2,L’=0,Q=2_81:2 0:1:6y 6y? Alladditional derivatives arezero. Hence by(38.13) and(38.22), with 2:0=0,yo=1,weﬁnd (38.23) am,=0+1=1, a10 =0; aO1 =2; 1 1 ¢12o=§'!(2)=1,l111=0» ¢1o2=§'!(2)=1- Substituting these values in(38.12), weobtain with 2:0=0,yo=1, (38-24) f(z,y) =1’+y’=1+2(y-—1)+re’+(21-—1)’, which isthesame as(c)above. Asecond method ofobtaining (38.24) forthis particular function 2:2+ya,onewhich ismuch quicker andsimpler, istoobserve that:02is already inpowers of:1:andthat (38-25) 2/"’=(2/—1)’+2y—1=(y—1)’+2(y—1)+1. Comment 38.26. Aneasy way toﬁnd Mintheabove example, without theneed of(c),istoobserve that (38-27) I1’+1/2|éIw’|+|z/’l- Iftherefore |x|<rand Iy—-1|<rsothat|y|<r+1,thenby(38.27), (38.28) |¢’+y’|<1'2+(r+1)’=21*+21'+1, which isthesame as(d)above. Example 38.29. Find bypower series methods ageneral solution, in powers of2:,ofthenonlinear equation (a) u’=1’—y’- Find aninterval ofconvergence oftheseries. Solution. Weshall seekaseries solution oftheform (b) y=a°+a;:c+a2x2+a3x3+a4x‘+-—-. Lesson 38 Sr-nuns SOLUTION ory’=f(z,y) 553 The derivative of(b),initsinterval ofconvergence is,byTheorem 37.2, (c) y’=a1+2a2x +3a3x2 -1-4a4:z:3 +---. Substituting (b)and(c)in(a),weobtain (d) a1+2a2:z: +31131:’ +4a4:c3 —|—---=:02-—-a°2 —2a0a1:z: —(2a0a2 —|—a12)x2 -—(2a0a3 —|—2a1a2):z:3 —--- Equating coefficients oflikepowers ofx,wehave (9) (11=—¢lo2, 01=-"1102; 2a2=—2a°a1, a2=-—aoa1 =a03; 3118=1—20002 —(112, as=30—2804 -'004) =11.‘—"04; 4114=-"2¢l0¢la -'201112, 114=il"2‘1o(‘§ I1104) "-2(—a02)(ao3)l =-8110 +(lo- Substituting (e)in(b),there results (fl 3/’=ao—802$ 'l'@0312 ‘l’(8_¢l04)$3 'l'(—tan +ao5)$4 +'‘'» which gives theﬁrstﬁveterms ofaseries solution of(a).Here anisthe arbitrary constant. By(37.26), itis,foraparticular solution, y(0). To ﬁnd aninterval ofconvergence oftheseries solution, weproceed just as wedidabove inExample 38.2. Here [(z,y) =:02—yz,2:0=0,yogOis arbitrary. The Taylor series expansion of2:2—yzinpowers of2:and (y—yo)is,see(38.25), (2)f(w,y) =$2—1/’=av’—[(21—1/o)2+21/o(y —1/0)+1/0”]- Since theseries isﬁnite itisvalid forallasandy.Wemay therefore choose fortherofTheorem 38.14, any value weplease. Foranarbitrary rand with |:v|<r,Iy-—yo]<r,(g)becomes (11)|f(w,z/)| =Ir’—y”|§Ir”!+|(y—2/0)’! +2yo|y—3/0|+yo’<1"”+1’+2w+yo’ =2(T2+yo")-l"1/02» which isanupper bound of:02-—1/2forall(x,y) ina2rX2rrectangle which has(0,yo) atitscenter. Itistherefore theMofTheorem 38.14. Hence, by(38.17), there isaninterval I,atleast equal to . . r(1) <min (r,——-a +yer) +M02), onwhich theseries (f)converges. 554- Smuas Mm-rrons Chapter 9 EXERCISE 38 1.Find aseries representation off(z,y) inpowers of2:andyif(a)f(z,y) = =1sin1/.(b)f(z,y) =11¢‘-2.Find aseries representation off(z,y) inpowers of(2:—1),(y—1)if (2)f(I,1/) =I1/1(b)f(1,1l) =111°81- Obtain terms toorder koftheparticular solution ofeach ofthefol- lowing differential equations, where kisthenumber shown alongside each equation. Usetwomethods. Also ﬁndaninterval ofconvergence ofeach series solution. 3-y’=1’~112,y(0)2-'=1’—112,11(1) oiQQQ-=y’—111,y(0) expansion inpowers ofyl~'>."‘.!"?r'?s~'Pr 1*?!“ =2:2+siny,y(0) =1r/2, k=3.Hint. Replace siny byitsseries (forundetermined coefficient method). 7.y’=:2:-1-e",y(0) =0,k=4.Hint. Replace e"byitsseries expansion inpowers ofy(forundetermined coeﬁicient method). 8.(1—2:-y)y' =1,y(1) =-2,k=4.Hint. Write (1—:1:—y)y'as [-—(:c —1)—y]y'(forundetermined coeﬂicient method). 9. 10. ll. 12. 13. 14-. 15.\‘¢\§\‘§\=<=0s(w+11), u(0) y11+ ,11( y’=1+$112.11(0)=I their series expansions.=sin(zy)—|—$2, y(0) =3,k=4. =108(wy). 11(1)=1,k=5-=1r/2,k =3. =2e"‘1 1)=1,k=3. y’=\/1+==y,11(0)=1,k=4-l,k=4.y=cos2:+siny,11(0) =0,lc=5.Hint. Replace cosx andsiny by Obtain terms toorder k,inpowers of2:,ofthegeneral solution ofeach ofthefollowing differential equations, where Icisshown alongside each equation. 16.y’=a:+y2, k=3. 17.y’=2:+51k=4.Hint. Write theequation asyy’=my+1. ANSWERS 38 3 5 2 3 1-<a>1(y—§+%—-~~)- (b)y(1+@+%+§,+--~)- 2-(a)(w—1)(z1—1)+(r—1)+(z1——1)+1- (b)11+(11-1)] (1—1)_E_;*1_)2+L’£_""_1Zi_....l 23l3.y=1——a:+:2:2——§a:3+§-a:4—---, |z|<0.069. 3 4 4.y= 1+(x—1)2—-(%+(L;—:l—)-—---, |¢-1|<0.04. 40 37‘s.y=2+4@+u"’+?»“+%+---, |¢|<0.029. Lesson 39A Seams SOLUTION orAFmsr ORDER Srsrnm 555 1 r _ 16.y=%+:::+Ea:3—|—~--, |2:|<56;-_}T)»rarb1trary,|a:| <8- 3 4 7.y=a;+z2+%+5%—|—---, |:c|=ﬁ»rarbitrary. B-z1=-2+"—;-5+,3—6e—1>’+%<1-1>’+;%<1—1>‘+---. 9.1] =3+§a:2+§z3—§x4:|:-~-. 10-11=1+t(w—1)2+1*z(w——1)“—1-11s(w—1)‘+----2 3 n.=;_g+%+~- 12.y= 1+2(1-1)+g(¢-1)=-+16%-1)3+---. 2 3 4 l3.y= 1+¢+%+%--12+---. wag+$2 2x314-11 =1+$+§+"§*+— 2 4 5 15.y=x+£i—;—4—;—0+--~,|z| <g-,rarbitrary. 16-1/=00+ 11021-|'(§-l" l103)$2-l" (3l10+ <104)13'l' '"1T - |22|< 1T8,1'l)1l5I‘8.l‘y. 1 003-12 3-1:033 7003'--154 l7.y=a0+Rx+ 2003 :c+ M05 1+ 24a07 1+-~ LESSON 39. Series Solution ofaNonlinear Diﬁerential Equation ofOrder Greater Than One and ofa System ofFirst Order Diﬂ'erentia.l Equations. LESSON 39A. Series Solution ofaSystem ofFirst Order Differ- ential Equations. InTheorem 62.12, westate asuﬁicient condition for theexistence anduniqueness ofasolution ofasystem ofﬁrstorder equa- tions d T1/ll’ =f1(t12/11?/21 '''121»), d % =f2(t;i'/1;?/21 '''1ya): ¢-----...--~~---¢---. d»% =fn(t1i'/11:‘/21 '''yll/11): satisfying theinitial conditions (3911) y1(io) =<11, 1/2(1o) =112,"'1 yn(io) =11»- 556 Sean-:s METHODS Chapter 9 Wenow state without proof asufficient condition fortheexistence ofa power series solution of(39.1) satisfying (39.11). Theorem 39.12. Ifeachfunction fl,f2,---,f,,ofthesystem (39.1)is analytic atapoint (to,al,a2,---,a,,),i.e.,ifeach function hasaTaylor series expansion inpowers of(t—to),(yl—a1),--',(vi-an),valid forIt—to]<r,|y1 —-a1|<r,---, |y,,—a,,|<r,and ifforevery point (t,yl,yg,---,y,,)inthis(2r)”+1-dimensional rectangle which has thepoint (to,al,a2,---,an)atitscenter, lfi(t1l/111/21 '''1?/n)l <My = 1121' '';n1 where Misapositive constant, thenaninterval Iexists onwhich there isone and only onesetoffunctions, y1(a:), y2(:c), ---,y,,(a:), each analytic at t=to,satisfying thegiven system (39.1) andtheinitial conditions (39.11), i.e.,eachfunction hasaTaylor series expansion inpowers of(t—-to),namely (39-14) 1/1(1) =111+ ¢111(5 "'to)—|—l112(t '"t0)2'1'¢113((_10)3 +'''1 1/2(1) =112+¢121(t _10)—|—l122(1"‘ (0)2-1-l12a(t —(0)3+'''1 2111(1) =11»+¢1»1(1—10)+ ¢1»2(t —(0)2+¢lna(t —(0)3+'''1 valid inaninterval about to.Thisinterval isatleastequal to (39.15) I:|t-to]<min(1,G-W-LLQW), where risgiven above, nisthenumber ofequations inthesystem (39.1) and Misgiven in(39.18). Theeoeﬂicients in(39.14) aregiven by _(.1') (39.16) an=%@ ' Forexample, thecoeﬂicient a12is,by(39.16), y1"(t0)/2!. Note that it agrees with thecoeﬁicient of(:0-—-mo)’ inTaylor series (37.27). The coeﬂicient ofanis,by(39.16), y2"'(to)/3! which agrees with thecoeﬂ'i- cient of(:2:—a:o)3 in(37.27). Remark. Ifwelookattheﬁrstorder equation dy/da: =f(z,y) asif itwere asystem ofoneequation dyl/dt =f1(t,y1), weseethat Theorem 38.14 isonlyaspecial caseofTheorem 39.12. Example 39.17. Find bypower series methods aparticular solution oftheﬁrstorder system d d (3') If =ytr ya =xys Lesson 39A Smuns SOLUTION orAFmsr Onnan Srsrsm 557 forwhich (b) 2(0)=1,y(0)=1. Solution. Comparing (a)with (39.1) and(b)with (39.11), wesee, with acandytaking theplaces ofylandyz,that fl= 3/tr f2=xi/1 t07:01 al =11 a2 :' ByTheorem 39.12, wemust ﬁndseries expansions off1andf2inpowers oft,(x—1),(y—1).These arerespectively (c) f1=vt=t(v—1)+t, f2=i¢?l=($"1)(Zl"1)-l-it-l-2/-"1 =(w—1)(v—1)+(w—1)+(v—1)+1. Since these series areﬁnite, they arevalid forallt,x,andy.Wemay therefore choose therofTheorem 39.12 arbitrarily. Hence when |tl<r, Ix—1|<r,|y—1|<r,wehave by(c), (<1) lfil<72+’. |f2|<T2+T+T+l=T2+2T+1. Forr>0,(r2+2r+1)>1'2+r.Therefore r2+2r+1istheMof (39.13). Hence, byTheorem 39.12, there isaunique pairoffunctions :::(t), y(t), each analytic att=0,satisfying thesystem (a)andtheini- tialconditions (b),i.e.,each function hasaseries expansion oftheform (39.14), valid inaninterval about t=0.By(39.15), thisinterval isat least equal to[here n=2,Misgiven bythesecond equation in(d),and risarbitrary] (e) I:|t|<min(1, Weﬁnd, intheusual manner, thatthemaximum value ofIoccurs when r=1.Hence Iwillbeamaximum if (f) I:|z|<11,=0.0625. Weshall ﬁndaseries solution of(a)bytheusual twomethods ofLesson 37B. First Method. BySuccessive Dijferentiations. Here theindependent variable ist,thedependent variables :0andy.Since theinitial conditions aregiven att=0,weseekseries expansions ofx(t)andy(t)oftheform (37.28), namely II III (4) 00xv)=1(0)+x'<0>1+”".,§°)1*+”3§°)1*+”4f“)1‘+---. 11(1)=11(0)+1/(0)1+ ””§°)1’+””,,:f°)1“+y(z)?)1‘+---. 558 Seams Mnrnons Chapter 9 Starting with each equation in(a)andtaking itssuccessive derivatives with respect tot,weobtain (11) w’=$11= Z!!! xm=vi. ’ v+iv’, 2v’+iv", "’= 321"+tv”',‘@‘€E“=9:=xv; =ivy’+vi’; my"+2y’:v’ +yx”; =2:y’”+3y":c’ +3y':c” +yx'”; ...,..................... Weknow from theinitial conditions that (i) 91(0)=1,y(0)=1. Toﬁndthevalues ofsucceeding coefficients in(g),weevaluate thede- rivatives in(h)intheorder ac’,y’,:0”,y",etc. These arewith t=0, x=11l/=11 (i) w’(0) :v”(0) w"’(0)x(4) (0)1-0=0,y’(0) 1+ 2+0=2,1/("(0) 3+0=3,y‘*>(0)--~~~s0=1-1=1; 1-1+1-0=1; 1+0-l-1=2; 2-l-0+3-l-2=7; --an», --¢¢--¢o¢¢¢-~~uuso0=1.v”(0) Substituting (i)and(j)in(g),there results t2 t3 t4 (11) $(t)=1-l-5-l"§+'§'l"'°1 1’1’71‘2/(t)=1+1+'§+'§+'ﬂ'+"'- which giveterms toorder fourofaseries solution of(a)satisfying (b), valid atleast intheinterval Iof(f). Second Method. ByUndetermined Coeﬂicients. Since theinitial condi- tions aregiven att=0,weseekseries expansions oftheform (1) 11(1) y(t)ao+a1t+a2t’+ast3+a4t‘+---, b@+b.1+btt’+b3t’+b.1‘+---. Differentiating (l)with respect totandsetting each resulting equation equal totherespective right sideof(a),weobtain (I11) 111+24123+3%!’+44141”+---=211. b1+2bgt -1-3b3t2 —|—4b4t3 —|—---=icy. In(m),replace xandybytheir values in(l).Hence (m)becomes Lesson 39B Seams Soumon orALINEAR Fnzsr Onnsn Srsrsm 559 (I1) a1+ 202i +30312 +4a4t3 +'''= t(bo +b1t+ bgtz +b3t3 +'''), bl+2b2z+3b3z2+4b4z”+--- = (Go +a1t+ agtz +a3t3 +'‘‘)(b() +b1t+ bgtz +''‘)- The pair offunctions in(I)willbeasolution of(a)ifwechoose thea’s andb’ssothat each equation in(n)isanidentity int.Performing the indicated expansions in(n)andequating foreach equation separately, coefficients oflikepowers oft,weobtain (0) <11=0, b1=@050; 2112=bo, 252=11051 +11150; 3113=bi, 35:;=11052 +11151 +41250; 404=b2, 454=“obs +@1152 +G251 +(labo- By(b)andTheorem 37.24, (P) <1o[= 1(0)] =1. be[=11(0)] =1- Hence from (0)and(p),weobtain, calculating thecoefficients intheorder @1151, <12,52,etc- (q) ¢11=0, b1=1; <l2=§b0=§, b2=§(1+0)=§$ ¢la=§b1=i, b3=§(i+0+§)=§§ a4=>ib2=i, b4=1i(i}i+0+‘§+?§)=27I- Substituting (p)and(q)in(1),weﬁnd t2 ta t4 (1') $(t)=1+"2'+§+'§+"‘» :2:37:‘y(t)=1+t+§+§+§z+"‘; which isthesame asthesolution (k)obtained previously. LESSON 39B. Series Solution ofaSystem ofLinear First Order Equations. InTheorem 62.3 westate andprove asuﬂicient condition fortheexistence anduniqueness ofasolution ofasystem oflinear ﬁrst order equations <39-2) %=111(1);/1+r12<»>y2+---+r1,,<¢>y,.+Q10), 2%/t2=f21(t)3/1 +f22(i)2/2 +'''—|—f21-(3)?/n +Q20)» %=f..1<¢>y1 +f,.¢<¢>y2 +---+f,.,.(¢)y,. +Q»(i), 560 Snmns Mnrnons chap”; 9 satisfying theinitial conditions (39-21) 1/1(to) ='11, 1l2(¢o) =02,'''7!I»(lo) =ll»- Wenowstate without proof asufficient condition fortheexistence ofa power series solution of(39.2) satisfying (39.21). Theorem 39.22. If,in(39.2), eachfunction f,-j,i=1,---,n,j=1, ---,nandeaehfurwtionQ,-, i=1,---, n,isanalyticatt= to,i.e.,if eachfunction hasaTaylor series expansion inpowers of(t—to),validfor It—to]<r,thenthere isaunique setoffunctions, y1(t), y2(t), ---,y,,(t), eachanalytic att=to,satisfying thegiven system (89.2) andtheinitial conditions \(3.9.21), i.e.,eachfunction hasaTaylor series expansion in powers of(t—to),namely (39-23) 111(5) =01+011“ "‘to)+a1=(l "$0)’+ll1s(t "t0)3+'''1 3/2(3) =<12+0210 —to)+l122(l —to)”+a2s(¢ —30):’+'''» y»(t)=as+a»1(¢-lo)+111.20-to)’+¢u.a(l-lo)“+---, validforIt—t0|<r.Thecoeﬁicients in(39.23) aregiven by $1.) (39.24) a,~,-= [For example, an=3/igfkl »an=3%, etc] Example 39.25. Find, bypower series methods, aparticular solution oftheﬁrstorder linear system da: d(a) E;=:1:cost, %=tzy, forwhich (b) 11(0)=1,1/(0)=—1- Solution. Comparing (a)and(b)with (39.2) and(39.21), wesee, with xandytaking theplace ofy1andya,that (C) fll =cos tr f22 =t2; t0=O: al=1; 02=—1- Since fnandfzzhave Maclaurin series expansions, valid forallt,it follows byTheorem 39.22, that there isaunique pairoffunctions :e(t), y(t),each analytic att=0,satisfying thesystem (a)andtheinitial con- ditions (b),i.e.,:e(t)andy(t)each hasaseries expansion inpowers oft valid forallt. Lesson 39B Snmss Sowrron orALmnan Fmsr Onnrza Srsrsm 561 There aretheusual twomethods available forﬁnding aseries solution of(a)satisfying (b).These have been described inExample 39.17. We shall usethemore diﬁicult one,inthiscase, ofundetermined coefficients. ByTheorem 39.22, thepairoffunctions :c(t), y(t)have series expansions of theform (39.23). With to=0,these become (d) :e(t)=ao+a1t+a2t2+ a3t3+---, y(t):bo+b1l+b2l2+b3l3+"'. Differentiating (d)with respect tot,andsetting each resulting right side equal totherespective right sideof(a),weobtain (e) a1_+ 2a2t —|—3a3t2 +4a4t3 +---=2:cost, bl—|—Zbgl —|—3b3l2 —|—4b4l3 +-''=t2y. In(e)replace reand ybytheir values in(d),and replace cost,see (37.42), byitsMaclaurin series expansion. There results (r) <1,+2a2t+3a3t2+4a4t3+--- 24 =(a0+a1¢+a2¢’+--->(1—§,+;iq—---)» b1+2b2t+3b3t2+4b4t3+---= t’(bo+b1t+b,t2+b3t3+---). The pair offunctions deﬁned in(d)willbeasolution of(a)ifwechoose thea’sandb’sin(f)sothat each equation isanidentity int.Performing theindicated operations in(f)andequating foreach equation separately, coeﬂicients oflikepowers oft,weobtain al =a0: bl=0; 2a2=al, 2b2=0, 3a3=a2- 31>,=bo, 4a4=a3- 41>,=b1. By(b)andTheorem 37.24, (h) ao=:c(0) =1,bo=y(0) =—l. Hence weobtain from (h)and(g),calculating thecoefficients intheorder als bl: a2; b2! etcw a1Z1! bl=0! a2:*1 b2Z01 a3= __ =0! b3= = _%1 a4== =“Mk: b4= = 562 SERIES Mnrnons Chapter 9 Substituting (h)and(i)in(d),weﬁnd . t’t‘<1) x<¢>=1+:+;—;,-+---.t3 l/(t)—'_1_§—_"': which giveterms toorder fourofaseries solution of(a)satisfying (b), valid forallt. LESSON 39C. Series Solution ofaNonlinear Differential Equa- tion ofOrder Greater Than One. InTheorem 62.22, westate and prove asuﬁicient condition fortheexistence anduniqueness ofaparticular solution ofannthorder differential equation dny 1u (n—1)% =f(z)?/vy all1'''1y ): satisfying asetofinitial conditions (39-31) 3l(1l=0) =1/0; I/($0) =111, 3/"($o) =1/2,''',y(n_1)(1¢o) =l/n-1- Theproof consists inshowing howevery nthorder differential equation with given initial conditions canbechanged toanequivalent ﬁrst order system with equivalent initial conditions. Hence Theorem 62.22 isaby- product ofTheorem 62.12 relating toﬁrst order systems. Analogously, thesufﬁcient condition wenow state fortheexistence ofapower series solution of(39.3) satisfying (39.31) stems from Theorem 39.12 foraﬁrst order system. Theorem 39.32. Ifthefunction fof(39.3) isanalytic atapoint (:e0,yo,y1, ---,y,,_1), i.e.,iffhasaTaylor series expansion inpowers of (w—to),(v—yo),(1/’—l/1):---,(v‘:“_‘;)— l/n-1); v<1lidf0r|1_— vol<1', Iy—yo|<r,Iy’—(y;1|)< r,---,Iy"+1 —y,,_1| <r,andiffor every point (:e,y,y', ---,y"")inthis(2r)" -dimensional rectangle which has thispoint (:e0,yo,y1, ---,y,,_1) atitscenter, lf(x:l/if/'1 '''11/(n_1))l <Ml where Misapositive constant, then there isaunique particular solution of (39.8) analytic atso=xosatisfying (39.31), i.e.,thesolution hasaTaylor series expansion inpowers of(:1:——2:0),namely <39-34> yo)=yet)+1/($o)($ —to+ o—M + ($—$o)3+"'; Lesson 39C S1-nuns SOLUTION or11Nontmmn Eouurou, Oannn n563 valid inaninterval about :1:=2:0.This interval isatleast equal to . rI.’I23"-130' <Inlfl <7‘, ), where risgiven above, nistheorder oftheequation (39.8) andMisgiven in(39.33). Example 39.36. Find byseries methods aparticular solution ofthe nonlinear equation (a) ylll =$2+I/2 forwhich y(1) =1,y'(1) =0,y"(1) =2. Solution. Comparing (a)with (39.3) weseethat n=3andf=2:2 —|—yz.Acomparison oftheinitial conditions with (39.31) shows that :00=1,yo=1,3/1=0,andyg=2.Theseries expansion offin powers of(v—1),(v—1).2/’,(v”—2),by(38-25). is (b)w"+y’=2+2(r—1)+2(y—1)+(w—-1)’+(v—1)’- Since theseries isﬁnite, itisvalid forall:1:and y.Wemay therefore choose anyvalue weplease forrofTheorem 39.32. Foranarbitrary r andwith Ia:—-1|<r,|y-—1|<r,(b)becomes (C) |:c2—|—y2|<2+2r+2r+r2+r2=2(r2+2r+1), which istheMof(39.33). Hence byTheorem 39.32, there isaunique particular solution y(z) of (a),analytic at2:=1,satisfying thegiven initial conditions, i.e., the solution hasaTaylor series expansion oftheform (39.34), namely (d) y(t)=1/(1)+1/’(1)(w —1)II 1 III 1 +”T(,)<w~ 1>”+L,¥<w— 1>=*+---. valid inaninterval about re=1.By(39.35), thisinterval isatleast equal to[here n=3,Misgiven by(c)andrisarbitrary] (8) I:|$—l|< ' Tomaximize I,welet 7‘ ‘flTaking itsderivative with respect torand setting theresult equal to zero, weobtain (g) 0=(r2+2r+1)—r(2r+2)=—r2-l-1, r=1. 564 Ssnms METHODS Chapter 9 Forthisvalue ofr,theinterval Iof(e)becomes (h) |a:—1|<415=0.025. ByTheorem 39.32, wenow know that theseries solution weshall obtain isvalid foratleast Ia:——1|<0.025. Weshall usetheusual twomethods toﬁndaseries solution. First Method. BySuccessive Diﬂerentiations. In(d),weknow bythe initial conditions that (i) 1/(1)=1, y'(1)=0, y”(1) =2- Toﬁndthevalues ofsucceeding coeﬁicients in(d),weevaluate (a)and itssuccessive derivatives at(1,1). These are,with 2:=1,y=1,y’=0, v"=2, / =$2+v’. v”’(1)=1+1=2; =2i+2w’. y“’(1>=2+0=2; =2+291/”+2(!/)2, 1/(“(1) =2+2'2+0=5; =2yy’"+en". v‘°’(1)=2-2+6-0-2=4;-¢-----------’----.¢¢¢¢¢|s------/5._.\-/ ‘=.‘§‘§<:e2eS Substituting (i)and(j)in(d),weobtain (k)2/=1+(r—1)2+%(w—1)“+1*z(w—1)‘ +s‘a(w~1)‘+rls(w—1)"+---, which gives theﬁrst seven terms ofaseries solution of(a)satisfying the initial conditions, valid atleast intheinterval (h). Second Method. ByUndetermined Coeﬂicients. Since theinitial condi- tions have been given interms ofx=1,weseek aseries solution ofthe form (1) 1/=90-l'¢l1($'"1)+<l2($'"1)2+<1a($"1)3+"' Itsnext three successive derivatives are (In) y’=<11+2w2(w —1)+3¢1s($ —1)’+4a4(1 —1)’ +5¢15($ *1)‘-l""; y"=2a,+603(1)}-1)+12a4(z: -1)”+20a5(a: -1)“+---, y"'=6a3+24a4(:e —1)+60a5(:e -1)’+---. Substituting (l)andthelastequation of(m)in(a),weobtain (11)6113+24(14(£C -1)+60a5(a: -1)’+--- =$2+[<10—|—l11($ —1)+¢l2($ _1)2 +'"l2- Lesson 39C SERIES Sommon orANONLINEAR EQUATION, Onosa n565 Itwillbeeasier toequate coefficients oflikepowers ofac,ifweﬁndthe series expansion of:02inpowers of(2:-—1).By(38.25), (0) a:2=1+2(a:—1)+(:c—1)2. In(n)replace 2:2byitsvalue in(0),perform theindicated multiplication ontheright, andthen equate thecoeflicients oflikepowers of(2:—1). There results (P) 503=@102+1, 24a; =2+2aoa1, 60a5 =1+2a0a2 —|—a12. ByTheorem 37.24 andtheinitial conditions, (q) a0=1, a1=0, a2=§=1. Therefore by(p)and(q) (F)¢ls=§=§, <l4=2'z(2)="1l2, a5="e16(1+2)=2l6- Substituting (q)and(r)in(l),weobtain (S)2/=1+(w —1)2+t(2= —1)3+1':(1- 1)‘+s1a(1— 1)5+---, which agrees with (k)toterms oforder ﬁve. Comment 39.37. Asexplained inComment 38.26, wecanﬁndM without theneed of(b),byobserving that |a:2+y2| §|a:2|+|y2|. Henceif |:e— 1|<r,sothat |:e|<r+1 andif |y— 1|<r, sothat I2/I<r+1.then|2’+v’| <(r+1)’+<r+ 1)”=2<r’+2r+ 1)asin(c)above. Example 39.38. Find byseries methods, aparticular solution ofthe nonlinear equation (2) v"’= (w—1)’+y’+y’—2 forwhich (b) 1/(1)=1.v’(1)=0, 2/"(1) =2- Solution. Comparing (a)with (39.3) weseethat n=3andf= (x—1)”+yz+y’—2.Comparing (b)with (39.31), weseethat 1:0=1,yo=1,1/1=0,yg=2.Theseries representation offinpowers of(x_ 1): (y_1): (ll, _O): (ii/H _2); is (c)f(w.v.v’) =(2—1)’+1+2(2)—1)+(21—1)’+v’—2. 566 Snares Msrnoos Chapter 9 which isaﬁnite series, andtherefore valid forallre,y,y’.Wemay there- fore choose rofTheorem 39.32 arbitrarily. Hence when |:e—-1|<r, |y—1|<r,|y’|<r,wehave by(c) (<1) If(2>,y,y’)| <T2+I-1!+2'+'2+1‘=212+3’+1, which istheMof(39.33). Hence, byTheorem 39.32, there isaunique particular solution y(z) of (a)analytic at:1:=1satisfying (b),i.e.,thesolution hasaTaylor series expansion oftheform (39.34), namely (e) 1/(2)=v(1)+v'(1)(w —1)II III I +”T(,‘)<x— 1>’+”T|)<2—1)“+---. valid inaninterval about 2:=1.By(39.35), this interval isatleast equal to[here n=3,Misgiven by(d)andrisarbitrary] (l) I!|$—1|< ‘ Tomaximize I,weletu=r/5(2r2 —|—3r—|—1),take itsderivative with respect torandsettheresult equal tozero. Wethusobtain (g)0=(2)2+3r+1)-r(4r+3)=-21’+1, r=1/\/2. Forthisvalue ofr,theinterval Iof(f)becomes (h) I:|x—1|<0.034. Weshall usetheusual twomethods ofﬁnding aseries solution of(a) satisfying (b). First Method. BySuccessive Diﬂerentiations. In(e),weknow by(b) (i) 1/(1)=1. y’(l) =0, 1/"(l) =2- Toﬁndthevalues ofsucceeding coeﬂicients in(e),weevaluate (a)andits successive derivatives at(1,1). These are,with at=1,y=1,y’=0,yr! =2’ (i)I/”=(2-1)2 +2/2+1/’-2.1/”(1)= 1-2=—1- 2/“’=2(w—1)+Zvv’+1/". v“’(1) =2-y(5) =2+2:,/yn +2(y/)2 +y///y y(5)(1) =2+4__ 1= -¢¢--¢----.....----’---------¢------ Substituting (i)and(j)in(e),weobtain (k)1/=1+a—1)*—s<i—1)“+a<2—1)*+a<x— 1>“+---. Lesson 39C Sniurs SOLUTION or.1Nonnmnm EQUATION, Onnsn n567 which aretheﬁrstsixterms ofaseries solution of(a)satisfying (b),valid atleast intheinterval Iof(h). Second Method. ByUndetermined Coeﬂicients. Since theinitial condi- tions have been given interms of:0=1,weseekaseries solution ofthe form (1) 1'/=¢1o+41($—1)'l'a2($"1)2+“3($—1)3+"' Itsnext three successive derivatives are (In) 1/(2)=<11+2az(2 —1)+3as(w —1)’+4<u(=v —1)“ +5aa(¢—1)‘+--'» I/'(¢) =2'12+6¢la(f¢ —1)+12¢14(1v —1)2+20as(f¢ -1)3+--', 1'/”'($) =603+24246” "1)+50056" _1)”+'''- Substituting (l)andtheﬁrstandlastequations of(m)in(a),weobtain (11)603—|—240469 "1)+601159? —'1)2—|—''' =(1 -1)2+l¢o+¢l1($"' 1)+¢lz($ "1)2'|-"'12 +111-l-2¢12($—1)-I-3¢1s(1>— 1):-I--~ —2- Equating coeﬁcients oflikepowers of(a:—1)in(n),wehave (0) 693=(102+01-2» 2404 =20001 +203, 60115 =- 1+211002 —|—G12 +303. By(b)andTheorem 37.24, weknow that (p) as=1, a1=0, G3=1. Therefore by(o)and(p), (<1) as=l(1-2)=‘*3, (14=214(2) =112: ¢s=316(1+2—‘l)=!14;~ Substituting (p)and(q)in(l),weobtain (I)1/a>=1+<x— 1)”-so=— 1)“+.1,<x- 1>*+aa-1)'*+---, which isthesame asthatobtained previously in(k). 568 SERIES Mnmons Chapter 9 EXERCISE 39 Obtain terms toorder koftheparticular solution ofeach ofthefollow- ingsystems ofdifferential equations, where kisthenumber shown along- sideeach equation. Usetwomethods. Also ﬁndaninterval ofconvergence ofeach series solution. da:1.Et- dz2.It dz:3.E E4'dz 5. Ed _=J+t %=e‘+@ wm=cmm=ck=¢ dy__g 1r__ 1_ _5-‘-{'22, Z(§)—2,]/(2)-—-1,k—4. %=¢~f,ﬂ®=QM®=Lk=£=1/sint, =t—|—a:2, . d=y+s1nt, Ey=x+cost, x(0)=0,y(0)=%»k=4. See(33.22), population problem. =ha:—lazy, %=kxy—py,where h,k,and pareconstants,dl and:2:(0) =1,y(0) =1,k=3. dz6.5 7.d=3+1’, 3%=yt,@(o)=-1,11(0)=1,k=4. -Z—':==a:yt, %=a:+t, a:(0)=l,y(0)=—1,k=4. Obtain terms toorder lcoftheparticular solution ofeach ofthefollow- ingdifferential equations, where kisshown alongside each equation. Use twomethods. Find aninterval ofconvergence. 8. yll II=Z2_'ll/21 =11 =0;k=6' 9-v=11/—(y')’, 14(0)=2,1/(0)=1,k=4- 10.1,"=¢2+siny,14/(0)=1/(0) =1,k=.5. ll.y” 12. yr]! 13.y" 14. ylll=cosa:—|— siny, y(0) =0,y'(0) =1,k=4. =yy’+wy,y(0)=0.1/(0) =1,1/'(0) =2,16=5- =1/l<>sy+r, y(0)=1,1/(0) =3,16=4- =U21°E$+ U’, =12 =0;y”(1) =11k=5' ANSWERS 39 3 4 3 Lz=¢+3+%+%+~»u=t+%+~»M<w- 2 2 2.,=1+(¢_;)+"8i*(,_1;) +:(,_:)3_2f<,_:)‘+...,6 2 48 2 Lesson 39—Exercise 569 12+4 1<+1>( 2"=1+?<‘*§)+JT" *3) + 3.2: 4-.2: 5.:c=1—|—(h 1/=1+(k 6.2: 7.:c=1 8.y 9.1; 10.y 2 l1.y =x+% l2.y =a:+a:2 13.y 14-. =1r=5-l"I+—12-1-12 #3 1r 14 1r T(‘-5) +ﬁ(‘"§> +""» l‘-§l<°°- $2 $5 2 3 4=§+%+"'» 1I=1'*t+g't“‘§t+‘H'i+"‘» r M<402+ 31+1) 2 1r 1r 1r21 1r4 ‘I’ 3 3=§t+¢ ‘FE! +"', y=§"l'¢+Z¢ +65 +Ei -l""', |t|<0°2 _k);+ t2 ha-3h2k+ 3hkp-hkz-if-I021)+21¢“£3+ 3, +, 2 _p)¢+ t2 hr’+3kp2-21?-3hkp+hzk+kzp-pa3 + 3! ‘+' =—1+t—¢’+§t”—%¢‘+---, y=1+i¢’+&t‘+--- 2 3 4 £2 ts t4 "%t'l'%¢‘l'it+"', y=—1+$'l"5"""6+ﬁ+"'- =1-—§:c2+%x4—§§-5a:6+---, |:c|<—————l»rarbitrary4(2r2 +21'+1) =2+1: —}x2+§:c3 —§:z:4—|—---, |:::|< »rarbitrary ex awn“4 5a: 2: r .+ﬁ_Z1B+..., |x|<F?ﬁ»rarb1trary. +—+ ~'-, Ixl<%r1‘8J‘bil3I‘8,X‘y. s_‘_¢,5:1: 1 .+ﬁ+i5—|—---, |x|< ,1arb1trary. =l+3:t+§a:3—-Q-x4+---, '”'< "<" U<x—1>’ <1»-1)‘ <x—1>“1+ 2+12'120 +"" t_1|<____L~__a,,<15[(r+1)210§(r+1)+T] 570 Smuas M1-rrnons Chapter 9 LESSON 40. Ordinary Points and Singularities ofaLinear Differential Equation. Method ofFrobenius. LESSON 4-0A. Ordinary Points and Singularities ofaLinear Dif- ferential Equation. Asremarked previously, power series methods are especially well suited forﬁnding solutions oflinear differential equations with nonconstant coefficients. These methods, however, cannot beapplied indiscriminately. Forexample, ifwetried toﬁndapower series solution of (40~1) wzy”+wy’+($2-—by=0 intheform (40-11) 9(1)=at+rm+we’+(ma+---, bythemethod ofsuccessive differentiations, wewould runinto trouble. By(37.26), a2=y"(0)/2!, andwhen as=0,weseefrom (40.1) that y” and, therefore, a2donotexist. Ifwetried tousethemethod ofundeter- mined coefficients, wewould ﬁnda0=0,a1=0,(Z2=0,---. The trouble arises because after division by:02in(40.1), thecoefficient ofy’which becomes 1/xisnotanalytic at:0=0,i.e.,itdoes nothave aMaclaurin series expansion inpowers of2:.Hence thehypothesis of Theorem 37.51 isnotsatisﬁed. Wedistinguish between points :00which satisfy thehypothesis ofTheorem 37.51 and those which donot by means ofDeﬁnitions 40.2 and40.22 which follow. Deﬁnition 40.2. Apoint :1:=:00iscalled anordinary point ofthe linear differential equation (40-21) y‘"’+Fn_1(w)y‘"_” +''-+F1(w)y' +Fo(fv)y =Q(w), ifeach function F0,F1,---,F,,_1, andQisanalytic atas=1:0.(Remem- berthismeans each function hasaTaylor series expansion inpowers of :0—-:00valid inaneighborhood of2:0.) ByTheorem 37.51, ifx=:00isanordinary point, then (40.21) hasa solution which isalsoanalytic ata:=1:0,i.e.,thesolution hasaTaylor series representation inpowers of(as—:00)valid inaneighborhood of:00. Deﬁnition 40.22. Apoint x=x0iscalled asingularity of(40.21), ifoneormore ofthefunctions F0(x), ---,F,,_1(a:), Q(x) isnotanalytic at:1:=1:0. ByDeﬁnition 40.22, thepoint 2:=0istherefore asingularity of(40.1). Fortheremainder ofthis lesson, weshall conﬁne ourattention toa second order linear equation (4023) 1/"+F1(w)y’ +F2(=v)y =0, Lesson 40A Onnmuw Pomr ANDSmoonuurr orLmrum Eqoxrron 571 where F1andF2arecontinuous functions ofanonacommon interval I. Itssingularities, ifthere areany,have been divided intotwokinds, regular singularities andirregular singularities. Deﬁnition 40.24. If:1:=:00isasingularity of(40.23) and ifthe multiplication ofF1(a:) by(:2:—x0)andofF2(x) by(x—a:0)2 result in functions, each ofwhich isanalytic at2:=2:0,then thepoint :0=x0is called aregular singularity of(40.23). Example 40.25. Show thatas=0and2:=1areregular singularities of (a) (w—1);)"+iy’—2y=0- Solution. Dividing (a)by(2:—l),weobtain 1 )_ 2 __(b) y"-lrﬁyy x_1?l—-0~ Comparing (b)with (40.23), weseethat 1 2 (0) F1($)=g-)-» F2($)="i' ByDeﬁnition 40.22, at=0and:0=1aresingularities of(b). Consider ﬁrstthepoint at=0.Following theinstructions inDeﬁnition 40.24, wemultiply F1by(:0—0)and F2by(as—0)”. There results respectively new functions 1/(:0—1)and ——2a:2/(:0 —1),each ofwhich isanalytic atx=0.Their Taylor series expansions areinfact (0) %=—(1+%+$2+"'), 2 -52-Z—T=2(:c2+a:3+x4+---), |:c|<1. Hence byDeﬁnition 40.24, :1:=0isaregular singularity of(b)and therefore of(a). Second weconsider thepoint 2:=1.Multiplication ofF1by(2:-—1) andofF2by(:1:—1)2give respectively (e) iand —2(:v —1), both ofwhich areanalytic at:1:=1.Their Taylor series expansions are respectively 1—-(at—1)—|—(:4:—-1)”—(:1:—1)3+---,and—2(x —1). Hence byDeﬁnition 40.24, :1:=1isaregular singularity of(b)andthere- foreof(a). Deﬁnition 40.26. If:0=:00isasingularity of(40.23) and ifthe multiplication ofF1(:c) by(:1:-—-2:0)and F2(:v) by(x—2:0)” result in 572 SERIES Mrrrnons Chapter 9 functions oneorboth ofwhich arenotanalytic at:1:=1:0,then thepoint 1:=2:0iscalled anirregular singularity of(40.23). Example 40.27. Show that x=0and:1:=1areirregular singulari- tiesof (a) (Z—1)”:/"+$5y'+2y=0- Solution. Division of(a)by(2:—1)2gives (bl y"+ V+(?€Wy=0- ByDeﬁnition (40.22), :1:=0anda:=1aresingularities of(b). Consider ﬁrst thepoint ze=0.Here F1= Multiplication ofF1by (re——0)gives 1/:c(:z: —-1)2,which isnotanalytic at:1:=0.Hence by Deﬁnition 40.26, at=0isanirregular singularity of(b)andtherefore of (a). Second, consider thepoint x=1.Multiplication ofF1by(0:——1) gives 1/:t2(:t —1)which isnotanalytic at:1:=1.Hence 2:=1isan irregular singularity of(b)andtherefore of(a). LESSON 4-0B. Solution ofaHomogeneous Linear Differential Equation About aRegular Singularity. Method ofFrobenius. If thelinear equation (40-3) y"+F1(¢)y’ +F2($)y =0 hasanirregular singularity at:1:=12:0,then theproblem ofﬁnding aseries solution istoodiﬁicult fordiscussion here. If,however, (40.3) hasa regular singularity at:0=2:0,then weshall describe amethod forﬁnding aseries solution, valid inaneighborhood ofx0.Itisknown asthemethod ofFrobenius. The series solution which Frobenius obtained, namely (40-31) 2/=(93"wolmlao +ll1($ _$0) —|—'l2(1P —f'>o)2—|—¢la($ —1410):;+"'1, 00#0. isknown asaFrobenius series. Note that when m=0orapositive integer, theseries becomes the usual Taylor series. However, fornegative values ofmorfornonintegral positive values ofm,(40.31) isnotaTaylor series. AFrobenius series, therefore, includes theTaylor series asaspecial case. Assume x0isaregular singularity of(40.3). Ittherefore follows by Deﬁnitions 40.22 and 40.24, that F1orF2orboth arenotanalytic at re=x0,butthat (re—2:0)F1 and (re—:c0)2F2 are. This means that F1 has(:0——x0)initsdenominator and/or F2has(re-—:c0)2 initsdenomina- Lesson 4-0B REGULAR Smounuurr. Mnrnon orFnonsmus 573 tor. Hence either F1(1:) =f1(1:)/(1: —1:0)orF2(1:) =f2(1:)/(1: -—1:0)2 or both F1andF2have these respective forms. Inanyevent, multiplication of(40.3) by(1:-—:e0)2 willtransform itintoanequation oftheform (40-311) (3-'$0)2y” -1'(I1'-$o)f1($)y' -l"f2($)1/ =0, inwhich both f1(1:) andf2(x) arenow analytic at1:=x0. The relevant theorem which willassure theexistence ofaFrobenius series solution (40.31) of(40.3) isthefollowing. Theorem 40.32. Let1:0bearegular singularity of(40311). Then (40311) hasatleast oneFrobenius series solution oftheform (40.31). Itis valid inthecommon interval ofconvergence off1(1:) andf2(1:) of(40311), except perhaps for1:=1:0,i.e.,ifeach Taylor series expansion off1(1:) and _f2(1:) isvalid intheinterval I:|x——1:0|<r,then atleast oneFrobenius series solution isalsovalid inI:I1:—-1:0]<rexcept perhaps for1:=1:0. Nolossingenerality results if,in(40.311), wetake 1:0=0since bya translation ofaxes wecanalways replace anexpansion inpowers of (1:—1:0)byoneinpowers ofx.With this understanding, werewrite (40.311) as (40-33) @211"+wf1(w)y' +f2(1)y =0, where thefunctions f1(1:) andf2(1:) areanalytic at1:=0.Hence each has aTaylor series expansion inpowers of1:valid inaneighborhood of1:=0. Let (40-34) f1($) =00+01$'1'02112 -l"'‘'1 f2($) =0o+0103'l'02$2+"'; betheir respective series expansions. The Frobenius series (40.31) and itsnext two derivatives are, with 130 =0, (40-35) 1/=¢”‘(¢o+we+(12132+---+aux"+---)=a01:’" +a11:'"+1+ a21:”‘+2 +----1-a,,1:"‘+" +---,a0#50, y’=a0m1:’”'1 +a1(m +1)1:’" -1-a2(m -1-2)1:’”*'1 -1---- -1-<l»(m -1"")$m+"_1 +''', y"=a0m(m -—1)1:""2 -1-a1m(m —|—1)1:""1 -1-¢l2("l -1-1)(m +2)?!” —|—''' +a..(m+v-1)(m+v)rv"‘+"" +---- The function y(z) willbeasolution of(40.33) ifitsatisﬁes theequation. Hence substituting (40.35) in(40.33) andreplacing atthesame time the 574 Snares METHODS Chapter 9 functions fl(1:) andf-2(1) bytheir series expansions in(40.34), weobtain (40.36) 1:h[a0m(m —1)1:""2 +alm(m +1)1:"“1 +--- +a,,(m —|—n-—1)(m +n)1:"‘+"_2 +---] -1"1li(b0 '1'01$'1‘02932 +-''') ><[a0mx"‘“‘ +a1(m+1)w"‘ +---+a..(m+n)=v’”+"_1 +---l +(¢o+clw+c2=v2+---) X(a01:"' +al1:"‘+1 +----1-a,,1:"‘+" -1----)E0. Expanding (40.36) and collecting coefficients oflikepowers of1:,there results (40.37) a0[m(m —1)—|—b0m+c0]1:"‘ +{m(m+1)m+b.(m+1)+c.1+a0lb1m+c.1}1’"+‘+la2[(m +2)(m+1)+(m(m+2)+60]+<11l01(m +1)+C1] +llolbzm +62]lw’"+2 +la3l(m+3)<m+2)+bo(m+3)+c.1+a.{b1(m+2)+¢.1 +a1[bz(m +1)+c2]+aolb-am +callrv"‘+3 +la..[(m+n)(m+ 11—1)+b<)(m+n) +60] +a...-1[b1(m +n—1)+vi] -l"an-2lb2("1 +71»""2)+62]+''' +aolbhm+c..131”"+---E0. Equation (40.37) willbeanidentity in1:,ifeach ofthecoefficients of1:", k=m,---,m+niszero. Since‘ wehave assumed a09'50,theﬁrst co- eﬂicient in(40.37) willbezero only if (40.38) m(m -—-1)-1-b0m —|—c0=0. This equation hasbeen given aspecial name. Itiscalled theindicial equation. Since itisaquadratic equation inm,ithastworoots. Let uscallthese roots mlandmg. These roots may be: * 1.Distinct, andtheir difference notequal toaninteger. 2.Distinct, andtheir difference equal toaninteger. 3.The same. Weshall consider each ofthese possibilities separately. Case 1.The roots mland mzoftheindicial equation are distinct and their difference isnot aninteger. Each oftheroots m=ml,andm=mgoftheindicial equation (40.38) willmake theﬁrst coeﬁicient in(40.37) zero. Weconcentrate ontheroot ml. Substituting itformintheremaining coefficients in(40.37) andsetting each equal to Lesson 4-0B SOLUTION ABOUT AREGULAR SINGULARI'l‘Y'-~CASE 1575 zero, willenable ustosolve each ofthese equations respectively foral, al,---,an,---interms ofa0.[Remember theb'sandc’sareconstants given by(40.34).] Since forthismlandforthissetofvalues ofal,---, a,,,each coefficient in(40.37) iszero, (40.37) isanidentity inre.Hence thesubstitution ofthismlandthissetofa’sintheﬁrstequation of(40.35) willmake y(1:) asolution of(40.33). Following thesame procedure outlined above, using thesecond root mg weobtain asecond setofvalues ofal,a2,---,an,---interms ofa0which with mgwillmake each coeﬂicient in(40.37) zero. Hence thesubstitution ofthissecond setofvalues ofthea’sandmgintheﬁrstequation of(40.35) willmake yasecond solution of(40.33). Comment 40.381. Not every equation oftheform (40.33) hastwo independent Frobenius series solutions. Some, asweshall show later, have only one. If,however, (40.33) hastwo Frobenius series solutions, then thefollowing relevant theorem, stated without proof, supplements Theorem 40.32. Theorem 40.39. ThetwoFrobenius series solutions of(40.33) are linearly independent. Each solution isvalid forevery 1:inthecommon in- terval ofconvergence offl(1:)andf2(1:) except perhaps for1:=0. Comment 40.391. Theorem 40.39 canalso bestated asfollows. Each Frobenius series solution willconverge forevery 1:,except perhaps forx=0,inacircle inthecomplex 1:plane, whose center isat0,and whose radius extends atleast tothenext nearest singularity of(40.33), i.e.,each solution isvalid atleast for0< <a,where aisthenearest singularity to0. Example 40.392. Find theinterval ofconvergence oftheseries solu- tionof (0) $231” +T'_,%c'5 1/’-'33/=0- Solution. Division of(a)byx2andapplication ofDeﬁnitions 40.22 and40.24 shows that1:=0isaregular singularity of(a). Comparing (a)with (40.33), weseethat 1f1(1i)=ﬁ=l.—-£132-I-Z64—1lia-l----,|I$|<l., f2(1:) =-3, —-co <1:<oo. Hence, byTheorem (40.39), each Frobenius series solution of(a)isvalid for <1,except perhaps for1:=0. Remark. You canverify that :l=iarealso regular singularities of (a).Since =l=iaresingularities, itfollows byComment 40.391, thateach 576 SERI1-:s METHODS Chapter 9 Frobenius series solution will converge forevery 1:,except perhaps for 1:=0,inacircle inthecomplex 1:plane ofatleast unit radius. This fact may help make clear why theinterval ofconvergence oftheseries representation of1/(1+1:2)is <1.Since thefunction iscontinuous forallreal1:,itwould seem that itought tohave aseries representation valid forallrealre. Example 40.393. Find aFrobenius series solution of (a) 121”+m(m+1):/’—(wz+1)v=0~ Solution. Division by1:2and application ofDeﬁnitions 40.22 and 40.24, shows that 1:=0isaregular singularity of(a). Wetherefore seek aFrobenius series solution oftheform (40.31), namely (b) ll=$”'(0o '1"01$—|—(12912 +'''),009"0- Ifwewish, wecandifferentiate (b)twice toobtain y’andy",substitute these values in(a)andthen equate thecoeﬂicient ofeach likepower of1: tozero. Butsince wealready didthiswork inobtaining (40.37), wemay aswell make useofthisequation. Touseit,weneed toknow thevalues oftheb'sandc’sinit.By(40.34), these arethecoeﬂicients intheseries representation offl(1:)and_f2(x). Comparing (a)with (40.33) weseethat (0) f1(w)=1+wandf2(1°)=-if-$2. both ofwhich arealready inseries form. Hence comparing (c)with (40.34), wehave (0) 00=3, 01==1, 00=“"4. 61=0, 62=""1- Allremaining b’sandc’sarezero. With these values ofb0andc0,thein- dicial equation (40.38) becomes (e) m(m—1)+§m—§~=0, 2m2-m—1=0, whose roots arem=1,m=-1}. Since these roots aredistinct anddo notdiffer byaninteger, themethod ofthiscaseisapplicable. Following themethod outlined above, wesubstitute theroot m=1inthere- maining coeﬁicients in(40.37) andsolve each foral,a2,---,interms of a0.Themost effective waytouse(40.37) istosetthecoeﬁicient of1:"‘+" equal tozero. When youdothis,keepinmind thatm=1,b0,bl,c0,cl,62 have thevalues in(d)andthatb2,b3,---,c3,cl,---arezero. Substituting these values inthecoefficient of1:"‘+", weobtain (f) ¢1..[(1+v)(1+ n-—1)+%(l+11)-ll +an-1[1(l +n—1)+01+a.._z[0 —1]=0- Lesson 4-0B Sontrrron Aaotrr AREGULAR Smoumu1rrr—C1isr. 1577 Solving foran,there results 2 (2) wi a.=—M.._1 +4..-» Formula (g)isarecursion formula. Itwillgivethevalue ofanforeach ng2.Before wecanuseit,therefore, wemust ﬁndal.Setting thesecond coefficient in(40.37) equal tozero, wehave, with m=1andthevalues ofb0,c0,bl,clasgiven in(d), (11) <11[(2)(1) +l¢(l+1)—4]+doll+0l=0.01=-ﬁve Wecannowuse(g)and(h)toobtain, when (i) n=2: 702=—201+¢lo=%0o+41o='g0o» ¢12=%a0; n=3: 27 27 2 41 82 _§¢ls= '-3412-l'<11 =-"$410 '"5¢1o= —ﬁ¢lo, (la="-E00. n=4: 328 9 571 5712204';—4G3+ﬂ2=§1'5Go+§5Go=§4-500, a4= a0. Substituting thevalue m=1,andtheabove values ofthea’sin(b),we have . 2 9 82 571 (J) U1=00-77(1—"5$+ﬁ172—§Z5$3+ﬁ)T7§6$4"""')' which aretheﬁrstﬁveterms ofoneseries solution of(a). Toobtain a second solution, weusetherootm=—§andproceed asabove. There- cursion formula, obtained from thecoeﬂicient of:e"‘+" in(40.37) becomes, with m=—§andthevalues given in(d), (k) <1..[(v—l¢)("-ii)+%(v"-l)+(—%)l +an-lln — +an-2l0 —ll=0) which simpliﬁes to 2?- an='_(n'_g)an-1_l'an—2) Asbefore, weﬁndalbysetting thesecond coefficient in(40.37) equal to zero. There results (H1) 01140-4) +(l#)(l) —51+¢1o(—l) =0.<11=-00- 578 Sann-:s Mrrrnoos chapter 9 Hence from (1)and(m),weobtain, when (11) "=2!a2=—t01+¢l0=tao+¢1o=%00i "=35'i'a3="‘%a2+a1='-i*10_a0=***4q0o, ¢la=—i‘§¢1o; n=4:10al =—§a;; —|—a2=§-§a0 +=§a0 =15139-a0, al,=-H-goo. Substituting m=—-Qand thevalues oftheabove setofa’sin(b),we obtain fortheﬁrst ﬁveterms ofasecond series solution of(a) (O) 1/2=aow‘”2(1 —w+21:’—4}-tr’+1-18¢‘ —-~-)- ByTheorem 40.39, thetwo series solutions arelinearly independent. Hence byTheorem 19.3 andComment 19.41, ageneral solution of(a)is alinear combination of(j)and(o),namely 2 9 82 571(p) y=cl1:<1—-51:+ﬁ1:2—%1:3+m)~:v4—---) _ 3 13 119+021: ”2(1-1:-l-51:2-—E1:3+§-661:4----)» where a0hasbeen incorporated into thearbitrary constants clandC2. Weobserve from (c)thatfl(1:) andf2(1:) arepolynomials. Hence, by Comment 37.53, their series representations arevalid forall1:.Therefore, byTheorem 40.39, each Frobenius series solution isvalid forall1:except perhaps for1:=0.Inthisexample thesolution (j)isvalid forallat.The second solution (o),because ofthepresence ofz‘1/2, isvalid forall1: except x=0.Hence thegeneral solution (p)isvalid for0<|1:|<oo. Note thatfor1:<0,thesecond series in(p)isimaginary. Wecanmake itrealbychoosing c2=ci,where cisreal. Case 2.The roots ml, "lgoftheindicial equation differ byaninteger. Ifthetworoots oftheindicial equation (40.38) differ by anonzero integer, wecanwrite them asmandm-1-N,where Nisa positive integer. Since m+Nisarootof(40.38), itsatisﬁes thisequa- tion. Hence, (40.4) (m+N)(m+N —1)+b0(m+N) +60 =0. Now compare theleftsideof(40.4) with thecoeﬂicient ofaninthelast term of(40.37). They both willbeexactly thesame ifnisreplaced byN. Thismeans thatifweusethesmaller rootmin(40.37) toﬁndasetofvalues forthea’swhich will make thecoefficient ofeach 1:"zero, wewill be stopped when wereach theterm inwhich al.;appears, since itscoeﬂicient willbezero. Hence wecannot solve thisequation foralvinterms ofprevi- ousa’sunless byaccident, theremaining terms intheequation alsoadd tozero. Inthiscase, theequation willbesatisﬁed foranyarbitrary value Lesson 40B SOLUTION Anotrr AREGULAR SmeU1.Anrrr—-CAsE 2A579 of01v.Wecanthen continue todetermine values ofsucceeding a’s,i.e.,of a1v.,.l, aN.,.2, ---interms ofa0anday. Iftherefore theroots oftheindicial equation (40.38) differ byapositive integer N,twopossibilities may occur. Each isconsidered separately in Cases 2Aand2Bbelow. Case 2A. The coeﬂicient ofaNin(40.37) iszero and theremain- ingterms inthecoefﬁcient ofx"'+N also add tozero. Inthiscase the larger rootm+Nwilldetermine, by(40.37), asetofvalues ofthea’sin terms ofa0;thesmaller rootmwilldetermine twosetsofvalues ofthea’s, oneinterms ofa0andtheother interms ofalv.However, theFrobenius series solution obtained bythelarger rootinterms ofa0willnotbelinearly independent oftheoneinterms ofa0obtained bythesmaller root. Hence thesmaller rootalone willgive, inthiscase, twoindependent solutions, whose linear combination willbeageneral solution of(40.33). Itwillhave twoarbitrary constants a0anday.These solutions willbevalid inthe same intervals given inTheorem 40.39. Example 40.41. Find ageneral solution of (a) I21”+av’+($2— y=0- [No'rr:. This equation isknown astheBessel equation ofindex Q.A discussion ofthegeneral Bessel equation ofindex nwillbefound inLesson 42.] Solution. ByDeﬁnitions 40.22 and40.24, itcanbeveriﬁed that 1:=0isaregular singularity of(a). Hence weseek aFrobenius series solution oftheform (b) y=:c"‘(a0 +alx+a21:2 +---). Comparing (a)with (40.33), weseethat (<1) f1(1) =1and f2(I) =-=1+$2, bothofwhich arealready inseries form. Hence comparing (c)with (40.34), wehave (<1) 1).,=1. C0= -i, C1= 0, 62= Allremaining b’sandc’sarezero. Therefore theindicial equation (40.38) becomes (e) m2__4=0) whose roots arem=1}andm=—§. These roots differ bytheinteger 1. Hence forthis example N=1.When m=—§, which isthesmaller 580 Snnms METHODS Chapter 9 root, weobtain, bysetting thesecond coefficient in(40.37) equal tozero (f) ¢11l(t)("-if) +=3—fl+110(0) =0,0111+0110=0- Thecoefﬁcient ofaliszero, buttheother term initsequation isalsozero. Hence thisCase 2Aisapplicable. (Nora. Since N=1,alv=al.We therefore could have anticipated from what wesaidabove, that theco- eﬂicient ofalwould bezero. Wecould notofcourse have anticipated that thecoefﬁcient ofa0would alsobezero.) Any values ofa0andalwill satisfy thelastequation in(f). Theremaining a’s,namely a2,a3,--- obtained byequating each subsequent coeﬁicient in(40.37) tozero, will riow besolvable interms ofanarbitrary a0andanarbitrary al. With m=—1§, therecursion formula obtained bysetting theco- efﬁcient ofthegeneral term 1:"‘+" in(40.37) equal tozero, becomes, with thehelp of(d), (g) ¢1»[(—i +")("-3 +11)+(-1+11)-111+an-1(0) +11»-2(1) =0) which simpliﬁes to (h) (n2——n)a,, =—-a,,_2, ng2. From itweﬁnd (remember a0andalarenow arbitrary) when (i)n=2:a2=-—=}a0, n=3:all=—<l~al, n=4=<u=~1‘m=r‘zao. "=5¢“5=—elc<13="1"irc01) "=01 06=“"s16“4="‘*ri"60o) "=7? 01=—zizas=“"5'o1I6a1- Substituting m=—-llandthevalues oftheabove a’sin(b),itbecomes . _ 1 1 1(1) 1'/=ao$ 1/2(1-§$2+ﬁ$4—'m$6+"') 2 4 6 which aretheﬁrst seven terms ofageneral series solution of(b). Weob- serve from (c)thatfl(x) andf2(:v) arepolynomials. Hence, byComment 37.53, their series representations arevalid forall1:.Therefore, byTheorem 40.39, each Frobenius series solution converges forall1:,except perhaps for1:=0.Here theﬁrst series converges forall1:except 1:=0,the second forall1:.The general solution (j)converges for0< <oo. Note that for1:<0,theﬁrst series isimaginary. Wecanmake itreal bychoosing a0=ciwhere cisreal. Case 2B. The coefficient ofaNin(40.37) iszero, but theremain- ingterms inthecoefficient ofx"'+N donotadd tozero. Inthiscase only thelarger root m+Noftheindicial equation (40.38) willdetermine Lesson 4-0B Soumon ABOUT AREGULAR SINGULARITY—CASE 2B581 asetofvalues ofthea’sinterms ofao.There willtherefore beonlyone Frobenius series solution of(40.33). Comment 40.5. Asecond independent solution of(40.33), ithas been proved, willbeoftheform (40-51) !l2($)=14(1)——bu!/1(¢¢)108$, I>0, where Nisthepositive integral diﬁerence between theroots oftheindicial equation (40.38), 1/1isaFrobenius series solution of(40.33) obtained with thelarger rootm+N,andu(:c)isaFrobenius series oftheform (40.52) u(a:) =:z:'"(bo +b1:c+b2Z2 +---). In(40.52), misthesmaller ofthetworoots of(40.38). Bysubstituting (40.51), (40.52), andthenecessary derivatives in(40.33), youwillﬁnd thaty2willbeasolution of(40.33), if(seeExercise 40,14) (40-53) wzu”+wfm'+fa"=b1v[2wy1’ +(fl—-1)1/1l- If,in(40.53), younowsubstitute uanditsderivatives asdetermined by (40.52), theFrobenius series solution ylanditsderivative, then the resulting equation willenable youtoﬁndthevalues oftheb’sin(40.52). Weshall, however, notdiscuss thismatter further since logarithmic solutions, asthese solutions arecalled, have limited applications. See Exercise 40,14-17. Example 40.531. Find asolution of (a) $211”—w(2—¢)y’+(2+w’)y=0- Solution. Division by2:2andapplication ofDeﬁnitions 40.22 and 40.24 shows that:1:=0isaregular singularity of(a).Hence weseeka. Frobenius series oftheform y=$m(Go +G123 +(12132 +'‘'). Comparing (a)with (40.33), weseethat (9) f1($) =-'2+Ivand f2(¢) =2-1'12, both ofwhich already areinseries form. Hence, comparing (c)with (40.34), weﬁnd bo= -2, bl= c0=2, c1=0, c¢=1. Alltheremaining b’sandc’sarezero. Theindicial equation (40.38) be- comes (e) m(m—1)—2m+2=0, m2—3m+2=0, whose roots arem =landm =2.These roots differ byanintegerN =1. 582 Sanrns Mnmons Chapter 9 Using thesmaller rootm=1andthevalues in(d),weobtain bysetting thesecond coefficient in(40.37) equal tozero, (f) a1(2 ——4+2)+a°(1 +O)=0, Oal+ao=0. The coefficient ofay=a1iszero, butthecoeﬂicient ofaoisnot. Hence thisroot m=1willnotlead toasolution. Asremarked attheoutset ofthislesson, only thelarger root inthiscase willgive asolution. Therefore using thelarger root m=2,weobtain bysetting thesecond coeﬁicient in(40.37) equal tozero, (s) a1(3-2—2-3+2)+2a0=0, a1=——a°. The recursion formula, obtained bysetting thecoeﬁicient of:z:"'+" in (40.37) equal tozero becomes, with thehelp of(d), (11) ¢1nl(2 +n)(1+n)-2(2+1»)+2]+(1+n)¢1n_1 +an_2 =9, which simpliﬁes to (i) (n2+n)a,, =——(n+l)a,,_1 —a,,._2, ng2. By(g)and(i),when (j) n= 2: 6112=——3a1 —ao=3ao—ao, a2=%1 n=3:12a3= ——4a¢-—a1= ——§a0+ao, a3= —g—g- Substituting m=2andtheabove values ofthea’sin(b),wehave 2 a (kl y1=llo5'?2(1"‘1$+%'-%—""')v which aretheﬁrst four terms ofaFrobenius series solution of(a). By (c),flandf2arepolynomials. Their series representations arethere- fore, byComment 37.53, valid forall2:.Hence, byTheorem 40.32, y;is alsovalid forallav. ByComment 40.5, asecond solution of(a)willhave theform [here N, thedifference oftheroots of(40.38), isone] (1) 1/2(w)=u(w)—b1y1(w)1<>s@, I>0, where u(:c) isaseries oftheform (m) u(:c) =:v(b0 —|—bla:-1-b2IC2 —|—---). SeeExercise 40,15. Lesson 40B SOLUTION ABOUT AREGULAR SINGULARI'l'Y—CASE 3583 Case 3.Roots ofindicial equation (40.38) equal. Iftheroots ofthe indicial equation (40.38) areequal, itisevident that only onesetofa’s andtherefore only oneFrobenius series solution of(40.33) canbeobtained from (40.37). Example 40.6. Find ageneral solution of (a) 1:21/" +xy’+xzy=0. (Norm. Thisequation isknown asBessel’s equation ofindex zero. Afuller discussion ofthegeneral Bessel equation ofindex nwillbefound inLesson 42.] Solution. Division by1:2and application ofDeﬁnitions 40.22 and 40.24, shows that as=0isaregular singularity of(a). Hence weseek a series solution oftheform (b) y=:z:"'(ao -l-alas-1-(Z2332 —|—---). Comparing (a)with (40.33), weseethat (0) f1($) =1»f2($) =$2, both ofwhich arealready inseries form, valid, byComment 37.53, for all2:.Comparing (c)with (40.34), weﬁnd bo= 1, Co=0, C1=0, C2=' 1. Alltheremaining b'sand c’sarezero. The indicial equation (40.38) therefore becomes (e) m(m—1)+m=0, m2=0, whose roots arem=0twice. Hence wecanexpect only oneFrobenius solution from these roots. Using therootm=0,weobtain from thesecond coeﬁicient in(40.37), (r) a1(l+0)+110(0)=0,a1=0. Setting thecoefficient ofa:"‘+" in(40.37) equal tozero, weobtain, with thehelpof(d),therecursion formula (E) <1»[(")(n —1)+(1)(")] +<1»-1(0) +11»-2(1) =0, which simpliﬁes to (11) nzan=—<1»_2, an=:%‘f3» n22. 584 Snares Mnmoos Chapter 9 From (f)and(h),when (i) n=2: a¢= -—-éao, n=3:a3=—§l§a1=0, n=4: a4=—ia2=%a0,42 22-41n=5: a5=—€éa3=0, __ _ 1 11L—6. a5——§04=— ao. Substituting in(b)these values ofthea’sandm=0,weobtain the Frobenius series solution . __ 2:2 :04 2:6 1:8 ) (1)1/1"a°<1 _22+22.42 _'22.42.62+22.42.62.82 ByTheorem 40.32 thisseries isvalid forall2:. Asecond solution of(a)canbefound bymeans ofthesubstitution (40.51), with N=0.(Remember Nisthedifference between theroots.) This solution isthecoeﬁicient ofc2in(k)below. Hence thegeneral solu- tionof(a)is 2 (kl y=61y1+62(%— $4+ $6+"' 1+.]l+...+£ 2 ..+(—1)”+1 w’ +---+1/ilosw)» w>0, where ylisgiven in(j). EXERCISE 40 1.Determine thesingularities ofeach ofthefollowing diﬁerential equations. Also indicate whether they areregular orirregular singularities. (a)(:1:—1)3z2y" —2(:c—1)a:y’ —3y=0. (b)(1—1)’r‘y” +2(1=——1)1u'—y=0-(c)(a:+ 1)”;/" +my’—(2:-l)y=0. Verify thattheorigin isaregular singularity ofeach oftheequations 2-6andthattheroots oftheindicial equation (40.38) donotdiffer byan integer. Find, bythemethod ofFrobenius, twoindependent solutions of each equation andintervals ofconvergence. 2-$211”+w(==+l):/'+ml=0-3.21:21/" +3:1/'—|—(21:-—-1)y=0. 4-Zwz/”+(1+1)u’+31/=0-5.2:c2y" -—-xy’+(1—22);; =0. 6-2(w’+1’):/’-—(I—31”)?!’ +11-=0- Lesson 40-Exercise 585 Verify that theorigin isaregular singularity ofeach oftheequations 7-9andthat theroots oftheindicial equation (40.38) differ byaninteger. Each equation, however, hastwoindependent Frobenius series solutions. Find these solutions andintervals ofconvergence. 7.12y” —-2:21;’—|—(x2—2)y=0. 8.x2y"+ (1+ :c3)a:y’ -—-y=0. 9-121/"+11/’+($2-—by=0- Verify thattheorigin isaregular singularity ofeach oftheequations 10and11,that theroots oftheindicial equation (40.38) differ byan integer, and that each equation hasonly oneFrobenius series solution. Find thissolution andaninterval ofconvergence. 10.22y” +2:y’—|—(x2—1)y=0. ll.xzy” +ray’+($2—-4)y=0. Verify that theorigin isaregular singularity ofeach oftheequations 12and13andthat there isonly oneroot oftheindicial equation (40.38). Hence, there isonly oneFrobenius series solution. Find thissolution and aninterval ofconvergence. 12.12y" —-3:cy'+4(:c+1)y=0. 13-11/”+(1—Z)!/'+$1!=0- 14.Prove that thesubstitution inequation (40.33) ofy-g(2:), yg'(a:), 1/2’’(1)as determined by(40.51) willyield (40.53). Hint. Usethefactthaty1(:c) isa solution of(40.33). 15.InExample 40.531, wefound onesolution of(a),namely 2 3 y1(1)=1’(1—¢+%-§—6—--~)- Byuseof(40.51), (40.52), and(40.53), ﬁndasecond solution of(a).Hint. See(1)and(m)oftheexample fortheform ofthesecond solution yg(:|:) andofu(a:). Substitute u,u’,u",yl,y1'in(40.53). Thefunctions f1andfg aregiven in(c). Equating coeﬂicients oflikepowers ofac,willdetermine, interms ofbo,those values ofb1,b2,---needed in(1)and(m). 16.Follow theinstructions given inproblem 15toﬁndthesecond solution of(a) ofExample 40.6. Itisgiven in(k).Hint. Each bwithoddsubscript iszero. Each bwith aneven subscript hastwoinitsnumerator. Break upthetwo into1+1.Two fractions thusresult. Onegives bqyl, where 1/1isthesolu- tion(j);theother gives thecoefﬁcient of62in(k). 17.With theaidofthehintgiven inproblem 15,ﬁndthegeneral solution of each ofthefollowing equations. (a)12y” —-3:2:y’—|—4(a:+1)y=0.Seeproblem 12forsolution y1(a:). (b)xzy” —z(2—5a:)y’ +(2:—6a:2)y =0. 18.Themethod ofFrobenius mayalsobeapplied toﬁndaparticular solution of thenonhomogeneous linear equation (40-61) @211”+‘1f1(1)y' +f2(1)y =Q(=v), where f1(:12),f2(:c) areanalytic ata:=0andQ(x) isafunction that canbe expressed asaFrobenius series. (40-62) Q(r) =1”(bo +(7133+62112+---)- 586 Snares Msrnons Chaplgf 9 Ifneither rootoftheindicial equation (40.38) exceeds n—1byapositive integer, then atrialsolution 1/,,(:r) willbeofthesame form asQ(x), namely 1/.»(==)==="(ao+a1+azx’+---)- Substitutins um). 1//(w). 1//'(r) intheoriginal equation andequating coeﬁicients oflikepowers of2:,willde- termine a0,a1,as,---.Foreach ofthefollowing equations ﬁndtheroots oftheindicial equation, then verify that these roots donotexceed n—1 byapositive integer, andﬁnally ﬁndaparticular solution. (a)xy”+y’—-22:1;=42+ 2’. Hint. Multiply theequation byxto determine n,f1(:r), andfz(:t). (b)2121/" +2:1/'+my=sinx+22:2cos2:.Hint. Replace sinxandcosz bytheir Maclaurin series expansions. Aseries solution, valid inaneighborhood of2=0,may notbeofmuch practical useforlarge values of2:since toomany terms intheseries maybe needed toobtain adesired degree ofaccuracy. Insuch cases, itisbestto make thesubstitution 1 1 du 1'lt=;; $=;r -a;='—;='-M2, ﬂ=4ldl=_,,=Q,dz: duda: du d2y dudy zdzydu as"-2“na"“ ma _adi! 4d21/-214 H-I-14 ms andsolve theresulting equation foryinpowers ofu.Asolution y(u) fora small value ofuinthenewequation willthen correspond toasolution y(z) foralarge value of1:intheoriginal equation. Using theabove substitutions, ﬁndaFrobenius series solution, ﬁrstinpowers ofu,then inpowers of1/2:, ofeach ofthefollowing. (a)2:c2(x -—-1);/'+ x(3z +l)y'-—-2y==0. (b)21¢“:/"+c’s’+2/=0- (c)$21/'+ w<5—32/’+ (7~91) =Z-+5-+%. 2/»<>nly- See problem 18. Also determine aninterval ofconvergence ineach case. Alinear differential equation Fz(:c)y” +F1(:c)y' +Fo(x)y =0issaidto have aregular singularity atsoifthesubstitution initof(40.61) results inanequation which hasaregular singularity atu=0.Show thateach of thefollowing equations (a)w‘:/”—31’:/’+ (1—w)v=0.(b)theLegendre equation (1—a:'*')y" -—-2:01/'+k(k—|—1)y=0, hasaregular singularity atw.InthecaseoftheLegendre equation, ﬁnd aFrobenius series solution ﬁrstinpowers ofu,then inpowers of1/x. Show thattheseries isconvergent for >1. Lesson 4-0—Exercise 587 21.Thefollowing equation t (40-7) =(1—-1)!/"+[Y—(H+B+1)¢l1/' —-aB'Y=0. where a,B,‘Yareconstants, isknown asGauss’s equation orasthehyper- geometric equation. Verify that: (a)2:=0isaregular singularity of(40.7). (b)Theroots oftheindicial equation (40.38) arem=0andm=1—‘Y. M+m+®M+m+® @““=n¢wImFﬁnWW~"=“*”"'where ao,a1,a-2,---,an,---arethecoeflieients intheFrobenius series y=:r"'(a0 +a1:r+ ag2:2+ a3:c3+---),an940.Hint. Substitute yand itsderivatives directly in(40.7) andequate tozerothecoefficient of:c""'"‘. (d)Solutions, using ﬁrsttheroot m=0andthen theroot m=1—‘Y, with as—1,arerespectively2 (40.11) 1/1=r<a.nm 1)=1+‘Z,-4w+95"-+T:.1’§;§_”3,",*—‘) % (+1)-~-(<1+n—1)B(B+1)~--(ﬂ+n—1) 1:"+"'+aa ~/(~/+1)---01+»-1) H+"" 'y7£0|—11"'2:—'3|"' w=,ap+e;:§n;1¢n,+m] =371—7F(a—7+1|B_7+1:2_7;x)r 7762|314|"'- Thenotation F(a—7+1,)3—‘Y+1,2—7;x)means, replace inthe 1/1equationof (40.71),abya —7+ 1,ﬁbyﬁ —'Y+1,'Yby2 —‘Y. (e)Both solutions arethesame when ‘Y=1.Hint. Note thatwhen ‘Y=1, therootm =1—-‘Y =0. (f)If‘Yisnotaninteger, thegeneral solution of(40.7) is (40-72) ll=cm+62112. convergent for <1except perhaps at2:=0.If‘Yisaninteger, then only oneofg1orygisasolution of(40.7). Theseries ontheright ofylin(40.71) isknown asthehypergeometrie series. ByTheorem 37.16, itdeﬁnes, ontheinterval <1,afunction, called thehyper-geometric function, which wehave designated by F(a,B,'Y;:c). IfaorBiszero oranegative integer, thehypergeornetric series terminates andthehypergeometric function isapolynomial. 22.Using theresults obtained inproblem 21,ﬁndaseries solution ofeach of thefollowing differential equations. (a):r(1—z)y" +(-3-—22:);/' —1y=0. Hint. In(40.7), ‘Y=-2-, a+B+1= 2,043 =4.Thereforea =4,5=§,‘Y =§. (b):c(1—:c)y” +(2-—-4:r)y' +-2y=0. Hint. In(40.7), ‘Y=2, a+B+1 ==4,015 =2.Thereforea =1,B =2,7=2. 588 Ssmss METHODS Chapter 9 23.Show thatthesubstitution (40.73) a:= (r2—r1)u+r1, u=ln, %=-—1--7T2-—T1 dit T2-——T1 ﬂ=ﬂ%=_..1...€£da: duda: T2—T1 du' d2y= d 1 dy = 1 dzydu dz? da:T2-—r1du T2—r1 duzdz: 1 dz intheequation (40.74) cl~mo—1'2):/"+2<1-13>;/+§7=0 willtransform itintotheGauss equation (40.7), namely (40.75) u(1—u)g+[§ -§u]%—§y =0. Aseries solution of(40.75) is,therefore, seeproblem 21, (40.76) y=c1F(a,ﬁ,'Y; u)+czul-7F(a —‘Y+1,5—‘Y+1,2—‘Y;u), where (40.77) 7=§£‘-ll“-Z. a+B+1=ll: a}3=c/a.¢l(T1—1'2) a Hence, by(40.73) and(40.76), asolution of(40.74) is (40.78) y=c1F(a,;3,'Y;l1—)7'2_T1 I—-T1 1;’ It—T1 -|'62<-;2—_T_1) F(d—'Y+1,B—'Y+1,2—'Y;;;-1;)‘ Thenotation Fd,ﬁ,'Y; H means replace xin(40.71) by5:-7-1 -T2—1'1 T2——T1 24.With theaidofproblem 23,ﬁndaseries solution ofeach ofthefollowing. (11)(w—2)(w+1):/”+%(w+1)?/’-in=0~ . 7Hint. r1=2,12 =-—-1,73 =—1,§= 51% =—g,a: =—3u+2. (b)(a:—2)(:r—1)y"+4xy'+2y =0. Hint. r1=2,r2=1,r3=0,a=l,b=4,c=2,:2:=—u+2. 25.Prove each ofthefollowing identities. (a)F(1,|3,/3; 2:)=% -Hint. Seeanswer toproblem 22(b). (b)F(¢1,l3,I3; 1)=(1"-1)“- (c):cF(1,1,2; —:c) =log(1+2:). (d):cF(§,§,§; 2:2)=sin"1 :22.Hint. Seeanswer toproblem 22(a). (8) 32) = _'$2)‘/2' Lesson 40—Answers 589 26.With theaidofproblem 23,ﬁndaseries solution ofthefollowing equation, known asTschebyseheﬂ"s equation, (40.79) (1:-—-1)(:t +1)y" +my’—nzy =0. Hint. r1=1,rg =-—l,r:; =0,a=1,b=1,c= —n2,:r= —2u+1, a=n,B=—n,‘Y=§. 27.With theaidof(40.71) andtheanswer given inproblem 26,ﬁnd the Tschebyscheff polynomials T0(a:), T1(:z:), T2(I), T3(x), T4(a:). ANSWERS 40 1.(a):4:=0regular singularity, 2:=1irregular singularity. (b)a:=0 irregular singularity, at=1regular singularity. (c)2:=—1irregular singularity. 2 s 2.y1=c1:2:1I2(1—x+%_%-|-~-~),alla:; yg=c2(1— 2z+§x2 —-f5:c3+ ---), allx. 3.y1=01:1/2(1—§a:+-335:r2—-9-1-513+---),alla:,' yg=cg:|:'1(1+2a: —-2a:2—|—§:|:3+ ---),:c 9'50. 3 4.y1=c1(1—3:r+2:rz—%+---),alla:; 2 3 yg=c2x1I2(1—%+%—%+---),allz.. 2 4 5.y1=c1a:(1+%+$+---),all:c; 2 4 yg=cg:2:1I2(1+%--l--E-t@+---)1allx. =(c1x”2+ c2:c)(1 —a:+2:2-—-23+ ---), <1. =c1r“(1+ tr-t412-£1‘-"'),1 #0; 7/2=czw’(1+t¢+ 21012-s1o$3—"'), all»3 6 8.3/1=c1:c<1-%+i%6----)»alla:; 3 0 yg=c2a:"1(1+£3——gé-+--~)»:c#0. e~—1 1<~>*11 9-y1'-131$ 5 -—"'!8,ll$, 2 4 U2=62$6_1I3|:l—1—:1_-?(g) —----:|»2;#0. 2 4 1°-”="”l1"i(i)+w+>e(i)+""l'“““-_2 11:2 1 :0411-11-63 1_§ 5 5 +"'r&ll$. 12.y(z) =c:r2(1— 4:c+4a:2 —155-;t3+§:r:4 ——-—-), allx.7"?‘§@s-I 590Ssams Mm-aons Chapter 9 15.y=bo[:2:(l+a:-—-§:|:2+§§x3- -—-)—-y1loga:], x>0. a17.(a)y=y1—|—c2[:r2(8a: —122:2+1,77% +---)+y1log2:],where 1/1is given inproblem 12. (b)111=c1r3(l -—41+92”-—---). 1/2=¢2l(1+ 41-$42-"‘2%3+ "-)+'5}!11108=vl-2 3 1s.<a)7. =¢”(1+§+%+%+---)- 2 coy.=1(1+§~%----)- 2 3 19-e)y=¢.(1+2u+3§‘-+‘—%‘-+---) 2 3 2 7 112 ll=¢1<l+;+ﬁ+%+"') _,,2 3 22 484 __ +‘Z’ (1+3::+15¢?+315:z:3+ Theuseries converges for <1.Hence the2:series converges for ]:t|>1. 1 1 1 “’”=‘*(‘-a+m-W+"')1/2 1 1 1+6256 Z950. c)7.= ~20. ,0_M)=ml.[1_1212-1)x-2+k(k—1)<k—2)<:-3)z_.+___2(-1) 22-2!(2k -1)(2-3) (—1)"l=(k —1)(k —2)(k—3)---(k—211+ 1)-.. +2»»!(21.- 1)(2k—3)---(2k——2n+l) ”2 (Iv-l-1)(k+ 2)(k+ 3)(k+ 4)- +22-2!(2k+3)(2k+5) "4+“ +<k+1)<k+2)<k+s)--~<k.+2n) x_..___]_2"n!(2k +a)(2r+5)---(21.+2n+1) 22-c)7=v1F(i.i.§; x)+m""”F<<).<).i; »)Z 312 5923 _1/2 =61 1+§+T0'+m+"' +6237 - <1»)71=c.r<1.2.2;») =v1(1+w+=v2+---) =,—°_‘—Z- Lesson 4-IA THE Lrosnnna DIFFERENTIAL Equxrron 591 2 You canverify that yz=cg2:"1F(l,2,2; 2:)=$5 isalsoasolution. Here oz=1,13=2,‘Y=2.Note thatinthisspecial case, thesecond and subsequent numerators anddenominators ofthesolution yzof(40.71) are both zero. 24-.(a)Thetransformed equation is d’ 17d3 "<1"“>a7Z+(§ —§“)ai+§” =°- Herea =3,5=—-},'Y = y=c1F(3.—i,%; 11)+¢2111—7'2F(%.—3,~¥;11)5/2 1/=01F(3.—i, 1}; +62 F(L-3,~i‘; - (b)U=¢1F(1,2,3; 2—11)+62(2—1)'7F(—6,—5,'"6; 2'"¢)- 26-y»(I) =01F(",—",§§ 1/2 _ +¢2(-1-5-2) F(n+i,—n+%,=§;L—2-l)- 27.To(:c) =1,T1(:c) =9:,Tz(:1:) =21:2—1,T3(a:) =4:3-—31:, T4,(:c) =82:4—82:2+1. LESSON 4-1. The Legendre Differential Equation. Legendre Functions. Legendre Polynomials P;,(x). Properties ofLegendre Polynomials Pk(x). LESSON 41A. The Legendre Dilferential Equation. The linear differential equation (41-1) (1-1v2)y" —21¢!/'+k(l¢+Dy=0, where kisarealconstant, occurs inmany physical problems. Itisknown as theLegendre equation, named after theFrench mathematician A.M. Legendre (1752-1833). Because ofitsgreat practical importance, this equation hasbeen studied bymany mathematicians andaconsiderable amount ofliterature isavailable onthesubject. You canverify byDeﬁnition 40.2, that 1:=0isanordinary point of (41.1). This equation cantherefore besolved bythepower series method outlined inLesson 37B. Dividing (41.1) by1—2:2andthen comparing theresulting equation with (37.5), weseethat (11.11) roe)=%'% =he+1><1+$2+Z4+--->, 2f1(:v) =—%= —2a:(l+x2+x4+---). Both series arevalid for <1.Hence byTheorem 37.51, (41.1) hasa 592Ssiuas Mwraons Chapter 9 series solution inpowers ofxwhich isalsovalid for|:c|<1.Wetherefore seekaMaclaurin series solution oftheform (41.12) y(z) =ao—l—alx+GQIC2 +a3:v3 +--- +aux”+a..+1x"+‘ +11».+2w"+’ +---- Successive differentiations of(41.12) give (41.13) y’(:c) =a,+2a2a: +3a3:c2 +--- +1w1..w"" +(n+l)11..+1w"+ (n+2)1a+zrv”+‘ +---, y”(x) =2a,+2-3a3:c +---+n(n—l)a,,a:"_2 +n(n+1)a..+1w"" +(11+1)(11+2)a»+2w” +---. Asusual ywillbeasolution of(41.1), ifthesubstitution initof(41.12) and(41.13) isanidentity inx.Making these substitutions, weobtain (41.14) (1-—-a:2)[2a2 —|—2-3a;;:c +---+n(n-—1)a,,a:"‘2 +n(n+ 1)<1»+11v"“+ (11+1)(n+2)a..+2=v" ---1 —2111411+21121:+3112.111" +---+M..w"“‘ ---1 +k(k+ l)[ao+¢m=+a2w”+---+¢m:"+---1 =0- Expanding andcollecting coefficients oflikepowers of:0,wehave (41.15) [2412 +Ic(k+1)a0] +[3!a3 —2a;-1-lc(k+1)a1]:z +[1204 -'5412+750°-1-llazllz +''' +[(11+1)(n+2)a»+z ""n(n+1)11»+WC+1)11»l1=" =0- Hence wemust choose thesetofa’ssothat (41.16) 2a;+ k(lc+l)ao = 2-3413- 2a1—|—Ic(k—l-1)a1= 3-4a4 -—2-3a2+lc(k+ 1)a2 = 4-5415 -—3-4a;;+k(lc+ 1)a3 = 5-6a; -—-4-5a4+Ic(Ic+1)a4 = (n+1)(n+2)a,,+2 ——n(n+l)a,,+Ic(k+1)a,,=0. Therefore, by(41.16),99999 rG2 =' iQ G9, a3= a1 =_ a1, Lesson 41B LEGENDRE POLYNOMIALS P;,(a:) 593 —-Ic Ic— la..._.6_%~:_1_> ..=_<__21>_;_1i> .. __k(k+1)(k—2)(k+3)a* 4' 0! (15:12-1;g¢+1)a3=_(11-3;ék+4)a3 _<1—1)<k+2)<k—a)c+4)a“ 5! 1' a6=2o- I§()k+1)a4= _(11-4gék+5)a4 —k(k+1)(k-2)(k+3)(k—4)(k+5)a6| O! -----.---¢¢¢¢¢--¢--..~---------.-.-------- a_(—1)"k(7° -1-1)(k—2)(7°+3)(7¢—4)(k+5)'''(70+2""1)a211 * 0! 0=(—1)"(k -1)(/¢+2)(k -3)(Ic+4)(k -5)(k-1-6)---(k+2n)al“+1 (2n+1)! ' Substituting (41.17) in(41.12), weobtain (41-18) y(z)=aoS).(w) +a1T1(rv), where aoanda1arearbitrary, and (41.2) Sm)=[1_k<k; 1)x2+k<k+ 1)<k4T2)(k+s) x.+___ <—1)"k<k+1)<k—2)<k+s)<k—4)---<1+2 -1)..+ (2n)l n $2+ jl, (41.21) Tk(.,=[,,_(k—13)§k+2),3+<k—1)<k+2g(k—3)(k+4)x.+___ +<-1)»<1 -1><k+2()2<I;;f))!<k+4) ---</¢+2n) $2.“+__ LESSON 41B. Comments ontheSolution (41.18) oftheLegendre Equation (41.1). Legendre Functions. Legendre Polynomials P,,(x). Thefollowing comments apply tothesolution (41.18). 1.The solution (41.18) isvalid for <1.Seeremarks immediately following (41.11). I? 2.When kiszero orapositive even integer, 2,4,6,---,oranegative odd integer, --1, —3,---,theseries in(41.2) terminates, i.e.,itbecomes a 594- Ssmns METHODS Chapter 9 polynomial in2:.Forexample, when k=2,thethird andallsucceeding terms in(41.2) arezero; when Ic=-—1,thesecond andallsucceeding terms arezero. 3.When kisapositive oddinteger, 1,3,5,---,oranegative even integer, —2, -4,---,theseries in(41.21) terminates, i.e.,itbecomes apoly- nomial in:11.Forexample, when Ic=3,thethird and allsucceeding terms in(41.21) arezero. When k=—2,thesecond andallsucceeding terms arezero. 4.Ifkisnotaninteger orzero, both series (41.2) and(41.21) arenon- terminating. ByTheorem 37.16, therefore, each deﬁnes, foreach Ic,a continuous function onI: <1.These functions arecalled Legendre functions. 5.Since aTaylor series isaspecial caseofaFrobenius series, theLegendre functions (oraLegendre function and apolynomial) deﬁned bythe series (41.2) and (41.21) are,byTheorem 40.39, linearly independent. And since each isasolution of(41.1), itfollows byTheorem 19.3 and Comment 19.41, that y(z) of(41.18) isageneral solution of(41.1) on I:|x[<1. Because ofComments 2and3above, wecanwrite polynomial solutions oftheLegendre differential equation (41.1) when kiszero oraninteger. Allweneed doisselect theproper series (41.2) or(41.21) forthegiven Ic. InTable 41.22, wehave listed solutions of(41.1) forafewinteger values oflcandforlc=0. Table 41.22 ASolution byIflc= TheLegendre Equation (41.1) Becomes (41.2) or(41.21) IS (1—=2):/”~—212/’=0 m(m)=1 (1—1%"—21y’+2y=0 1/1(1)=11 (1—w2)y"—Zry’+62=0 :1/2(r)=1—312 (1—11%” —212/’+12y=0 ys(I) =I——%=v3 (1—x2)y" —2:cy'+20y=0 y4(a:) =1-10x2+ -33i:t4 vPOOtQ|—1O Ifwesolve thelastequation in(41.16) fora,.+2, weobtain (41.23) 4...=@ 4.,n=0,1,2,3,--- Since thenumerator of(41.23) isequal to—(lc —n)(lc +n+1),wecan write (41.23) as (41.24) 4,,”=- 11.. Lesson 4-1B LEGENDRE POLYNOMIALS P;,(x) 595 Therefore by(41.24), iflcisaninteger, and 22k——1(41.25) n=Ic——2: ak=— ak_2, 4210-3ak_2=— Gk_4, Therefore by(41.25), (41.26) G],_2 =— Gk. (k——2)(k -3) k(k—1)(k —2)(k —-3)aM=——mF3T“*=7WﬁTmwIw% Substituting (41.26) in(41.12), weobtain polynomial solutions p;.(:v) of Legendre’s equation (41.1) oftheform [write y(z) indescending powers ofx,namely y(z) =---a;.:v'° +a;.__2:v"_2 -1-a;._4:c"_4 —|—---] (41.27) _1c—no—mc—ac—oc—oJ4+H],2-4-6(2k-1)(2k-3)(2k-5) 15:01]-221"’: where a).isanarbitrary constant. Letustake forittheparticular value (41.3)_(21):_1-2-3-4---(21¢- 1)(2k)_ 1-3-5---(2k— 1)_“'=*2'=(k!)2 * 112-4-6---(214) _ kl (The reason forselecting this value ofa).willbemade clear inLesson 41C-A.) Todistinguish between thepolynomial solutions p).(:c) of(41.27) and those weshall obtain bygiving a).thevalue (41.3), wereplace pk byPk.Hence by(41.3) and(41.27) k(k—-1)(k—2)(k—-3) _+2-4(2k—-1)(2k—-3) “k4 _ko—nc—mc—ac—oc—eka+H_2-4-6(2k-—1)(2k——3)(2k—5) ” (—1)"k(k—1)(k—2)---(k—2 +1) _..+2»n1(214-1)(2/4-3)---(214-2:+1)”k 2 1.-=0,1,2,-~ 596 Seams Marnons Chapter 9 Bydeﬁnition, s!=1-2-3---s.Hence, (41.311) (s+1)(s+ 2)---(s+n) r-1»-1NJIQ0909 69 an-8(8+1)--'(-*1+n)= (8+n)!_ By(41.2.11) (41312) (24+2)(24+4)---(28+2n) =2"<s+1)---<s+n) = By'(4.311) and(4l.312), (41.2.13) (28+1)(2s+3)---(24+2n-1) _(2s+1)(2s +2)(2s +3)(2s +4)---(2s—|—2n—1)(2s +2n) _ (2s+2) (23+4)--- (2s+2n) (2s+2n)! _ (2s)! ___s!(2s -1-2n)! _ _2"(s +n)!_2"(2s)!(s —|—n)! s! In(41.313), lets=k-—n.There results (41314) (21-211+1)(2k-2n+3) ---(211-1)= Bythedeﬁnition ofkl, kl Substituting (41.315) and(41.314) inthecoefficient of:v"‘2" of(41.31), weobtain forthecoefﬁcient (41316) (—1)"k! 2"(2k-2n)!Ic! _(-1)"(k!)2(2k -21.):_" (14-2n)!2"n!(k-n)!(2k)l“n!(2k)!(k -2n)!(k-n)! Since (41.316) isthecoeﬂicient ofthegeneral term of(41.31), wecanwrite (41.31), with 1-3-5---(2k-—1)/kl replaced byitsequivalent fac- torial form asgiven in(41.3) after ah,as In ___ n __ (41.31?) P;,(x)=‘fl% :0"-2»_1|:-0 Thesymbol [Ic/2] means thelargest integer ink/2. Forexample, ifk=3, thelargest integer ink/2isone. Hence toevaluate P3(x), replace kby3 in(41.317)]_-land sum thetwoterms obtained with n=Oandn=1. Ifk=4,t'elargest integer inlc/2is2. Lesson 41B L1-zormmn POLYNOMIALS P;,(:a) 597 By(41.31) or(41.317), when (41.32) 1.=0:P°(:z:)=1. lo=1:P1(a:) =x. 1=2;1-12(4)= (42-=é(342-1). k=3:P3(:z:) = (223— =é(5:03 -—-31:). 1--5-7 214=4¢ P.(@)=3+ﬂ-(44-6%+%) =,3;(3514-30.42+3). Thepolynomials in(41.32) areknown asLegendre polynomials. Ifintherecursion formula (41.24), weletn=—k—3,——k-—-5,etc., weobtain, inplace of(41.27), (41.33) q;.(a:) =a_),_1[x"°_1 + x"‘_3 c+nc+ac+ac+o 4- H]_+ 2-4(21¢+s)(21+s) ”5+ Since a_;,_1 isanarbitrary constant wemay take forittheparticular value 2"(m)’(41.34) a_;,_1 = _ k!2-4-6---(210) _ k! -1-2-3---(21¢)(2k+1)' 1-3-5---(2k+1)' lcaé-—1, —-2,--- Thefunctions X[x-In-1 + g%2) x—k-3 o+no+ac+aw+o __H]+ 2-4(‘2k+3)(2k+5) ”E6+ ’ 1441-1,-2,---, obtained from (41.33) byreplacing a_;,_1 byitsvalue in(41.34) are known asLegendre functions ofthesecond kind. 598 Seams Mrrnons Chapter 9 LESSON 41C. Properties ofLegendre Polynomials P;(x). The Legendre polynomials areimportant primarily intheinterval I:|x|§1. Thegraphs oftheﬁrstﬁveLegendre polynomials intheinterval 0§2:§1 areshown inFig.41.4. 1.0 _ ‘*P.(x) - P1(x) 0.5- P405) .'5 0.5 Figure 41.4- Ithasbeen proved that aLegendre polynomial ofdegree nhasexactly ndistinct realzeros intheinterval —1§:1:§1.Ifwehadextended the graphs shown inFig.41.4 toinclude theinterval -1§:1:§0,wewould have found that P0(a:) hasnozeros, P1(:c) hasonezero, P2(:z:) hastwo zeros, P3(:z:) hasthree zeros, and P4(x) hasfour zeros intheinterval —1éasé1. Intheremainder ofthislesson weshall develop anumber ofproperties ofLegendre polynomials which arevalid intheinterval I: §1. llA.The Legendre polynomials P,,(x) arethecoeﬂicients oftin theMaclaurin series expansion of(1—-2xt-1-t2)_1/2, i.e., (41.41) (1-24¢+:2)-1/1' =P0(a:)+P1(x)t +15(4):’ +--- +P..(w)1" +---, §1sltl< 1' Proof. Weshall merely give theformal steps oftheproof, without attempting tojustify each ofthesteps used. Arigorous proof isbeyond thescope ofthistext. TheMaclaurin series expansion of(1+u)",called Lesson 41C Pnormvrms orLEGENDRE POLYNOMIALS P;,(x) 599 abinomial series, is (a)(1+u>”=1+ku+i'-°5}—1lu”+--- +k(k—1)..;L§k—-n+1)u,,+___, valid forthose values ofuandkforwhich theseries converges. IfIo=—§, theseries converges for|u|<1.Letusnow write theleft-hand member of(41.41) as[1+t(t—2:c)]'1/2. Hence by(a),ifIt”—2a:t| <1, (b)[1+c(z- 2:c)]_1/2= 1+%¢(2@- t)+2-1,g¢’(2¢- 0*+--- + t"(2x_t)"+... =1+%(2a:t—t2)+§%t2(4:z:2-—4:ct+t2)+--- + t"(2x_t)"+... _ From thelastexpression ontheright of(b),weseethatthecoeﬂicients oft°,t,andt2arerespectively (0) 1,é(2:0)=:0,-%+5%(4102)=é(sf_1). Acomparison of(c)with (41.32) shows thatthese coeﬂicients arerespec- tively P0,P1,P2.Thecoeﬁicient ofthegeneral term t"isthesumofthe coefficients oft"inthelastterm ontheright of(b)andofcoefﬁcients oft“ inpreceding terms. Hence thetotal coeﬂicient oft"is (d) ex)"—1"§§.§{(;,' 9"1)T3)("Q1)<2x>""’ +1- (Tl'-' * (2x)ﬂ_4 __'__ The lastterm in(d)canbewritten as (6) 1-3-5---(2n—5)(2n-—3)(2n-—1)_(n)(n—1) 2"-2 (2n-—3)(2n -—1)(n)(n —1)(n—2)! X(n_2;?‘ "'3)2n—4xn-4 =1-3-5---(211 —1)[n(n—l)(n —-2)(n— 3)x,,_4]_n! _2-4(2n -—1)(2n —-3) 600 Smums Mmnons Chapter 9 Hence (d)becomes, with itslastterm replaced bytheexpression onthe right of(e), 1.3.5...(2 _1)(f)i__7”i’___ Acomparison of(f),which isthecoeﬁicient P,,(x) oft"of(41.41), with thepolynomial P;,(:c) of(41.31) shows thatwith lcreplaced byn,theﬁrst three terms ofeach arealike. Ifwehadused more terms ofthebinomial series (d),wewould have obtained additional terms in(f)inagreement with (41.31). Nolealsothesimilarity ofthecoeﬁicienl in(f)with thevalue ofahasgiven in(41.3). This value wasinfactgiven toakinorder toob- taintheidentity (41.41). B.Values ofP,,(0), P,,(1), and P,,(—1). If:0=1,theleftside of (41.41) simpliﬁes to (a) (1-2:+:2)-1/2 =(1-l)-1. Itsseries expansion is (b) 1+¢+z’+z’+---+¢"+---, |:|<1. Comparing thecoeﬂicients ofthisseries with thecoeiﬁcients oftheseries ontheright of(41.41), weseethat, with 1:=1, (41-42) Po(1) =1, P1(1) =1, P2(1) =1,'‘wP1-(1) =1- Ifas=0,theleftsideof(41.41) becomes (¢) (1+#2)-1' 2, whose series expansion is (<1) 1—&¢’+&-&¢‘—---+(—1>" ¢2"+---, It!<1- Acomparison ofthisseries with theright sideof(41.41) shows that, with 1:=0, (41-43) P1(0) =0, Pa(0) =0,''',P2»-1(0) =0- (41~44) 130(0) =1, P2(0) =—%, P4(0) =i'2!''', (Zn —-1) I-' N6:#1s3%Pz,.(0) =(~1)" Lesson 41C Pnormvrms orLsormnnn POLYNOMIALS P;,(a:) 601 If:1:=—1,theleftsideof(41.41) becomes (e) (1+2t+l2)‘1/2 =(1+z)"1 =1—z+z’—t=’+t‘— ---, |t|<1. Comparing (e)with theright sideof(41.41), weﬁnd, with :1:=—-1, (41-45) Po(—1) =1,P1(-1) =-1, Pz(-1) =1, Ps(—1)= —1,"', Pn(_1) =(—l)"- C.Recursion formula forP,,(x). Differentiating (41.41) withrespect tot,Weobtain (a) —-§(1-2a:l+l2)_3/2(--21: +2:)=P,+2P2:+---+nP,,¢"-1 +---. Multiplying (a)by1—-2xl—|—l2,there results (b) (:2:—-t)(1~—2a:t+la)“/2 =(1——2a:t+t2) ><[P1++2P2l+---+(n—1)P.._1¢"*” +nP,,¢"—‘ +(11+1)P»+1t" +---]- By(41.41), wecanwrite (b)as (9) $lPo+P1l+' ''+Pnl”+' '‘l_[Poi-l'P1i2+ '''+P»-1i”+' ' =[P1+2P2t+---+ (n+1)P,,+1l"+---] _2,,[pl,;_|....+npnt"_|_...] +lP1i2+"'+("'- 1)Pn-1i”+"‘l- Equating thecoefficient ofl”onboth sides oftheequal sign, weobtain xpn —Pn-1 = (n+1)Pn+1 _217nPn +(n_1)Pn—11 which simpliﬁes totherecursion formula (41-46) (H+1)P'n+l(x) —(211+1)wPn(I) +nP1._1(1) =0, n=1,2,--- Itiseasily veriﬁed, by(41.32), that (41.461) P1(x) -—a:P0(x) =0, and, by(41.46), that when (41.47) n=1: 2P2(ac) -—3xP1(x) +P0(:c) =0, n=2: 3P3(:c) —5zP2(:c) —|—2P1(a:) =0, etc. Weleave ittoyouasanexercise toverify, bymeans of(41.32), that theformulas in(41.47) areindeed true equations. 602 Snnnss METHODS Chapter 9 D. Rodriguc’s Formula. Another compact formula forexpressing theLegendre polynomials P,,(a:) is 1|. 1d ,,(41.48) P,,(x)=WIE;($2-1). Itisknown asRodrigue’s formula. Proof oftheFormula. The binomial series expansion of(ac—1)” canbewritten as " 1 _1n= ___1 n___l7'_;____ n-Ic- <a> cc>2“>,,(n_,,,,w Hence ,1__”_,, n! ,,_ Taking nsuccessive derivatives ofa:2"'2",, weobtain forthenthderivative, (0)%(@”"-2") =(211-2k)(2n -2k-1)---(n-21¢+1)¢""'°, 2k§n, =0, n<2k. Bydeﬁnition ofthefactorial function, (c)canbewritten as dn 2n—2k _(2"—27¢)! n-2k =O, n<2k. Hence by(b)and(d), d" ,,‘"2‘ ,. !(2-21¢)!,._ <8)Z:F(‘”2_ 1)1;,(*1) 2" Therefore by(41.48) and(e), n/2 __ <f>P-<w>=2'<—1>" ”""'° amga\ o320/'\_,,-_2-»_—;i_ = <1-)= 2"Ic!(n -—k)!(n ——2k)! Acomparison of(f)with (41.3l7) shows they arealike. (Interchange n andloineither equation.) E.Orthogonal Property ofLegendre Polynomials. Deﬁnition 41.5. Asetoffunctions fl,f2,---,issaid tobeorthog- onal onaninterval I:a§:1:§b,if,forevery two distinct functions Lesson 41C PROPERTIES orL1-zormnns Pommomxns P;,(a:) 603 oftheset, b (41.51) /f,,,(:c)f,,(x) dx=0,m96n. Weshall now prove that thesetofLegendre polynomials isorthogonal ontheinterval I:-1§as§1,i.e.,weshall prove 1 (4l.511) /_1P,,,(:r)P,,(:t) da:=0,m75n. Proof. Each Legendre polynomial P,,,(x) satisﬁes theLegendre equa- tion(41.1). Therefore (a) (1—w2)Pm”(w) —-2xPm'(w) +m(m+1)P»-(w) =0, which canbewritten as (b) i[(1-:v2)P'(@)1+ m(m+1)P(1)=0.drc '” "' Multiplying (b)byP,,(:v), naém,andintegrating between thelimits —1, 1,weobtain (C)1 1 fP,.(=v)§{<1—x”>P..'(x>1-11 +m(m+1)fP.<x>P..<x> dz=0. _1 I _1 Theﬁrstintegral in(c)canbeintegrated byparts inaccordance with theformula (d) /udv=uv——/vdu. Intheﬁrst integral of(c),letu=P,,(:v) anddvequal therestofthein- tegrand. Then by(d),thisintegral becomes 1 1 (e) P,,[(1-:c2)P,,,’] _1-L1(1-:c2)P,,,’P,,'dx. You caneasily verify that theﬁrst term in(e)iszero. Therefore, by(e), (c)simpliﬁes to 1 1 (f) -/1(1-z2)P,,,’P,,'dx +m(m+1)IIP,,P,,,drc=0. Ifwehad started with P,,(x) in(a)instead ofwith P,,,(.r) andfollowed thesteps outlined above, wewould have obtained thesame result asin(f) 604 SERIES Mnrnoos Chapter 9 with mandninterchanged. Hence also 1 1 (2) -I1(1-x2)P,,'P,,,’ dx+n(n+1)/l1P,,,P,,dz=0. Subtracting (g)from (f),there results, with m96n, 1 (h) (m2-1-m—n2-—n)/‘-1 P,,,(a:)P,,(:c) dx=0. Since m2+m—n’—n= (m—-n)(m+n+1) and maén,we may divide (h)bythisterm toobtain (41.511). F.Other Integral Properties ofLegendre Polynomials. Theﬁrst integral property weshall prove is, 1 (41.52) /1R,,,(x)P,,(a:) dz=0, where R,,,(x) isapolynomial ofdegree mlessthan n. Proof of(41.52). Let (a) Rm(w) =do+we+@2932+~--+am-1w"“‘ +amw". Thepolynomial R,,,(x) canalways bewritten asalinear combination of theLegendre polynomials Po,P1,---,P,,,,i.e.,wecanalways Write (bl Rm($) =ac+01$'1'0-21132 +'''+am-1I$”'_1 +amilim =Cope -1-C1131 +''-+Cm—1Pm—1 +CmPm- The truth ofthisstatement canbedemonstrated asfollows. Pmisthe only polynomial ontheright of(b)which hasaterm in:c"'.Hence wecan choose C,,,sothat thecoeflicient ofthe:c"‘term ofP,,,isequal toam. Next wesum thetwocoefficients of:c""'1 inP,,,_1 andP,,,,andgive C,,,_1 avalue sothatC,,,_1 times thissumequals a,,,_.1, etc. In(41.52) replace R,,,(x) byitsequal lastexpression in(b). Wethus obtain 1 1 l (C) Co‘/’1P0PndZ+C1-ll1P1Pnd13+"'+Cm‘/‘1PmPndIB, m<n. By(41.511), each term in(c)iszero. Hence (41.52) follows. Thesecond integral property weshall prove is 1 2(41.53) /Q1[P,,(x)]2da: =275- Lesson 4-1-Exercise 605 Proof of(41.53). Squaring both sides of(41.41), weobtain (a) (1—2:ct+t2)'l = P,,(1:)t":|2 . 1|.—0 Integration of(a)with respect to1:between thelimits -1and1gives 1 1 G (b) f1(1-211+1*)-1dx=/1[ZP,,(1)1"]2d1._ _ n-0 When wesquare theintegrand ontheright of(b)andperform theinte- gration, theonly terms, by(41.511), which arenotzero arethose inwhich thesubscripts ofP,,(1:) arethesame. Hence (b)simpliﬁes to 1 Q 1 (C) II(1-21¢+1’)-1d1:=Z12";I(P,,(1)]2d1._ "=0 _ Integration oftheleftside of(c)results in(remember tisaconstant in thisintegration andlog(1+l)=l—-§l2-1-§t3-—---) (<1) _1 _ -‘__1 <*~ﬁ_1 1+1 2tlog(1 2:ct+t)_l- 2tlog(1+t)2-tlog1_t a s 12 l l l =2h+§+§+r+"it2 t4 t2n =2[1-1-'5-FE-l"""l'm-1'-l-"-:|» |t|<1. Substituting thelastexpression of(d)fortheleftsideof(c),wehave (e) ﬁ . 1 i t2n/,1 1l—0 2n + 1 ‘II-0 -1 Equating coeﬂicients ofl2"in(e),weobtain foreach n, 1 c) L§awWa=%%q- EXERCISE 4-1 1.Find thegeneral solution ofeach ofthefollowing Legendre equations. (a)(1—12):!/”—Zn/’+21/=0-(b)(1-—w"’)y”——2111/’+61/=0- 2.iind apolynomial solution oftheLegendre equation (1-—1:2);/” ——2zy’ 30y=0. 3.Find theLegendre polynomials P5(:c), P6(1:), P7(a:). 606 SERIES Maruons Chapter 9 4- 5 6 7 8 9 (=1) 12. 13. 1410. llFind thegeneral solution of(1-—a:2);|/' -2zy’—|—ﬂy=0. Verify bydirect substitution in(41.32) that Po(1) =1.P1(1) =1.P2(1) =1.Ps(1) =1,P-1(1) =1; Po(-1) =1,P1(—1) =—1,P2(~—1) =1,Pa(—l) =-1.1’-1(—1) =1; Po(0) =1,P1(0) =0,P2(0) =—1,Ps(0) =0,P4(0) =i- Verify, byuseof(41.32), theaccuracy oftheequations in(41.47). Express 1:3asalinear combination ofLegendre polynomials. Express 1:4+21:3+21:2—-1:—-3as1linear combination ofLegendre polynomials. Assume thatafunction f(z), deﬁned intheinterval (—1,1), canberepre- sented byaseries ofLegendre polynomials, i.e.,assume f(z)=¢oPo(1)+ c1P1(w) +¢2P2(1)+ -~-- Show that ifthisseries, after multiplication byP;,(:::), canbeintegrated term byterm, then thecoeflicients co,c1,---,aregiven by 1 ck=2%‘/_l f(:c)P;,(a:) dz. Hint. Multiply (a)byP;,(:c), integrate from --1to1,then use(41.511) and(-11.53).Prove thatP21-(-22) =P2»(z);Pg,.+1(—-1:) =——P2,,+1(It). Prove each ofthefollowing identities: (a)1:P,,'(a:) -P,._1’(1:) =nP..(a:). Hint. Let u=(1-—211+ t’)"’2 =Po(w) +P1(w)t+ --~+P,.(r)t"+ ---- First show (:1:-t)(6u/81:) isequal tot(8u/dt). Then perform theindi- cated operations andequate coefficients oft". (b)P,.+1'(a:) -—(n+ 1)P,,(1:) =1:P,,’(a:). Hint. Differentiate (41.46), then substitute forP,._.1'(1:) itsvalue asgiven in(a)ofthisproblem. (c)($2—-1)P,.'(1:) =nxP,,(a:) —-nP1.-1(w). Hint. Multiply (a)ofthis problem by1:;replace n+1in(b)byn,thentakethedifference between thetworesulting equations. Verify thattheformula inproblem l1(a) isvalid forP2,P3,P4. Show thatthehypergeometric series F(ls—|—1,-k,1; isasolution oftheLegendre equation (41.1). Hint. Write (41.1) as(:1:-1)(1:+1)y” +21:y'—k(lc+ l)y=0.Then follow theprocedure given inExercise 40.23. Here r1=1, 1‘g=—1, r3=0, a=1, b=2, c=—k(k—|-1),1:= —2u -1-1.Thetransformed Gauss equation (40.7) becomes 2 1-<1—u)%+<1—2u>f—Z+k<k+1>y =0. whose solution isF(k-1-1,—-k,1;u). Verify each ofthefollowing. 1 1 <1)/_11P3<w>1’d» =1-(b)/_11P.<1>1’dw =$- Lesson 4-l—Exercise 607 15.Thefollowing differential equation (41.6) y"-—-21:1/' +2ky =0,kreal, isknown astheHermite equation. Show thatitssolution, valid forall1:,is (41.61) 2 2 22 4 23 (5 y1.(a:)=ao1—-ﬁkx +Il0(lt—2)1 2 —|—a1[:c -gut -1)@3+%(k —1)(k -3)? 23 -ﬁ(1=- 1)(k—-3)(k—-5)z7+---] (*l)n2n 2n =a0[1+Z-E57-1.(1=-2)(1.-4)+---+(1-2n+2)41n=l ' (-)"” 2-.1+ah+i@fﬁ%o—ne—o~o—%+nz+]n=1 ' 16.Find polynomial solutions of(41.6) fork=0,1,2,---,7.Hint. Note that fork=0orapositive integer, oneoftheseries in(41.61) terminates. Ans. yo(1:) ya(1) Z/5(1) 111(1)“O: a1(x _5'13): l/4(x) ¢l1(1 '-$13+1*51¢5). 1/6(1!) a1(:c -—-21:3+§1:5—1-8-51:7).a(1(1 —21:2), ao(l -—41:2+§1:4), a0(1 ——61:2+41:4-15516),1/10¢) =air, 1/2(1) 17.Thefollowing polynomials, known asHermite polynomials, H0(¢) H101) H201) Hs($) H4($) H5($) H6(1?) H7(a:)11 21:, 41:2—2, 81:3—-121:, 16334 —4812 —|—12, 3215 —16013 —|—I201, 6416 —48014 —|—72012 -—-120, l28It7 ——13445 +336013 -—-16802;, canbeobtained from thesolutions given inproblem 16,bychoosing appro- priate values forananda1.Show thatthese Hermite polynomials canalso beobtained from theformula (41.62) H,.(1) =(-1)"¢“’ i4"’, 11=0,1,2,---dx" 18.Insolving (41.6), thefollowing recursion formula results, (8') an+2 =2(n——k) o+nm+m“” 608 Serums METHODS Chapter 9 Following exactly thesteps from (41.24) on,obtain polynomial solutions oftheHermite equation (41.6), oftheform (i.e.,indescending powers of1:) <1»)hk($)=ah1—i —i—?-— (0) 19 20. 1 2. 3. 4-. 7.1:3k(k-1)_k(k-1)(k-2)(k-3)"“ lk 22“k2+ 21 Z24 k(k—1)(k—2)---(k -5) "-6 _ 3! '26 _|_... (-—l)"k(k _1)---(k-211+1)J-2" + n! 22"[k/2] n k—2n(-1)k!=Gkg %1 k=0,1,2,"'. In(b),a),isanarbitrary constant. Ifwetakeforitthevalue 2",weobtain theHermite polynomials —1"k! _,,Hk(1)= ,((%W<21>" 2,k=0,1,2,----ZME3 Verify thattheabove formula (0)alsogives theHermite polynomials shown inproblem 17above. Inthesummation in(c)ofproblem 18above, andin(4l.317), ncannot be larger than [Io/2]. Why? Show thatthecoeﬂicient oft"intheseries expansion ofe2""‘° inpowers of tisH,,(:c)/n! Hint. e2""" =e2“e"". Write theseries expansion ofeach term ofthisproduct, multiply both series, then show thecoeﬂicient oft" isH,,(z)/n!. ANSWERS 41 (a)y=a0(1_<1;1)x2+<1+1)<1;2)<1+s>I4+___)+m_ <1»)1/=<10(1—3:”)+a1(»-£1“ +<2—1)<2+2;§2 -s)<2+4> x5+___)_ y=:c——3§:c3 +3Q1:5. P5(:c) =§(63:c5 —702:3 —|—15:0); P6(a:) =13,-;(231z5 -—31514 +10512 —5); P-1(1) =-ll;-(42917 —69315 +31513 ——352:). y=a0(1_ 1<1;!r1)12+1<1+1><14T2)<i+3>x4+___) (—)(+) +a1_i___12.T§_1,,3 _(;-1><a+2)<#—s><1+4)z5+___)_! =§P3<x>+2101(1). 5 Lesson 42A THE Bsssnr. Drrrsnmmrar Eqmrrron 609 8.-595P4(a:) +§P3(a:) +§-fP¢(a:) +§P1(1)-¥§P0(a:). 19.Forn>[k/2], k—2nisanegative integer and(lo—2n)!isundeﬁned for negative integers. LESSON 4-2. The Bessel Diﬁerential Equation. Bessel Function oftheFirst Kind ];,(x). Differential Equations Leading toaBessel Equation. Properties of];,(x). LESSON 42A. The Bessel Differential Equation. Another differen- tialequation which arises frequently inphysical problems andabout which much hasbeen written isknown astheBessel equation, named inhonor oftheGerman mathematician F.W.Bessel (1784-1846). Itis (42-1) $21/"+wy’+(12—k’)y=0, where kisapositive constant orzero. Youcanverify bydeﬁnitions 40.22 and40.24 that :1:=0isaregular singularity of(42.1). Hence weseek a Frobenius series solution oftheform (4211) 2/=w"‘(a<> +rm+an’+---+anr"+---)- Comparing (42.1) with (40.33), Weseethat (42.12) f1(a:) =1and f2(x) =—Ic2 -1-x2, both ofwhich arealready inseries form. Comparing (42.12) with (40.34), wehave bg= 1, C0= —k2, C1=0, C2= Allremaining b'sandc’sarezero. Theindicial equation (40.38) therefore becomes (42.14) m(m —1)+m—-k2=O,m2=k2, whose roots are;1=lc. Asweshall discover, when kisnot0oraninteger, there willbetwoFrobenius series solutions of(42.1). Iflciszero oran integer, there willbeonly oneFrobenius series solution of(42.1). The second independent solution Will, byComment 40.5, bealogarithmic solution andhave theform (40.51). Using theroot m=kandthevalues in(42.13) weobtain, bysetting thesecond coeﬁicient in(40.37) equal tozero, (42.15) a1[(lc+1)k+(k+1)-191+011.,=0,(21+1)a1=0. Since Icisapositive constant orzero, thesecond equation in(42.15) will beanidentity only if (42151) a1=0. 610 Seams l\I1—:Tnons Chapter 9 The recursion formula, obtained bysetting thecoefficient ofa;"‘+" in (40.37) equal tozero, becomes with m=lcandwith thehelp of(42.13), (42-16) a..[(7° +")(/C+n—1)+(if+H)—7021+ 0+an_2(1) =0, which simpliﬁes to (42.17) an=- ng2. Since, by(42.151), a1=0,weseefrom (42.17) that (42.18) a3= a5= a7= =a2,,+, =0,n=0,1,2,---. Foreven values ofn,wehave, by(42.17), (42.19) 2 112=—i =——(i2** ‘10,4+41¢ 1+k -1 -(=92 G)‘ 112 .114-'=—* = 110=, ‘lo,16+8k 8(2-l-I6) (1+k) 2.(2+k)(1+k) a_._;"4;=____if)4E,,6‘ 36+12k 2-12(3+k)(2+k)(1+k) ° =‘E ac, 3!(3+Ic)(2+l¢)(1+Iv) ,,=i<—1>"i>*’Z___,,2"n!(1+Ic)(2+k)---(n—l—lc) °" Substituting in(42.11) theroot m=lcandtheabove values ofthea’s, weobtain 2 4 <4“)”'*=“"”'°l‘*_i.._. ...1 _<§)6_|_... 3l(1+ 7¢)(2+k)(3+k)2<__l)n 2 21l- which isonesolution of(42.1), valid forlcg0. Similarly, byfollowing theabove procedure fortheroot m=——k, kg0,weobtain [equivalent toreplacing kby—lcin(42.2)] <42-2””-*=“"“"°l1-1-5 " '" +410—k)<2(:113;~ --<4—k)(§)2n+ '' Lesson 4-2B Bsssm. Frmcrrons orTHEFmsr KIND J;,(x) 611 which isasecond solution of(42.1), valid forIt9-51,2,3,---.(For these values ofk,oneofthedenominators in(24.21) iszero.) Since the functions flandfaof(42.12) arepolynomials, byComment 37.53, their series representations arevalid forallx.Hence byTheorem 40.39 the series solutions, (42.2) and(42.21), converge forallxexcept perhaps at :0=0. Note that theabove results support thestatement wemade after (42.14). IfIc=0,thetwosolutions (42.2) and(42.21) are,asyoucan verify, identical, andthere istherefore onlyoneFrobenius series solution of(42.1). And ifk=1,2,3,---,there isagain only oneFrobenius series solution of(42.1), namely (42.2). Forallother positive values of k,yk,andy_;,are,byTheorem 40.39, twolinearly independent solutions of(42.1). Hence ifIc¢0,1,2,3,---,thegeneral solution of(42.1), byTheorem 19.3andComment 19.41, is =613/k +cﬂy-1111 k76or1:2:31'''1 convergent forallan960,where y),andy_;,aredeﬁned by(42.2) and (42.21) respectively. LESSON 42B. Bessel Functions oftheFirst Kind J;,(x). In(42.2) anisanarbitrary constant. Letuschoose foritthevalue 1G0 =Mr ,0Q (We showed inLesson 27E, that klisdeﬁned forallvalues ofkexcept negative integers.) Thefunctions resulting from (42.2) when aoisgiven thevalue (42.3) aredesignated byJ;,(a:)andarecalled Bessel functions oftheﬁrst kind ofindex k.Hence, with k!(lc+1)replaced byits equal (k+1)!,weobtain, by(42.3) and(42.2), In1 1 2 1 4 W“)"=‘“’= *(Wm +2% ‘W4-lT>!(§)6+"' valid forallIcg0. If,in(42.31), wereplace Itby——Ic[equivalent tochoosing ao=5%; 612Seams Mrrrnons Chapter 9 in(42.21)], weobtain (42.32) _..<.—..1.>._"2)“ ..+mm-Ia)!(2+ 'validforallk 9'51,2,3,---. Weshowed inLesson 42A, thatifIaiszeroorapositive integer, then y),of(42.2) istheonly Frobenius series solution oftheBessel equation (42.1). However, ifIoisnotzerooraninteger, then thefunctions y),and y_;,of(42.2) and(42.21) aretwolinearly independent solutions of(42.1). Since J;,(:r)andJ_;,(x) deﬁned in(42.31) and(42.32) arethese same functions ykandy_;,multiplied byanappropriate constant, itfollows thatifkisnotzerooraninteger, J;,andJ_;,arelinearly independent solutions of(42.1), each valid forallavexcept, perhaps for:0:=0.Hence wecanassert, byTheorem 19.3andComment 19.41, that (42.33) y(z)=c1J;,(:c) +c2J_;,(:c), k950,1,2,3,-~- isalsoageneral solution of(42.1), valid forallx;£0,where J;,(x)and J_;,(x) aredeﬁned by(42.31) and(42.32). Ifhowever la=0,1,2,3,---,then (42.31) istheonlyFrobenius solu- tion of(42.1). Inthiscase, asecond solution of(42.1) willhave the logarithmic form shown in(40.51). Theoneweshall givebelow* isdesig- nated by—N;,(x) andiscalled aBessel function ofthesecond kind ofindex k.Itcanbeshown thatif,in(42.1), wemake thesubstitution 1/2(%) ="(=v)-Jx(=v) 10$$1#1>0, itbecomes (seeExercise 42,2) 2:214" +zu’+(222—k2)u =2:cJ;,’(:z:), where 'll.($) =Z_k(b° +121$ +bg$2 +'''), k> Asecond solution of(42.1), ifIaisapositive integer, isthen (42.34)_ 1k—1 _ __ 2n—k 1so _N"(’”) ‘22, :1 +22% nl(n+k)! 2nI: ><lH»+(1+%+~-~+m>l(%> i—J;,(a:) logx, x>0, ‘There areother forms oftheBessel function ofthesecond kind. Lesson 42B Bassar. Frmcrrons orrnaFmsr Kmn J;,(a:) 613 where 11 1 H,,= 1+5-1--3-+---+5, ifn #0, =0, ifn=0. Hence forpositive integer lc,thegeneral solution of(42.1) is (42.35) y(z) =c1J;,(a:) -1-c2N;,(x). Remark. Asecond solution, when It=0,canbefound attheend ofExample 40.6. See(k),page 584. Note thatthus farwehave nosolution J_;,(x) when lcisapositive integer. The solution J_;,(:z:) of(42.32) isundeﬁned fork=1,2,3,---.Toﬁll inthisgap, wedeﬁne J_;,(:r), k=0,1,2,3,---,bytherelation (42.36) J_;,(x) =(-1)"J,,(¢;), k=0,1,2,3, ---. Thejustiﬁcation forthisdeﬁnition isoutlined below. Justification for(4-2.36). By(42.31), so __ n 2n-Hr: (42.37) J;,(x)=Z ,1=0,1,2,3,---n=-0 By(42.32), wewrite formally on (__1)n x2n—k (42371) J_;,(:z:) =Z ,1=0,1,2,3,---,n-0 even though theseries isnotdeﬁned forn<k.[For these values ofn, (n—lc)isanegative integer andasweshowed inLesson 27E, (n—Io)! isundeﬁned fornegative integers.] Wealso showed inLesson 27E that (Ic——1)!=I‘(k)—>ooasla->0,—1,—2,---.Since anegative integer factorial appears inthedenominator ofeach term oftheseries (42.371) forwhich n<k,weareledtodeﬁne each term, forwhich n=0,1, 2,---,Ic—1,tobezero. Hence wecanwrite (42.371) as an Z_1)n x2n-Ia J__]¢(1li) ='—2 ,I€= 0,1,2, n=-h Letp=n—lc.Hence when n=lc,p=0.This means wecanstart oursummation in(42.38) withp=0instead ofwithn=k,ifatthesame timewesubstitute p+Icforitsequal n.Therefore (42.38) canbewritten ES an (__ 1)p—|-It x29-}-k J_|¢(Z) =5‘ ,k=0,1,2,-.-. 614- Snares Marnoos Chapter 9 Since thesummation in(42.39) isover p,wemay remove (—1)" from in- sidethesummation. Itthen becomes, after simpliﬁcation, 1""<-1)" 42"" (42.4) J_,,(x) =(-1)1;)71:,-m(§) ,1=0,1, Comparing thesummation in(42.37) with that in(42.4), weseethat both arethesame. Hence (42.36) follows. Themost frequently encountered Bessel functions oftheﬁrstkind are JQ(x)andJ1(:c). By(42.31), these are 2 1 4 1 6 (42.41) .I.,(1)=1-;i,+(ﬁ,;-,-(§w,%+--- 2 <—1>"x"+W%+“'; §1__l§”.j+_l__it_i__L§f+...+__£in___£+... .(42.42) J,(1)= 2( 2122 213124 314126 'nl(n—|—1)!22" ) Thegraphs ofthese twofunctions areshown inFig.42.43. 1.0 ‘ '-70(1) 0.5— J,(x) 0 l l I 1 2 4 5 7 8 -0.5 [- Figure 42.43 Tables ofvalues exist forBessel functions oftheﬁrst kind J;,(:r), just asthey doforsin:0,log:1:ore’.Thus ifyouhadtoevaluate J1(2), you could use(42.42) with ac=2,orlook upitsvalue inatable. Itsvalue is 0.5767. Therefore by(42.36), J_1(2) =—0.5767. Ifyouhadtoevaluate Lesson 4-2C Equsrrons Wnrcn Lam T0ABrzssar. Eomvrron 615 J_1,2(3), you could use(42.32) with k==§,:1:=3orlook upitsvalue inatable. Itsvalue is—0.4560. LESSON 42C. Differential Equations Which Lead toaBessel Equation. We give below examples ofdifferential equations whose solutions canbeobtained bytransforming each intoaBessel equation bymeans ofasuitable substitution. Example 42.5. Find asolution of _ 2 (a) u”+(1+ u=0,Itreal. Solution. The substitution, (b) u=141/21/,ul= %x—1I2y +Z1/2y/’ un = _ix-—3I2y +x—1/2y! +$1/2y!/, willtransform (a)into theBessel equation (Q) $21/"+wz/'+(12—k’)y=0- Since y=J;,(a:)isasolution of(c),itfollows bytheﬁrst equation in(b) that (<1) u=w"”J1(¢) isasolution of(a). Example 42.51. Use theresult obtained inExample 42.5 toﬁnd a solution of (a) u"—l-(1-—%)u=0. Solution. Ifin(a)ofExample 42.5 above, wechoose k=1,it becomes the(a)ofthisexample. Hence by(d)above, asolution of(a)is (b) u=2:1/2J1(a:). Example 42.52. Find asolution of (a) u"—|—2:211=0. Solution. Here twosubstitutions willbeneeded. The ﬁrst substitu- tionis (b) u=:01/2y. 616 Sanras Msrnons Chapter 9 Substituting in(a)thevalue ofuasgiven in(b)andthevalue ofu"as given inthelastequation of(b)ofExample 42.5, and then multiplying theresult by1:3’2,weobtain (0) w’y"+my’+(rv‘—by=0- Thesecond substitution is $2 1/2(d) w=§» :t=(2w) , %l?:-=1:=(2w)1/2. With thehelp of(d),weobtain @_:111".’_ 1/2E(e) at_dwdx_<2”) dw' (ﬂy_ 1,,daydw _,,,dwdyw~<2”)ma+<2“’> andzy d =<2“)m+ai.' Substituting (d)and(e)in(c),itbecomes <1’ d d<1) ow)’J,-,%+2w3%+2wi+1(2w>” -111=0. which simpliﬁes to 2 <3) w’§w—Z+w%+(w’—1*.>y=0- Since (g)istheBessel equation (42.1) with k=1},itssolution is (h) 1/=J1/4(w)- Replacing wbyitsvalue in(d)andthen substituting theresulting value ofyin(b),weobtain 2 (i) u=$1/“J1/4’ which isasolution of(a). Comment 42.53. Themethod used tosolve equation (a)ofExample 42.52 canbeapplied tothemore general equation (42.54) u"+ba:"'u =0. Theﬁrstsubstitution isthesame as(b)ofExample 42.52, namely (a) u=xl/2y. Lesson 42C Eouxrrous W1-non Lmn 'roABsssar. EQUATION 617 The second substitution, however, ismore complicated than theonein (d)ofExample 42.52. Itis (1.) w=75%./.112. Thesubstitution of(a)in(42.54) andof(b)intheresulting equation, will yield theBessel equation <1’ d 1(c) w2d7i;+wdTil’)+(w2—(7n:-5)-2)y=0. Itssolution is (<1) 1/=J11(m+2)(w)- Replacing wbyitsvalue in(b)andthen substituting theresulting value ofyin(a),weobtain (42-55) "==$1,211/o»+¢> xmq) , which isasolution of(42.54). Note that when m=2,b=1,(42.54) reduces to(a)ofExample 42.52, and(42.55) reduces toitssolution (i). Example 42.56. Find asolution of (a) u"+9a:u=0. Solution. Ifin(42.54) weletb=9,m=1,itreduces to(a)above. Hence asolution of(a),by(42.55) is (b) u=1v”2J1/3(2$3/2)- Comment 42.57. Themethod used tosolve equation (a)ofExample 42.52 canbeapplied toastillmore general equation than (42.54), namely (42.58) xzu” +(1—2a):z:u’ +(b2c2x2‘ +a2—k2c2)u =0. Theﬁrstsubstitution (4) '14=11“!/1 willchange (42.58) into theequation 2 (b) :02%+:1:3-;+(b2a:2° -—lc2)c2y =0. Verify it.The second substitution (c) w=bx“, 618 Snares Mrrrnons Chapter 9 willchange (b)into theBessel equation dz d(<1) w*ﬁ+wﬁ+(w’—k’)y=0. Itssolution is (6) 1/=J1¢(w)- Hence by(a),(e),and(c),asolution of(42.58) is (42.59) u=a:“J;,(bx°). Comment 42.6. If,in(42.58), wemake thesubstitutions __ __m-1'2 ____L_ (a) a-Q, c- 21 Ic-m+2, andreplace b2by4b/(m-1-2)2,itreduces to (b) :z:2u" +b:r"‘+2u =0, u"+b:c"'u =0. Asolution of(b),by(42.59), isthen (C) u=$1/2J1/(m+2) \/$1) ' Note that thesecond equation in(b)isnow_the same as(42.54). Note toothat their respective solutions (c)and (42.55) are,asthey should be, alsothesame. Hence (42.54) isonly aspecial caseof(42.58). Byassigning different values toa,b,c,and lcin(42.58), wecanthus obtain many different equations whose solutions willbegiven by(42.59). Example 42.61 .Find asolution of (a) :z:2u” +am’+(:1:—- u=0. Solution. Ifin(42.58), welet (b) a=0, b=2, c=%, itreduces to(a). Hence asolution of(a)by(42.59) is (c) u=J,,(2\/5). Comment 42.62. IfItisnotzero orapositive integer, then, by(c) above and(42.33), (d) u(w)=c.J1.(2\/5) +c¢J_1.(2\/5) isageneral solution of(a)ofExample 42.61. Ifkisapositive integer, Lesson 42D Przorarrrrrzs orBnsssr. Funcrrons orFrasr KIND J;,(x) 619 then by(42.35), ageneral solution of(a)is (9) 14(4) =61-7k(2\/5) -l-621)/1=(2\/5)» where N;,(x)isgiven by(42.34). Example 42.63. Find asolution of (a) 2:214” —|—:z:u'—(:02—l-k2)u =0. Solution. Ifin(42.58), welet (b) a=0, b2=i2=-—1, c=1, itreduces to(a). Hence asolution of(a)by(42.59) is (<1) 14(I)=J1.(iw)- Comment 42.64. Thefunction obtained bymultiplying (c)ofExam- ple42.63 above, bytheconstant i"°isalso asolution of(a)ofthis example. Itisusually written asI;,(:z:). Hence (42.65) I;,(x)=r'*J,.(."2). The function I;,(:r) of(42.65) iscalled amodiﬁed Bessel function of theﬁrst kind. _ Comment 42.66. Welistonemore equation which can, byasuitable substitution, bechanged intoaBessel equation. Itis (42.67) xzu” +a:(1—22:tan:t)u' —(actan:0+k2)u =0. The substitution u=y/cosx will transform (42.67) into the Bessel equation d2 d(a) :v2$+x(7;:~+(a:2—-k2)y=0. Since asolution of(a)isy=J;,(x),asolution of(42.67) is 1 LESSON 42D. Properties ofBessel Functions ofthe First Kind Ju(x)- A.Properties ofthe Zeros ofBessel Functions Jg(x). Welist below, without proof, anumber ofproperties ofthezeros ofBessel func- tions J;,(:c)oftheﬁrstkind. 620 Sanras Mrrrnons Chapter 9 1.If2:1and2:2aretwozeros ofJ;,(a:), then intheinterval I.’:01<:1:<1:2, there isazero ofJ;,_1(:c) andJ;,+1(a;). 2.TheBessel function JQ(x)hasazeroineach interval oflength 1r. 3.Each Bessel function J;,(x), k=1,2,---,hasaninﬁnite number of realpositive zeros intheinterval I:0<:1:<oo. 4.IfIt>§,then thedifference between twoconsecutive zeros ofJ;,(x) isgreater than 1r. 5.Ifk>§,then thedifference between twoconsecutive zeros ofJ;,(x) approaches 1r,as:1:approaches inﬁnity. 6.The ﬁrst positive zero ofJ;.(x)isgreater than k. 7.ABessel function J;,(:z:)hasonlyrealzeros. B. Integral Property ofBessel Functions ];,(x). Theorem 42.7. Letr1,1'2,---,bedistirwt positive zeros ofaBessel function Jf(z),where kisaﬁxed realnumber. Then 1 (42.71) /L):cJ;,(r,<r:)J;,(r,-2:) d2:=0,if1‘;¢r,-, =2lJh'(1‘i)l2. if1‘:=Ti- Proof. InExample 42.5, weproved thatu(:z:)=cc‘/2J;,(:c) isasolution _1,2ofu"+(1+ u=0.Inasimilar manner, itcanbeshown that_ 2 (a) u1(x) =x”2J;,(r,<z:) isasolution ofu”+(r.-2+ u=()_ . . 1-41¢’u2(x) =11:1/2J1,(1',<lJ) isasolution ofu"+(r,-2+—,h;7—> u==0. [Oryoucanobtain these solutions byletting a=§,b=r,c=1in (42.58). Itwillthen reduce toxzu” +(rzxz +1——Ic2)u =0.Division byac’willgiveittheform ofthedifferential equations in(a).Asolution by(42.59) willthen be (b) u=:c1'2J;,(r:c).] , Hence, by(a),_ 2 (0) "1"+(Tia + "1=0. _ 2 uz”+(T52 + "2=0- Multiplying theﬁrstequation in(c)by‘M2,thesecond by—u1andadding thetworesulting equation, weobtain (d) 142141’, —141142" =(r,-2—-r,-2)u1u2. Lesson 4-2D Pnormvrxms orBESSEL FUNCTIONS OFFmsr KIND J;,(a:) 621 Integrating (d)between thelimits 0andac,andrecognizing that theleft sideof(d)is(d/d:c)(u2u1’ —ulug’), there results (9) lu2"1' —u1142'lii =(T12—H2)/0 H1112 div- By(a)anditsderivatives, (f) 111(0) =0,142(0) =0, ui’=r.~r"2J1¢’(r.~r) +if”2J1¢(r.-1:), uz’ =r,-2:‘/ZJ;/(r,~a:) +§x_1/2J;,(r,-ac). Substituting (a)and(f)in(e),andthen simplifying theresult, weobtain (g) ,, £li[TiJ]¢(Tj23)Jk'(T,'il2) -TjJk(T,'1Z)J|¢'(T,'3I)] =(Tjz —T,'2)[0 (l3Jk(T,'Ili)Jk(Tj23) dﬂi. Ifac=1,i.e.,iftheinterval (0,x) istheinterval (0,1), then (g)becomes 1 (11)"aJ1¢(1'j)Jk'(T-") —T1-7k("i)Jk'("i) =(U2"H6]; 111k("='$)-7 k("a'1?) d1- Byhypotheses r,-andr,-arezeros ofJ;,(x). Therefore, J;,(r,-) =0, J;,(r,-) =0andtheleftsideof(h)vanishes. Hence if1',-,r,-,aretwodistinct zeros ofJ;,(:2:), wecandivide (h)byr,-2—r,-2toobtain theﬁrst equation in(42.71). Wenow prove thesecond equation in(42.71). Diﬂerentiating theequa- tion in(g)with respect to1',-,weobtain (i)$l1'i1vJk' (":13-71¢’ (T1'11)—-71¢(T¢$)Jn' ('11?) ""17$-7k(T¢1)Jk"("i$)l =21,-/0xJ;,(r,-x)J;,(r,-:z:) dx+(r,-2-T,-2)%/0xJ;,(r,-:c)J;,(r,-ac) dz. Ifr¢=r,~,then (i)simpliﬁes to (1)$lI1‘iJr/2(T¢I) "-Jk(7'\73)Jk’(7'|'$) —HI-7k(1‘¢1)Jk"(T¢$)l I =21,-foxJ;,2(r,-:0) dx. And iftheinterval is(0,1) sothat :1:=1,then (j)becomes [remember r,- isazero ofJ;,(:c); therefore J;,(r,-) =0] 1 (k) 1','J],'2(T,') =27;’/0 £lZJk2(1‘,'£B) 622 Serums Mrrrnons Chapter 9 Hence 1 (1) 1;1J1t2(Tt<v)dI¢ =‘Q-71='2(T.'), which isthesecond equation in(42.71). Comment 42.72. Because oftheﬁrstequality in(42.71), thesetof functions J;,(r,:c), J;,(r2x), ---,issaidtobeorthogonal onI:0§:1:§1 with respect totheweight function x.Compare with Deﬁnition 41.5 foranorthogonal setoffunctions. EXERCISE 4-2 1.Find general solutions ofeach ofthefollowing Bessel equations. I I Z -" = . E35”3”53i””3iE”:‘1?”=3'<c>1*i"+wi'+(Z2—oi=(d)wzy”+my’+($2—by= 2.Prove that thesubstitution in(42.1) ofy2(a:) =u;,(a:) —J|,(a:) log2:, :0:>0,transforms the equation into a:2u;,” +arm,’+(22—k2)u;, = 21.1,,’ (z). Hint. Make useofthefactthat J;,(z) isasolution of(42.1). 3.Verify, bydirect substitution in(42.31), that J1(2)==0.5767; J-1/2(3) = —0.4560. Hint. Factor out(lo)!andusefactthat (—§)l -\/F. Also verify that J0(0.3) =0.9776; J1(0.2) =0.0995. 4.Prove that (a)ofExample 42.5istransformed into(c)bythesubstitution (b)-5.Verify theaccuracy of(c)ofExample 42.52. 6.Verify theaccuracy of(a)ofComment 42.66.PP Find asolution ofeach ofthefollowing equations 7-14, byusing the appropriate formula given inLesson 42C. 7.y”+9:z:2y =0.Hint. In(42.54), b=9,m=2. H 38.y +(l+-1-6?-)y=0. 49-y"+(1—9?5>y =0. 10.y”+4:c3y =0. ll.12y" +2:y’+(2:—1)y=0. 12.22y” +my’—($2—|—4)y=0. 13.xzy” —:1/'+2:211=0. 14-I:1/”+§y’+ (w—f3)u =0- 15.Byassigning various values toa,b,c,andkin(42.58), obtain atleast three different differential equations andtheir solutions. 16.Show that thesubstitution u=2e‘/2, e‘=142/4, willtransform theequa-2 tion y"+(e‘—m2)y =0into theBessel equation uz3%+u%+ Lesson 4-2—Exercise 623 17 18 (a) 19. 20.(uz—4m2)y =0.Since asolution ofthesecond equation isy=Jg,,.(u), asolution oftheoriginal equation isy(z) =Jg,,,(2e"/2). Hint. Follow the procedure used inmaking thesecond substitution after (c)ofExample 42.52. With thehelp ofproblem 16,ﬁndasolution ofeach ofthefollowing differ- ential equations. (a)y"+(e‘—9)u=0-(b)1/”+(e‘-—by=0-(0)1/"+(e"-—t)y=0- Assume that afunction f(z), deﬁned ontheinterval (0,1), canberepre- sented byaseries ofBessel functions, i.e.,assume f(¢) =¢oJ1.(To$) +61Jk(7'133) +62-/1=(?‘21v) -l-''‘1 where ro,r1,T2,---,arethedistinct, positive zeros ofJ;,(2:)andkisaﬁxed realnumber. Show thatthecoeﬁicients co,01,62,---,aregiven by 1 mwm%=2Lemnmeu Hint. Multiply (a)by:cJ;,(r,-:0), integrate from 0to1,then use(42.71). Prove each ofthefollowing identities. d . .(a)E[a:"J;,(a:)] =at"-I;,_.1(1).Hmt. Multiply (42.37) byac"andthen take itsderivative. d_ _ ..(b)E[1"J,.(¢)1 =-1"J,.+1(¢). Seeh1nt1n(a). (C)J,/(1)+ka:_1J;,(a:) =J,._,(¢). Hint. Carryoutthediﬁerentiationin(a)andthen divide by2:". (d)J;/(a:)k-— ka:"1J;,(a:) =—J;.+1(x). Apply hint in(c)to(b). Divide by2:‘. (e)Jt_1(a:) -J;,+1(:::) =2J;,’(a:). Hint. Add (c)and(d). mJHw+nme=%mn (g)3%Jo(a:) =-J1(a:). Hint. Setls=0in(a)above, andthen make useof(42.36). 2 (h)#[Jr¢(:c)] Ei;[J1.-2(1) -—-2J;.(a:) +J;,+2(Z)]. Hint. Differentiate (e). Then use(e)again toﬁndJ;.+1’(2:) andJ;._1'(x). Substitute these values intheJ;,’'(1)equation. (1)J1/2(1)-E \/Esin ft.Hint.(Q)!=W/2. (1)J_1/2(£t)E \/gcosft.Hm.(-1.)!=ﬁ. Show thatthecoefﬁcient oft"intheseries expansion ofe("/2)l"'(1”)1 inpowers oftisJ,.(a:). Hint. e(=/2>l"<1")1 =e“/2e““/2‘. Write theseries expansion ofeach term ofthisproduct, multiply both series, then show that theco- eﬂicient oft"=J,.(:z:). 624 Snares Mrrrnons Chapter 9 ANSWERS 4-2 1.Substitute: (a)It=lin(42.2) and(42.34); (b)k=2in(42.2) and(42.34); (c)lc=1}in(42.2) and(42.21); (d)lc=Qin(42.2) and(42.21). 1.y=i"2J,,.,(g<i’). 11.y=J2(2\/E). s.y=¢"’J,,.(¢). 12.y=ma). 9.y=£t1I2J5/5($). 13.y=1.141). 2 10.y=:v”2J1/5(§:v5I2). 14-.y=:cm'J1/2 . 11-(a)11=Jt<2e"’). 1/=J.<2e"’>. <@>y=J.,.<2e"*>. LESSON 4-3. The Laguerre Differential Equation. Laguerre Polynomials L|,(x). Properties ofL;,(x). LESSON 43A. The Laguerre Differential Equation and ItsSolu- tion. Thedifferential equation (43.1) xy”+(1—x)y'—|—Icy=0,kreal, iscalled theLaguerre equation, after E.Laguerre (1834-1866). Itisof interest onlywhen lcisaninteger andtheinterval isxg0.Youcanverify byDeﬁnitions 40.22 and40.24 that:0=0isaregular singularity of(43.1). Hence weseek aFrobenius series solution oftheform (43.11) y=:z:"‘(a° +ala:+a2:c2 +---). Multiplying (43.1) byasand comparing theresulting equation with (40.33), weﬁndthat (43-12) f1(w) =1-w, f2(w) =kw, both ofwhich arealready inseries form andvalid, byComment 37.53, forallx.Hence, byTheorem 40.32, aFrobenius series solution of(43.1) will bevalid forallas,except perhaps at1:=0.Comparing (43.12) with (40.34), weﬁnd bo= 1, bl= ‘*1, C0=0, C1= Allremaining b'sandc’sarezero. Theindicial equation (40.38) therefore becomes (43.14) m2—m+m=0,m2=0, whose roots arem=0twice. Wecantherefore expect only oneFrobenius series solution of(43.1). Lesson 43B Tm-: Lxoumma Ponrnonnar. L;,(z) 625 Using therootm=0andthevalues in(43.13), weobtain, bysetting thesecond coefficient in(40.37) equal tozero, (43.15) a1+aolc=0, a1=—lca0. The recursion formula, obtained bysetting thecoefﬁcient ofx’”+" in (40.37) equal tozero, becomes with m=0andthehelp of(43.13), (4316) anin(n —1)+nl+an——li_-(n —1)+ =0: which simpliﬁes to —1—Ic(43.17) an=@—-—%- a,,_1. By(43.17) and(43.15), 1—k —-kl-—k kIc——1G2 =-T01 = G0= , 2-1. —k(k-not-2) “*=—§'=*“2='""a%;».='*"“°»3—Is Ic(k—1)(k-—2)(k—3) “‘=7T“’= “°' a_(-—1)”k(k-— 1)(Ic—2)---(Ic—n+1)a"_ 22.32.42...n2 °v (—1)"k'= ao, 1|.=O,1,2,"'. Substituting m=0andtheabove values ofthea’sin(43.11), weobtain (43.19) 1/|.(w)=at(1—km+'“('“2'2" D122-W‘'é,1?g°,_ 21¢“ + ,4+... (——1)"k! ,, which isaseries solution of(43.1), valid forallx.Asecond solution will have thelogarithmic form shown in(40.51). LESSON 43B. The Laguerre Polynomial Lk(x). If1::=O,1,2, 3,---,theseries (43.19) terminates. The resulting polynomials, with ao=kl,areknown asLaguerre polynomials and aredesignated by 626 Scams Mm-nons Chap“; 9 L;,(:c). Hence, by(43.19), with an=kl, C>l\’J»—-v-¢(43.2) Lo(x) = L1($) = "4'3, L2(:r) = —4x—|—x2, L3(:z:) = —18:1:+9x2——$3, L4(a:) =24——961':+72:02 —162:3 —|—:0‘. L = (k')2 f 'xn " "n_=°(n!)2(k -n).' Wemay place (k!)2 outside thesummation sign, since thesummation is over n.Graphs oftheﬁrst four Laguerre polynomials areshown in Fig.43.21. Y It “12 113(1) ~9 L,(x) 6 3 '/L006) 1 l 1 l >-i I-2-1 0 1 2 4 5 6 7 X --3 _6 Ll(x)/T 1 Figure 4-3.21 Ithasbeen proved that aLaguerre polynomial ofdegree nhasexactly nrealzeros intheinterval 0<rt<oo.Note inFig.43.21 that L0(x) has nozeros, L1(:r) hasonezero, L2(x) hastwo zeros, and L3(:r) hasthree zeros intheinterval a:>0. Wegive below twoequations which canbetransformed into Laguerre equations byaproper substitution. 1.The substitution (43.22) u=e""y Lesson 43C Paornarrns orLAGUERRE POLYNOMIALS L;.(a:) 627 intheequation (43.23) xu”—|—(1+:c)u’ +(lc+1)u=0 willtransform itintotheLaguerre equation (43.1). Since y=L;,(x) isa solution of(43.1), itfollows by(43.22) that (43.24) u=e_"L;,(x) isasolution of(43.23). 2.Similarly, thesubstitution (43.25) u=e_‘/2x1/2y intheequation (43.26) 1."+(é+3% -2).1=0, willtransform itinto theLaguerre equation (43.1). Since y=L;,(x) isa solution of(43.1), itfollows by(43.25) that (43.27) u=e_"2:c'/2L;,(:c) isasolution of(43.26). LESSON 43C. Some Properties ofLaguerre Polynomials Lk(x). A.Analog ofRodrigue’s Formula fortheLegendre Polynomial. (43.3) L,,(:r) =e’%(x"e_'). Proof. Weshall show ﬁrst that theformula isvalid fortheﬁrst three Laguerre polynomials. By(43.3), (a) L0(:c) =e’:v°e_' =1, L1(:c) =ex%(xe"”) =e"(e"' —ace“) =1-—av, L(x)=exii(x2e"") =e”i(2:ce"' —z2e_”)2 dz? dz: =e"(2e_' -—4a:e_‘ +:c2e"‘) =2-4x+$2. L xda 3-:1: zd2 2-1 3-:3(:z:)=e%(:re )=e;i-gt-;(3xe —a:e) =e‘%(6xe_’ ——6x2e_” -1-:c3e_') =e”(6e_’ —-18:ce_" —|—9:z:2e_’ —:c3e_") =6——18:z:+9x2—:c3. You canverify that each oftheformulas in(a)agrees with those in(43.2). 628 Seams Mrrrnons Chapter 9 The proof of(43.3) forthegeneral case follows. First verify that ,, ” ! ,,_(43.31) [r<x>ge>1‘ ’=,T(,,"r;,-,f<'"’<eg‘ ho), i.e.,thecoeﬁicients ofthenthderivative off(x)g(z) arethesame asthe coefﬁcients inthebinomial expansion of(x+1)”.In(43.31), letf(z) =1:" andg(x)=e“‘. Itthen becomes dn n-2 n n —z n-(43.32) %(axe)= (.1)<”>(@ )<"K When (43.33) n=1:ii(a:")=nx"_‘, d2 n=2:E(a:”) =n(n-—1):c"_2, s n=3:31-3(a:") =n(n——l)(n —-2)x"_3, n=k: if-6,;(x")=n(n——1)(n-2)---(n-k+1):z:"_'° n! ,,_ =(Tn ““-Also dn-It _ n_ _ (43.34) %(e’)=(-1) '=e=. Substituting in(43.32) thelastequality of(43.33), theequality (43.34) andthen multiplying theresult bye’,wehave (43.35) exin(:r"e_’) =(n!)2 itiﬁ :z:"_'°.dz" kuolc![(n —-k)l]2 You canverify that theright side of(43.35) also willgive theﬁrst four Laguerre polynomials in(43.2). Itis,infact, another form ofwriting the polynomial solutions oftheLaguerre equation. Weshall now show that theright sideof(43.35) andofthelastequation of(43.2) areequivalent. Inthesummation of(43.35), letp=n-—lc.Therefore k=n——p andwhen k=0,p=nandwhen lc=n,p=0.Hence theright side of(43.35) canbewritten as 4336 Mi (-17) P<-> ‘"-> Therefore, by(43.36), with n=lc, ‘Z 0 It (43.37) (lc.)2‘M_P),(1),),:1:. Lesson 43C Paornarms orLaouasaa POLYNOMIALS L),(x) 629 The summation in(43.37) isnow thesame astheright side ofL;,(a:) of (43.2). B.Integral Property ofLaguerre Polynomials. Theorem 43.4. LetL0(x), L1(x), L2(x), ---,beLaguerre polynomial solutions of(43.1). Then (43.41) fe_'L,,,(x)L,,(:c) da:=0,ifm#n. 0 Proof. Letumandu,,betwosolutions of(43.26). Therefore 1 2 1 1 (3') um,I+<Z}E+ —Z>um=01 2 l 1 Multiplying theﬁrst byu,,,thesecond by—u,,, andadding theresulting equations, weobtain I___ 1/= n_m(b) mum'umu. (x)unm- Integrating between thelimits 0,oo,andrecognizing thattheleftsideof (b)is(d/dx) (u,,u,,,’ —u,,,u,,’)“, there results (<1) Ilimlum.’ —u...u..’l'8 =(H—"Of éumun dw,mr‘11-—>Q 0 By(43.27), solutions of(a)andtheir respective derivatives are, (d) u,,,=e_‘/2:01/2L,,,, um,=__%e-1/22,1/2Lm +is-1/2,,-1/2Lm +6-e/2x1/2Lm/, u,,=e_‘/22:1/2L,,, unr=_%e-=/2x1/2L" +£8-1/2,,-1/2L" _|_e-1/23,1/2L"/_ Inserting theabove values in(c),weobtain (e) E33,[e-‘wL.<w>L.'(f> —e-’xL..<»)L..'<x>1'a =(n——m)/0e_’L,,,(x)L,,(:c) drc. Theleftsideof(e)iszerowhen :1:=0,and, by(27.113) (d),itapproaches zero forpositive hash—>co. Byhypothesis msfn.Therefore (e) simpliﬁes to (f) /0e_”L,,,(1:)L,,(:c) dz=0. 630 Seams Ma'r1-Ions Chapter 9 Because of(f),thesetofLaguerre polynomials issaid tobeorthogonal onI:0§:1:<oowith respect totheweight function e"“’. Compare with Deﬁnition 41.5 foranorthogonal setoffunctions. EXERCISE 4-3 1.Find aseries solution ofeach ofthefollowing differential equations. (a)11/"+(1—¢)y’+ it=0-(b)11/’+ (1—w):/'+ 1-211=0- 2.Verify that thesubstitution (43.22) in(43.23) gives theLaguerre equation (43.1). 3.Verify that thesubstitution (43.25) in(43.26) gives theLaguerre equation (43.1). 4-.Useformula (43.3) toﬁndL4(a:). 5.Find aseries solution ofeach ofthefollowing. (a)xy”+(1—|—:2:)y’+y=0.Hint. See(43.23). I (b)11/”+(1+¢)y'+21/=0- (0)$11"+(1+I)y+$1!=0- 6.Find aseries solution ofeach ofthefollowing. 1 1 1 .(8.) y"+ +'2; '— y=0.Htﬂt. See <1»)1/"+($+% —91/=0-(C)y"+(§+;c1- -911 =0. 7.Assume thatafunction f(z), deﬁned ontheinterval (0,w), canberepresented byaseries ofLaguerre polynomials, i.e.,assume (*1) f(fv)=¢oLo(1) +¢1L1(1) +¢2L2(1) +--- Show thatthecoeﬁicients co,01,62,---aregiven by ck= 'Ke_zL|,(1)f(:t) dz: /0e‘[L,.(¢)1’ at Hint. Multiply (a)bye"‘L;.(:v), integrate from 0tow,then use(43.41). ANSWERS 43 1.(a)replace ItbyQin(43.19). (b)replace lcby1.2in(43.19). 5.(a)y=e“Lo(:c) =e“. (b)y =e"L1(:c) =e“(1 —2:). (c)y=e"'L1/g(2:).6_(a) y=e—z/22:1/2L0(x) =8-:/2x1/2_ (b)y=e-1/2x1/2L1(x) =e-=/211/2(1 _x)_ (0)1/=¢"‘/2$”2L1/2(¢)- Chapter 10 Numerical Methods Introduction. Inmany practical problems involving differential equa- tions, what isoften wanted isatable ofvalues ofasolution y=y(:z:), satisfying given initial conditions, foralimited range ofvalues of2:near theinitial point :00. For example, wemay want values ofy(z) when at=:00+h,:00—|—2h,1:0+3h,etc., where his0.05 or0.1or0.2,etc. Even when asolution y=y(z) ofadifferential equation canbewritten interms ofelementary functions, itmay attimes beeasier toobtain this limited table ofvalues bythenumerical methods weshall describe inthis chapter, rather than from theanalytic solution itself. This statement is especially true when thesolution isanimplicit one. Aswehave remarked onnumerous occasions, implicit solutions areusually such complicated expressions thatitisalmost impossible toﬁndtheneeded function g(x) which itimplicitly deﬁnes, ortocalculate values ofyforgiven values of2:. Moreover, inagreat many problems, asolution ofadifferential equation cannot beexpressed interms ofelementary functions forthevery good reason that thedifferential equation does nothave anysuch solution. F‘or example, theequation y’=1/\/x3—|—1does nothave asolution interms ofelementary functions. Byanumerical solution ofadifferential equation, weshall mean a table ofvalues such that foreach 2:there isacorresponding value ofy(z). Inthissense, even anexplicit solution interms ofanelementary function, such asy=sin1:orinterms ofanonelementary function such asy=J0(1), isanumerical solution. Foreach x,wecanlook inatable andﬁndavalue ofsin:0oroftheBessel function JQ(x). You willsoon discover that thework involved incomputing atable of values even when amoderate degree ofaccuracy isneeded islaborious and tedious. However, with thecurrent increased availability ofhigh- speed computing machines, itispossible tohave them make many burden- some calculations foryou. Butitwillstillbenecessary foryoutoknow how andwhat tofeed these machines. Inthefollowing lessons ofthischapter, weshall explain various methods bywhich anumerical solution ofadifferential equation canbeobtained. 631 632 NUMERICAL Marnons Chapter 10 Tokeep thearithmetical calculations within reasonable bounds, andalso tobeable tocheck theaccuracy ofourresults, wehave selected simple differential equations forourexamples, ones which canbesolved explicitly interms ofelementary functions. You must keep inmind, however, that weareusing them only toillustrate amethod. These same methods can beemployed toﬁndnumerical solutions ofmore complicated equations. Weillustrate themethods weshall develop forﬁnding anumerical solu- tionofaﬁrstorder differential equation byapplying them toanequation oftheform y’=f(z,y) forwhich y(:c0) =yo.Weassume inourdiscus- sionthat aunique particular solution ofthisequation, satisfying thegiven initial condition, exists. (For criteria which willgive asufficient condition fortheexistence ofthisunique particular solution, seeTheorem 58.5. If thecriteria oftheexistence theorem aretoodifficult toapply, apractical man willusually know from hisexperience and from thenature ofthe physical problem which gave risetothedifferential equation whether a solution exists.) The methods weshall develop forﬁnding anumerical solution ofadif- ferential equation have been divided into three categories. Inonecate- gory, weinclude those methods which need only thegiven equation y’=f(z,y) andtheinitial condition y(a:0) =yoinorder tostart thecon- struction ofatable ofvalues ofyforgiven values ofx.They aretherefore called appropriately starting methods. Inasecond category, weinclude those methods which need more values ofythan only theinitial condition y(x0) =yobefore they canbeused. These methods aretherefore called appropriately continuing methods since they canbeused tocontinue theconstruction ofthetable only after theneeded preliminary values have been obtained bystarting methods. Inathird category weinclude those methods whose only purpose istocorrect values ofyobtained by starting and continuing methods. These methods aretherefore called appropriately corrector methods. LESSON 4-4-. Starting Method. Polygonal Approximation. Inthislesson weshall show, byamethod called thepolygonal method, howtostart theconstruction ofatable ofapproximate values ofy(:r0 +h), y(x0 +2h), ---,where hisaconstant andy(z) istheunique particular solution of (44-1) y’=f(r,y) satisfying theinitial condition (44-11) I/($0) =.310- Weproceed asfollows (seeFig.44.12). By(44.1) and(44.11), wedetermine Lesson 44 STARTING METHOD. POLYGONAL Arraoxrmrron 633 y=y(r)\‘ }E2 E] (x2) y2) "‘<"*""’E["°'““°”\ W.) W+h)_ y<x.+2h)Ey<x.+h)§°_= ;v(r)5)y(xo)Ey0 J'(11)=J/1 2 2 xo x0+h=x, xO+2h=x1+h=x2 Figure 4-4.12 thevalue ofy’at(:r0,yo). The equation ofthetangent totheintegral curve y(z) atthepoint (a:0,y0) istherefore (44-13) 1/—1/($0) =(4—$0)1/'($0)- This tangent line will intersect theline at=20—}—hinapoint whose ordinate is[in(44.13), replace soby2:0+h], (44-14) I/($0+h)=y(wo) +y'($o)h- Inourtable, wecannow record thevalue ofy(a:0 +h)obtained by (44.14). Itisanapproximation totheactual value ofy(:r° +h).The error inthiscomputation isshown asE1inFig.44.12. Forconvenience wewrite ($1,;/1) forthepoint [:00+h,y(x(, —|—h)]. At [:r1,y1] werepeat theabove procedure asif(:t1,y1) were actually onthe integral curve. Weﬁnd theequation ofalinethrough ($1,;/1) having a slope obtained by(44.1), with :1:=:01,y=yl. Ifwecallthis slope y’(:::1), then theequation ofthislineis (44-15) y—y(w1)=1/’(r1)(x —11)- Itsintersection with thelineat=xo+2hEx1+his (44-16) !l($o +2h)E1/($1 +h)=3/(I1) -l"2/'(5'31)h- Inourtable, wecan therefore now record theapproximate value of y(:r0 +2h). Theerror inthecomputation isshown asE2inFig.44.12. Continuing inthismanner, weﬁnd (44-17) Zl(13o +3h)E1/($2 -l-h)=1/($2) +y'(I2)h, 634 NUMERICAL Mrrrnoos Chapter 10 which isanapproximate value ofy(:::0—|—3h).Andingeneral, weﬁnd (44-18) y(w..+h)=y(t))+2/'(w..)h, which isanapproximate value of3/[:00 +(n-1-1)h]. Example 44.2. Bymeans ofthepolygonal method, ﬁndapproximate values when 2:=0.1,0.2,0.3oftheparticular solution ofthedifferential equation (a) y’=w”+1/, for-which y(0) =1.Take h=0.1andh=0.05. Solution. Comparing theinitial condition with (44.11), weseethat 2:0=0,yo=1.By(44.18), with h=0.1and1,,taking onthevalues 0,0.1,0.2,weobtain (b) 1/(0+0.1)=y(0-1) =y(0)+y'(0)(0-1), (c) y(0.1 +0.1) =1/(0.2) =y(0.1) +y’(0.1)(0.1), (d) y(0.2 +0.1) =3/(0.3) =y(0.2) +y’(0.2)(0.1). By(a)and theinitial conditions, 3/(0) =0—|—1=1.Therefore (b) becomes (e) y(0.1) =1+ 1(0.l) =1.1. By(a),when at=0.1,y==1.1,weﬁnd y’(0.1) =(0.1)2 +1.1 =1.11. Therefore (c)becomes (f) y(0.2) =1.1+1.l1(0.1) =1.211. With .1;=0.2, y=1.211, weﬁnd by(a),y'(0.2) =(0.2)2 +1.211 = 1.251, andby(d) (g) y(0.3)=1.211+1.2s1(0.1) =1.336. Innumerical solutions, itisusually desirable toconstruct atable in which allrelevant computations aresystematically recorded. Forthe above example, thetable hastheappearance ofTable 44.21. Theactual Table 44.21 1...=1)<:.)= 1/<1.)= hm.)=y<x.+h)=Actiil))alies°f 0.0 1.0 1.000 0.1 1.1 1.000 0.1 1.1 1.110 0.111 1.211 1.106 0.2 1.211 1.251 0.125 1.336 1.224 0.3 1.336 1.360 Lesson 44 STARTING M1=.'r1-101). Pomroonzu. APPROXIMATION 635 solution of(a)satisfying y(0) =1isy=3e”—1:2—2:0—2.Byitwe obtained theﬁgures inthelastcolumn ofTable 44.21. With h=0.05, ourtable ofvalues becomes, by(a)and (44.18), Table 44.22. Table 4-4-.22 1» Uh») y’(w-) hi/'(1»)Actual Values ofum+h) W") 0.0 1.0000 0.05 1.0500 0.1 1.1026 0.15 1.1582 0.2 1.2172 0.25 1.2801 0.3 1.34721.0000 1.0525 1.1126 1.1807 1.2572 1.34260.0500 0.0526 0.0556 0.0590 0.0629 0.067 11.0500 1.1026 1.1582 1.2172 1.2801 1.34721.0000 1.0513 1.1055 1.1630 1.2242 1.2896 1.3596 L 1.3— 1.2— 1.1— 1.0Actual °°1uti°n Polygonal approximation h=0.1 _ \Polygonal approximation h=0.05 Figure 44.23I I I I I L 0 0.05 010 0.15 0.20 0.25 0.30 Agraph oftheactual solution andofthepolygonal approximations are shown inFig.44.23. 636 NUMERICAL Mnrnons Chapter 10 General Comment onErrors inaNumerical Computation. The question ofdetermining theerrors inatable ofnumerical values isan extremely complex one. There areingeneral four types oferrors. 1.Arithmetical errors made bytheindividual orduetothemisbehavior ofacalculator. 2.Rounding oﬁ’errors duetostopping with acertain decimal place. 3.Formula errors duetotheuseofanapproximating formula toobtain a numerical answer. 4.Cumulative errors. Ateach stepinalengthy process anerror occurs that iscarried along tothenext stage. Weshall assume that theerror duetotherounding offofadecimal can becompensated forbyretaining asuﬂicient number ofdecimal places at each step, sothat theaccuracy desired inthelast tabulated value of y(:z:0 +nh)will notbeaffected even inthemost unfavorable circum- stances, asforexample when wehave todrop the49in0.3265349 inorder toround offthedecimal toﬁveplaces. If,however, toomany steps are required toreach y(x0 +nh), itmay notalways bepossible toattain this desired objective, butthen thelossinrounding offatonestep may be offset byagain atanother. Asregards theother types oferrors, weshall comment oneach ofthem attheappropriate time. Comment 44.3. Comment onError inPolygonal Method. 1.Inthismethod, westart with apoint andaslope that agree with the solution of(44.1) satisfying (44.11). Butsince thestraight linedrawn at thispoint (a:0,y0) may notbetheactual integral curve y=y(z), aformula error isintroduced intheﬁrst step. Itisrepresented byE1inFig. 44.12. Ateach successive point (x2,y2), (:c3,y3), ---,atleast twoerrors areintro- duced, astarting error andaformula error, sothat thecumulative error may soon become large. Hence this method isuseful only iftoogreat accuracy isnotrequired orifhisvery small. 2.Comparing (44.14) with theTaylor series formula, see(37.35), yo.+1»)=ya.)+1/<@.>h+ If+--- "<>.."+‘(X> .. —|—11% h+ h+1, weseethat (44.14) contains theﬁrst two terms ofaTaylor series. Its remainder orerror term istherefore, see(37.36), (44.31) E= h2, Lesson 4-4 STARTING METHOD. POLYGONAL APPROXIMATION 637 where Xisavalue of:0intheinterval under consideration ofwidth h.Let (44.32) E(x0 —I-h)=error incomputing y(:r0 +h)by(44.14). Then, by(44.31), (44.33) E_(@,,+h)=chz, where c=y”(X)/2!. Letusnow divide thehinterval inhalf andcom- pute ;/(rco —I-h)intwosteps. Ifwehave some basis forbelieving that for asmall h,y”(x) changes rather slowly inthishinterval sothatthevaria- tioninthevalue ofy”(X) ineach h/2interval isnegligible, then wecom- mitasmall error inusing thesame cof(44.33) foreach half interval. Therefore, by(44.33), theerror incomputing 1/(mo —I-h/2) forahalf interval h/2isapproximately (44.34) E(1.,+=C =‘ii=%,E(@,,+h). hThis means that theerror iny($0—|—2)isapproximately equal toone- fourth theerror of1/(xo +h).Hence theerror incomputing intwosteps . . h h .thevalue ofy(a:0+h),which wewrite asy[Geo + +5],Wlllhave . . . h .aninherited error inthestarting value ofy(mo—I— plus itsown for- mula error. Since each error equals one-fourth theerror in1/(xo —|—h),the h h. . .total error iny[(xo —I- —I—5],i.e.,theerror incomputing y(xo —I-h), intwo steps, isapproximately equal toone-half theerror iny(:ro +h) computed inonestep. Hence (44.35) E[(x., ++=gE(@,,+h), approximately. letY(x0 -1-h)betheactual value ofthesolution. Then (4436) Y($0 +h)'"?/($0 +h)=E050 +h)! Y($o+h>—y[(x..+§)+§]=E[(x.,+2-) Subtracting thesecond equation in(44.36) from theﬁrst, weobtain (44.37) h h h h 1/$0-F5 +5 -1/(Ind-h) :E($o+h) "E 1o+§ +5‘ 638 NUMERICAL Mnrnons Chapter 10 Substituting (44.35) intheright sideof(44.37), thefollowing equations result. (44.38) <4) ++=1/[(4.++—yo.+h>. (b)%E($o +h)=3/|i($o + + "3/($0 +h)- Theﬁrstformula saysthattheerror inthevalue ofy(a:o+h)computed intwosteps isequal tothedifference invalues ofy(a:0+h)computed in twosteps andinonestep. Formulas (44.38) give usameans ofestimating errors ateach step. Their accuracy does notdepend onaknowledge ofthesize ofy”(X), which wedonotknow, butonly onthevariation ofy”(X) over asmall interval. Wehave assumed thisvariation tobenegligible, anassumption which isnotunreasonable ify"(:c) changes slowly over h.Forexample, from Tables 44.22 and44.21, with :00=0,h=0.1, (a) y[<a:o + + =1/(0.05 +0.05) =y(0.1) =1.1026, y(xo —|—h)=y(0.1) =1.1. Therefore by(44.38)(a), (b) E(0.05 +0.05) =1.1026 —1.1=0.0026, which istheapproximate error ofy(0.1) computed intwo steps. The actual error byTable 44.22 is1.1055 -—1.1026 =0.0029. 3.Acheck onerrors, which practical people frequently use,andwhich seems towork, istomake allcalculations over again with anhhalfthe sizeoftheoriginal one. Iftheresults obtained with thesmaller hagree with those obtained with thelarger htolcdecimal places, after being properly rounded off,then itisassumed that their common numerical value haslcdecimal place accuracy. Forexample, byTables 44.21 and 44.22, y(0.3) computed with h=0.1and with h=0.05 agree toone decimal place. Hence weassume that y(0.3) =1.3hasone-decimal accuracy. 4.This method does notgive acheck onarithmetical errors. However, ifthere islittle agreement between thevalues obtained byusing hand h/2, allarithmetical computations should bechecked. Ifthearithmetic iscorrect, hshould bereduced. Comment 44.39. Wehave headed thislesson “Starting Method." It canalsobeused asacontinuing one. Forinstance, wecan,inExample 44.2, usethismethod toﬁndy(0.4), 3/(0.5), etc. However itsloworder of accuracy makes itapoor continuing method. Inthenext andsucceeding lessons, weshall present better ones. Lesson 44—Exercise 639 1. 2. 3. 4 5 6. 7. 8. 9. 10. ll.EXERCISE 44 Add toTable 44.21, values ofywhen at=0.4and0.5. Also addactual values obtained from thesolution y=3e‘—2:2—22:—-2. AddtoTable 44.22, values ofywhen 2:=0.35, 0.4,0.45, 0.5.Alsoaddactual values obtained from thesolution y=3e’—2:2—2:2:-—-2. Reconstruct Table 44.21 byapplying error formulas (44.38) tocorrect the entries ateach step before proceeding tothenext one. Forexample, by (44.38)(b), with 2:0=0,h=0.1, E(0.1) =2[y(0.05 +0.05)-y(0.1)] =2(1.102c -1.1)=0.0052. Hence thecorrected value ofy(0.l) =1.1—I-0.0052 =1.1052. Record thisnewvalue of1/(0.1) inyour reconstructed Table 44.21. Starting withthis corrected value ofy(0.1) =1.1052, compute y(0.2) inonestepandtwosteps. Then apply error formulas (44.38) tocorrect y(0.2), etc. Compare with previous results andwith actual values ofthesolution. Find approximate values when 2:=0.05, 0.1,0.15, 0.2oftheparticular solution oftheequation y’=:0:—|—yzforwhich y(0) =1.Take h=0.05. Inproblem 4,compute y(0.l) andy(0.2) with h=0.1. Compare with the value ofy(0.2) obtained in4.Intheabsence ofasolution, howmany deci- malplace accuracy could youassume inthevalue ofy(0.2)? Hint. See Comment 44.3—3. Byuseoferror formulas (44.38), correct thevalue ofy(0.1) obtained in problems 4and5.Using thiscorrected ﬁgure, proceed toﬁndy(0.2) intwo steps andinonestep. Then correct y(0.2). Find approximate values when x=1.1,1.2,1.3oftheparticular solution of thedifferential equation 1/=1:2+1/2forwhich 11(1) -1.Take h=0.1. What istheapproximate formula error in3/(1.1)? [Hint Calculate 1/(1.1) intwosteps andthen useerror formula (44.38).] Intheabsence ofasolu- tionorerror formula, howcould youestimate theerror iny(1.3) ? Find anapproximate value when a:=1,oftheparticular solution ofthe equation y’=1/(1+1:2)forwhich 1/(0) =0.Useh=0.2andproceed asfollows. Find ;i/(0.2) inonestepandintwosteps. Correct y(0.2) bymeans of(44.38). Then ﬁndy(0.4) inonestepandintwosteps. Correct y(0.4) by means of(44.38). Continue inthiswayuntil youreach y(1). Show howthis value ofy(1)canbeused toapproximate -ir.Compare withactual value of-ir. Hint. Thesolution ofy’=1/(1+:22)forwhich 1/(0) =0isy=Arctan2. Therefore y(1) =-ir/4sothat1r=41/(1). Follow theprocedure outlined inproblem 8toﬁndanapproximate value when a:=1,oftheparticular solution oftheequation y’=yforwhich 1/(0) =1.Show how thisvalue ofy(1) canbeused toapproximate e. Compare with actual value ofe.Hint. Thesolution ofy’=1/forwhich 11(0) =lisy=e‘.Therefore y(1) =e. Find anapproximate value when 2:=2,oftheparticular solution ofthe equation y’=1/:2:forwhich 1/(1) =0.Take h=0.2andfollow thepro- cedure outlined inproblem 8.Show howthisvalue of11(2)canbeused to approximate log2.Hint. Thesolution of1;’=1/2:forwhich y(l) =0is y=log2:.Therefore y(2) =log2. Byformulas (44.14) and(44.31), wehave y(¢o+h)=1/(lo) +hi/(wo) +E. where E=y"(X)h2/2 andXisavalue of:2:intheinterval (:c0,zo +h). Eistheformula error duetostopping withthe1/(1:0) term inaTaylor series. 640 NUMERICAL IVIETHODS Chapter 10 Ifweknew which value of2:tochoose forXinthisinterval, wewould know theexact value ofE.Butwedon't. However, ifMisthemaximum value of|y”(:c)| intheinterval (:c0,:i:0+ h),then §Mhz/2. Wecanthus establish anupper bound oftheerror, inaninterval ofwidth h,inusing formula (44.14) tocompute 1/(xo+h).Callthiserror E1andcallM1the maximum value ofy”(a:) inthis interval. Designate byE2,E3,---,En, theerrors ineach additional interval ofwidth h,andM2,---,M,,the maximum value of|y”(:2:)| ineach such additional interval. Then theupper bound ofthetotal error Eforallintervals is \2 <4-14> |E|§|E1|+lE2|+~-+lE..|§%(Mi+M2+~-~+M»)- LetMbethelargest ofthenumbers M1,M2,---,M,..Then, by(44.4), 2 (4441) |E|gh5nM, where Misthemaximum value of|y"(:i:)| intheinterval (xo,10—I—nh). (a)Useformula (44.41) tocompute theupper bound oftheerror inthe value oflog2ifinproblem 10wehadused h=0.2without corrections. Hint. h=0.2,n=5,M=max. |y”|=max.|—1/2:2] =1inthe interval (1,2). (b)What isthelargest value ofhthatcanbeused toinsure thattheupper bound oftheerror inthecomputation oflog2inproblem 10is0.005, ifnocorrections were made? Hint. By(44.41), wewant anhsuch that h2nM/2 §0.005. Remember hn=1. ANSWERS 44 y(0.4) =1.471, y(0.5) =1.634. Actual values: 1.51547,1.69616. y(0.35) =1.4191, y(0.4) =1.4962, y(0.45) =1.5790, y(0.5) =1.6681. Actual values 1.43470, 1.51547, 1.60244, 1.69616. 1/(0.2) =1.2237, y(0.3) =1.3557, y(0.4) =1.5107, y(0.5) =1.0902. 1/(0.05) =1.0500, y(0.1) =1.1076, 1/(0.15) =1.1739, y(0.2) =1.2503. y(0.1) =1.100, 3/(0.2) =1.231. Canassume only zerodecimal place accuracy ifrounded offtoonedecimal. y(0.1) =1.1152, y[(0.l +0.05) +0.05] =1.2598, y(0.1—I~ 0.1) =1.2496, y(0.2) =1.2700. y(1.1) =1.2000, y(1.2) =1.4650, y(1.3) =1.8236, E(1.1) =0.0312. Would need tomake allcalculations over again with h=0.05, seeCom- ment 44.3—3. y(0.2) =0.1980, y(0.4) =0.3815, y(0.6) =0.5415, y(0.8) =0.6756, y(1) =0.7860. Actual value: 1r=3.14159. y(0.2) =1.2200, y(0.4) =1.4884, y(0.6) =1.8158, y(0.8) =2.2152, y(1) =2.7026. Actual value: e=2.71828. y(1.2) =0.1818, y(1.4) =0.3355, y(1.6) =0.4688, y(1.8) =0.5864, y(2) = 0.6917._ Actual value: log2=0.6931. (a) §0.1. (b)h=0.01. Lesson 45 ANImrnovnmnnr orrnnPOLYGONAL METHOD 641 LESSON 4-5. AnImprovement ofthe Polygonal Starting Method. Lety(z) beaparticular solution of (45-1) y’=f($,y) satisfying theinitial condition (45.11) y(a:0) =yo. Inthepolygonal method, wefound theapproximation, see(44.14), (45-12) 2/(930+h)=1/($0) +y'(rvo)h- Itispossible toimprove thisestimate using amethod similar totheone oftheprevious lesson. Inthis previous lesson, wefound 1/(xo +h)of (45.12) bydrawing, at(:v0,yo), atangent linetotheintegral curve y(z), anddetermining itsintersection with theline:0=:00—I-h.Itismarked RinFig. 45.13. The error in3/(:00 -I—h)isshown asE1. The point Pin lEz .>'=>'(x) E1 _I h R[xo +h,y(xo+h)]E m—y(xo+ Rlxo +hi.'>'(xo) +.Y'(xo)hl P[x.+3-.yo.)+y'(x.)Q] Q(1o.;Vo) m=y'(x°) J’(x0 +(1)571 J’(10‘I’ J’(x0)EJ’o x,, xo+g x12xo+h Figure 45.13 theﬁgure isthemid-point ofthesegment ofthistangent linebetween Q and R.The coordinates ofPwere obtained byusing themid-point for- mula ofanalytic geometry. Substituting thecoordinates ofPin(45.1), weﬁnd (45-14) I/'($0-I" =flivo —|—gr 3/($0) -1-2/($0) ' 642 NUMERICAL Mnrnons Chapter 10 Theequation ofthelinethrough (:co,y0) with slope (45.14) is (45-15> 5=yo.)+<4-an’(4.+ =yo.)+<4-@.>r[x.,+ ya.)+4/<4.) Itsintersection with theline x=2:0+his (45.15) ya.+5)=yo.)+1»/[4,+ 5.+Qy'<».>]- Weshall now prove that theright side of(45.16) isequivalent tothe ﬁrstthree terms ofaTaylor series instead ofonly theﬁrsttwoasin(45.12). Hence theerror intheapproximate value ofy(:z:0 +h)of(45.16), shown asE2inFig.45.13, usually willbesmaller than theerror E1.* Proof. By(38.12) and(38.13), aTaylor series expansion ofafunction oftwovariables is,with :2:=:00+a,y=yo-I-b, (=0f(ivo+¢1,!lo+b)=f($o,yo)+a +b +"'- Therefore witha=b=Qy’(:co), 0»)fl».+Q14.+Q50.)]=100,4.)+Q +Qy'(w.)‘3i9‘,;‘—f°—)+---. Substituting (b)in(45.16), and replacing f(a:o,y0) byitsequal y'(a:0) of (45.1), weobtain (0) 1/(wt+h)=um)+hy'(ro)2 +2[ax +1/(110) ay + Bydifferentiation of(45.1)—we assume thederivatives exist—we obtain 1/"(X)‘By(37.36), theerror orremainder term E’;=-ii hfiftheseries includes the III ﬁrsttwoterms ofaTaylor series; theerror term E;=LEE haiftheseries includes II theﬁrstthree terms. Hence E;<E’;if@ hi‘<Q Ii’,i.e.,ify’”(X2)<2y’'(X1). Forasmall h,3/hislarge. Lesson 45-—Exerci se 643 Hence (9) 3/"($10) =iilzaf/‘Q -I"al£%)% 1/'($o)- Therefore by(e),wecanwrite (c)as 2 <0 ya.+h)=yo.)+ht/($0)+Qy"<4.>+---. Acomparison of(f)with theTaylor series (37.27) shows, with :0=2:0+h, thattheﬁrstthree terms ofeach arethesame. 3!Nofrn. Useofthismethod istherefore permissible only if£5/Z and 8f(w2/).T; 6£I31.8l at(a:0,y°). Example 45.2. Find, bythemethod ofthis lesson, anapproximate value, when :0=0.1,oftheparticular solution of (8) y’=w’+y forwhich y(0) =1. Solution. Comparing theinitial condition with (45.11), weseethat 2:0=0,yo=1.Therefore by(a),y’(0) =0+1=1.Hence with no=0,h=0.1,(45.16) becomes (b) y(0.1) =1+0.1f(0.05, 1+0.05). Here f(z,y) =1:2+y.Therefore (5) f(0.05, 1.05)=(0.05)” +1.05=1.0525. Substituting (c)in(b),weobtain (<1) y(0.1) =1+0.1053 =1.1053. Theactual value ofy(0.1) is1.1055. Inthepolygonal method using two steps, weobtained avalue of1.1026. Note thegreater accuracy ofthe above method. EXERCISE 45 Apply themethod ofthislesson tosolve theproblems which follow. 1.Starting with y(0.1) =1.1053, compute y(0.2) and1/(0.3) ofExample 45.2. Take h=0.1. Compare with thevalues found inLesson 44andinExer- cise44,3. 2.Find y(0.1) ofExample 45.2intwosteps, i.e.,ﬁrstﬁndy(0.05) andthen y(0.05 +0.05). Compare with actual value ofy(0.1) =1.1055. Dothe same fory(0.2). 644 NUMERICAL METHODS Chapter 10 3. (45.3) 4 5 6. 7. 8 9. 10.Following thesteps inComment 44.3—2, develop error formulas, comparable tothose in(44.38), forthenumerical method ofthislesson. Hint. Since formula (45.16) isequivalent totheﬁrstthree terms ofaTaylor series, the remainder orerror term, by(37.36), isy”'(X)h3/3!, where Xisavalue of xintheinterval under consideration ofwidth h.Ans. E[(1@+%)+%l=%{i[(1@+%)+%]~i»@+h>JiE(ro+h)=§[11[(10++—1/<10+Io]- Find approximate values when 2:=0.05, 0.1,0.15, 0.2oftheparticular solution ofthediﬁerential equation y’=z+y,forwhich y(0) =1.Take h-O05 Inproblem 4,compute 1/(0.1) andy(0.2) with h=0.1. Compare with the value ofy(0.2) obtained in4.Intheabsence ofasolution interms ofele- mentary function oranerror formula, how many decimal place accuracy could youassume inyour value ofy(0.2). Hint. SeeComment 44.3—3 Solve theequation andcompare with actual value of1/(0.2). Find approximate values when 2:=1.1,1.2,1.3oftheparticular solution oftheequation y’=2:2+112forwhich y(l) =1.Take h=0.1. What istheapproximate formula error iny(1.1)? Hint. Calculate 3/(1.1) intwo steps andthen apply error formula (45.3). Find approximate values when 2:=0.2,0.4,0.6,0.8,1oftheparticular solution oftheequation y’=1/(1+x2)_for which y(0) =0.Takeh =0.2. Usethisvalue ofy(1)toapproximate 1r.Hint. SeeExercise 44,8. Compare results. Find anapproximate value when 2:=1oftheparticular solution ofthe differential equation y’=yforwhich y(0) =1.Take h=0.1. Usethe value ofy(l)toapproximate e.Hint. SeeExercise 44,9. Compare results. Find anapproximate value when 2:=2oftheparticular solution ofthe equation y’=1/:2:forwhich 1/(1) =0.Take h=0.25. Usethisvalue of y(2)toapproximate log2.Hint. SeeExercise 44,10. Compare results. Byformulas (45.16) andtheerror term asgiven inproblem 3,wehave y(wo+h)=y(wo)+hflwo +2» yo+g1/’(wo)] +E. where E=y’"(X)h3/3! Following theprocedure outlined inExercise 44,11, show that theupper bound oftheerror Eintheinterval (mo,1:0+nh)is 3 (45.31) |E|§%nM, where Misthemaximum value ofy”'(:c) intheinterval (:::0,xo +nh). (a)Useformula (45.31) tocompute theupper bound oftheerror inthe value oflog2asfound inproblem 9above. Hint. h=0.25, n=4, 2 .. M=max|y'”| =max F=2ininterval (1,2). (b)What isthelargest value ofhwhich canbeused toinsure thattheupper bound oftheerror inthecomputation oflog2islessthan 0.005? Hint. By(45.31) wewant anhsuch that h3nM/3! <0.005 andremember nh=1. Lesson 46 STARTING METHOD—TAYLOR Snnrns 645 ANSWERS 4-5 1.y(0.2) =1.2237, y(0.3) =1.3586. ' 2.y(0.05) =1.0513, y(0.05+ 0.05) =1.1055, y(0.15) =1.1630, y(0.15 +0.05)=1.2242.4.y(0.05) =1.0525,y(0.1) =1.1103, y(0.15) =1.1736, y(0.2) =1.2427.5.y(0.1) =1.11, y(0.2) =1.2421. Canassume twodecimal place accuracy. Actual value: y(0.2) =1.2428. 6.y(1.1) =1.2313, y(1.2) =1.5506, y(1.3) =2.0106, y(1.05) =1.1077, y(1.05 +0.05) =1.2334; E(l.1) =0.0028. 7.1/(0.2) =0.1980, y(0.4) =0.3815, 1/(0.6) =0.5415, y(0.8) =0.6757, 1/(1) =0.7862. 8.y(0.1) =1.105, 1/(0.2) =1.2210, 1/(0.3) =1.3492, 1/(0.4) =1.4909, y(0.5) =1.6474, 1/(0.6) =1.8204, 1/(0.7) =2.0115, y(0.8) =2.2227, y(0.9) =2.4561, 1/(1) =2.7140. 9.1/(1.25) =0.2222, y(1.5) =0.4040, y(1.75) =0.5578, y(2) =0.6911. 10.(a) §0.0209. (b)h=0.12totwodecimal places. LESSON 46. Starting Method—Taylor Series. InLesson 44,wefound anumerical solution ofthedifferential equation (46-1) y’=f(r.y), forwhich (46-ll) y(wo)=yo, byamethod which isequivalent tousing aTaylor series toterms ofthe ﬁrstorder; inLesson 45byamethod which isequivalent tousing aTaylor series toterms ofthesecond order. These methods suggest that greater accuracy may beachieved if,forastarting method, weuseaTaylor series toterms oforder greater than two. There are, however, two practical diﬁiculties totheuseofaTaylor series. 1.The function f(z,y) may nothave aTaylor series expansion over the interval inwhich asolution isdesired. Forexample, ify’=f(z,y) = \/Z+1/2,then y”andhigher derivatives donotexist at2:=0. 2.Iff(x,y) hasaTaylor series expansion, itmay beextremely diﬂicult to obtain thederivatives needed informula (37.27). Forexample, try taking afewderivatives off(x,y) =\/:r3y +2:3/3. Ifthese two diﬂiculties arenotpresent, then aTaylor series isindeed agood starting method. By(37.27), aTaylor series hastheform (46.12) yo.+h>=yo.)+1/<w.>h+ 1.’+ 1“ (4) \ + h4+... 64-6 Numrznrcxr. Mm-nons Chapter 10 Itsays ineffect that ifoneknows thevalues ofthefunction y(z) andits derivatives atapoint :0=2:0,then onecanﬁnd thevalue ofthefunction for aneighboring point hunits away. Bydirect substitution in(46.12), wecan therefore ﬁnd y(:c0 +0.1), 3/(xo +0.2), y(x0 +0.3), etc., provided the series converges forthese values ofx. LESSON 46A. Numerical Solution ofy’=f(x,y) byDirect Sub- stitution inaTaylor Series. Weillustrate thismethod bymeans of anexample. Example 46.2. Find approximate values when 2:=0.1,0.2,0.3,0.4 ofaparticular solution ofthedifferential equation (a) y’=$2+2/. forwhich 3/(0) =1. Solution. From (a)weobtain (b)u’=r’+y, 1/”=2x+y’. 2/"'=2+1/”, um=y”’- Hence when :1:=0andy=1,weﬁndfrom (b) <01/<0>= 1.y"<0>=0+1=1, y'"<0>=2+1=3,y“’<0>=3- By(46.12), with 2:0=0,wehave (<1)you=1/<0)+y'<0>h+”';‘,‘”1*+”';§°)1“+”(:f°)1*+---. Substituting in(d),theinitial condition andthevalues found in(c),we obtain hz ha h‘ (9) Z/(h)=1+h+§+§+§+'--. Bydirect substitution in(e),wehave, using only terms toh‘/8, (f)1/(0.1) =1+0.1+§(0.1)’ +§(0.1)3 +§(0.1)* =1.1055125. y(0.2) =1+0.2+%(0.2)2 +g(0.2)3 +§(0.2)‘ =1.2242000. y(0.3) =1+0.3+§(0.3)* +§(0.3)* +§(0.3)* =1.3595125. y(0.4) =1+0.4+§(0.4)2 +§(0.4)=* +§(0.4)* =1.5152000. LESSON 46B. Numerical Solution ofy'=f(x,y) bythe“Creeping up” Process. Bythedirect substitution method, more andmore terms oftheseries must beincluded, ashincreases, inorder tomaintain ade- sired degree ofaccuracy. These terms, ifthederivatives off(:z:,y) are complicated functions, may bediﬁicult toobtain. Amore accurate Lesson 46B SOLUTION nr“CREEPING Ur"Pnocass 647 method ofusing aTaylor series with thesame number ofterms istokeep hﬁxed and “creep up”tothevalue of3/(xo —|—nh)insuccessive steps. We shall demonstrate bytheexample below how this method works. You willsoon discover thatthegreater accuracy ispurchased ataprice—more labor. Example 46.21. Solve theproblem ofExample 46.2 bythe“creeping up”process. Solution. Using ourbasic equation (46.12), with h=0.1andso equal successively 0,0.1,0.2,0.3,weobtain (a)y(0+0-1)=1/(0-1) I, I” =ya»+y'<o><0-1) +”T(,‘”<01)’+lgfi<01)“(4 +%.).(0_1)4+...’ 1/(0.1 —|—0.1) =14/(0.2) H =1/<0-1)+1/<o.1><0-1) + <01)’ "'(0.1) ("(0.1)+y? <01)“+”—,,-,- <0-1)‘+---. y(0.2+0.1)=y(0.3) ' H =y(0.2)+y’(0.2)(0.1) +”—g-3,3 <01)’ +”"',%(0.1>*+y%<0-1>*+---. y(0.3+0.1)=1/(0.4) ' i =y(0.3)+1/(o.3)(0.1) +9% (0.1)’ _.|_ (0_1)3 +3/(nlﬂi) (0_1)4 _|_..._ Theﬁrstequation in(a)isthesame as(d)ofExample 46.2with h=0.1. Hence by(f)ofthatexample (5) 3/(0.1)=1.1055125. By(b)ofExample 46.2, thevalues ofthederivatives ofy(z)when :1:=0.1 andy=1.1055125 are (5) 1/(0.1) =(0.1)’+1.1055125 =1.1155125, y”(0.1) =0.2+1.1155125 =1.3155125, y"’(0.1) =2+1.3155125 =3.3155125, 1/“(0.1) =3.3155125. 648 Nuunnrcru. Marnons Chapter 10 Substituting (b)and(c)inthesecond lineof(a)gives (d) y(0.2) =1.1055125 +(0.1)(1.1155125) + (1.3155125) —|— (3.3l55125) + (3.3155125) =1.2242077. When at=0.2and1/(0.2) =1.2242077, weﬁndfrom (b)ofExample 46.2, (5) y'(0.2) =1.2642077, y”(0.2) =1.6642077, y"’(O.2) =3.6642077, y“>(o.2) =3.6642077. Substituting (d)and(e)inthethird lineof(a),itbecomes (f) y(0.3) =1.2242077 +(0.1)(1.2642077) —|—(0.005)(1.6642077) _ + (3.6642077) +%1(3.6642077) =1.3595755. And when :1:=0.3,y(0.3) =1.3595755, weﬁndfrom (b),Example 46.2, (g) 1/(0.3) =1.4495755, y”(0.3) =2.0495755, y"'(0.3) =4.0495755, y‘4’(0.3) =4.0495755. Substituting (f)and(g)inthefourth lineof(a),weobtain (h) 1/(0.4) =1.3595755 +(0.1)(1.4495755) —|—(0.005)(2.0495755) +gig(40495755) +%l91 (40495755) =1.5154727. Comment 46.3. The particular solution of(a)ofExample 46.2 for which y(0) =1is (i) y(z) =3e‘—2:2—2x—-2. Wecantherefore compare theactual values ofywith those obtained by Lesson 46B SOLUTION‘ BY“CREEPING Ur”Pnocnss 64-9 thedirect substitution method andbythecreeping upprocess, seeTable 46.31. Anexamination ofthetable discloses that thedirect substitution Table 46.31 Direct Substitution Creeping UpProcess See(r)of See(6),(r),(1.)ofObfjfxjé :.£‘(‘)1I‘I‘1°(i)Example 46.2 Example 46.21 1/(0-1) = y(0.2) = 1/(0-3) =1.1055125 1.2242000 1.35951251.1055125 1.2242077 1.35957551.1055128 1.2242083 1.3595764 y(0.4) = 1.5152000 1.5154727 1.5154741 method hasgiven sixdecimal accuracy fory(0.1), four for3/(0.2), and three for1/(0.3) and y(0.4), after being rounded offtothese respective number ofplaces. Ontheother hand, thecreeping upmethod hasgiven sixdecimal accuracy fory(0.1), 3/(0.2), 1/(0.3), andﬁvedecimal accuracy for1/(0.4). Comment 46.4. Comment onError inTaylor Series Method. 1.ByTheorem 37.34, theremainder orerror term ofaTaylor series is _y(”+')(X) "+1(46.41) E-(n+ 1)!h , where Xisavalue ofxintheinterval under consideration ofwidth h. InExample 46.2, westopped with y(4)(:t). Therefore by(46.41), thelimit oferror inusing h=0.4is,with n=4, 5 (46.42) |E|gQglM=0.000085M, where Misthemaximum value of|y(5)(:t)| intheinterval (0,0.4) ofwidth 0.4. Inthecreeping upprocess, thelimit oferror ineach calculation is, by(46.41), withh=0.1, <01)" (46.43) IE]§ M. The total upper limit oferror forfour steps duetoformula only, i.e.,the limit oferror inaninterval ofwidth 0.4bythecreeping upprocess, ex- cluding cumulative errors, istherefore, 5 (46.44) Eg4Q5-1}M=0.000000333M. 650 NUMERICAL METHODS Chapter 10 Acomparison of(46.44) with (46.42) shows approximately how much more accurate thecreeping upmethod is. 2.Let (46.5) E(2;0 +h)=theerror incomputing y(xo +h)byuseofformula (46.12) including terms toh‘. Then by(46.41), with n=4, (46.51) E(5.,+h)=ch‘, where c=y(5)(X)/5! Letusdivide thehinterval inhalf and compute y(:c0 -1-h)intwosteps. Ifwehave some basis forbelieving that y(5)(:t) changes rather slowly inthissmall hinterval, sothat thevariation inthe value of3/5)(X)ineach h/2interval isnegligible, thenwecommit asmall error byusing thesame cof(46.51) foreach halfinterval. Therefore, by (46.51), theerror incomputing 1/(xo +h/2) forahalf interval h/2is approximately 5 (46.52) E(5.,+=5 =size?=3l2E(5., +1.). hThis means that theerror iny(:60+ isapproximately equal toone thirty-second theerror in1/(xo+h).Hence theerror incomputing in . . h htwosteps thevalue ofy(:c° +h),which wewrite asy[(20 + + 1 . . . . h .willhave aninherited error inthestarting value ofy(:00+ plus its own formula error. Since each error equals onethirty-second theerror in1/(:00 +h),thetotal error iny(a:o +h)computed intwosteps, i.e.,the h herror iny[<:z:0 + + ,willbeapproximately equal toone-sixteenth theerror iny(x° +h)computed inonestep. Hence (46-53> EKZ.++-=,—‘6E<w.+h). approximately. LetY(:v0 +h)betheactual value ofthesolution. Then (46-54) Y($0+h)—1/($0+h)=E'($o+h). Y<w.+h) —y[(w.+§)+§] = Subtracting thesecond equation in(46.54) from theﬁrst, weobtain (46.55) 1/[(2.+-2)+§]—yo.+1)=E6.+h)—E[(w.+§) Lesson 4-6B SOLUTION Br“Cizi~:r:i>mo UP”Pnocnss 651 Substituting (46.53) intheright sideof(46.55), thefollowing equations result. uMoyKa+@+Q—ua+0 =w46+9+3-4®+9+3 !!|:(x0 + + —1/($0 -1-h)=E650 —|—h)_‘116'E($0 +h)-and Simpliﬁcation of(46.56) gives G%0EKm+®+g=%hKm+9+g—M%+M> Md ,,,,_nH 1i_ . o -151/($0'|'2)—|—2] !!($o +ll): The ﬁrst formula says that theerror inthevalue ofy(:z:0 +h)computed intwosteps isequal toone-ﬁfteenth thedifference invalues of1/(220 +h) computed intwosteps andinonestep. Formulas (46.57) give usameans of_estimating errors ateach step. Their accuracy does notdepend onaknowledge ofthesizeof3/5)(X) which wedonotknow, butonly onthevariation ofy(5)(X)/5! over a small interval. Wehave assumed thisvariation tobenegligible, an assumption which isnotunreasonable ify“)(a:) changes slowly over h. Forexample, from thecolumn headed “creeping up”process inTable 46.31, wehave with so=0,h=0.2, (5)yKx0++=y(0.1+ 0.1)=y(0.2)=1.2242077, andfrom thedirect substitution column, (b) y(:c0+h)=y(0.2) =1.2242000. Hence bytheﬁrstequation in(46.57), (5)EK5, ++=E(0.1+ 0.1)=,1,(1.2242o77 -1.224200) =0.0000005, which istheapproximate error of3/(0.2) computed intwo steps. The actual error is1.2242083 ——1.2242077 =0.0000006. 3.Acheck onerrors which practical people frequently use,andwhich seems towork, istomake allcalculations over again with anhhalf the 652 NUMERICAL METHODS Chapter 10 sizeoftheoriginal one. Iftheresults obtained with thesmaller hagree with those obtained with thelarger htolcdecimal places, after being properly rounded off,then itisassumed that thecommon numerical value haskdecimal place accuracy. This method is,ofcourse, meaningful only ifthecreeping upprocess isused. 4.This method does notcontain acheck onarithmetical errors. How- ever, ifthere islittle agreement between thevalues obtained byusing h andh/2, allarithmetical computations should bechecked. Ifarithmetical computations arecorrect, hshould bereduced. Comment 46.58. Wehave headed this lesson “Starting Formula- Taylor Series. ”Itisevident that itmay also beused asacontinuing method toﬁnd 1/(0.5), y(0.6), etc. There is,however, apractical objec- tion toitscontinued use. Toinsure adesired degree ofaccuracy, more and more terms oftheseries will beneeded asthedistance from the initial point :00increases. This means wemust evaluate higher andhigher order derivatives. Inmany practical cases, these derivatives, asremarked earlier, may become extremely complicated ormay beeven soright from thestart. Hence, wemust seek other starting and continuing methods which donotdepend onourability toobtain derivatives. If,however, it isnotdifﬁcult toobtain many derivatives, then aTaylor series isnotonly agood starting method butisalsoagood continuing method. Itcanbe used aslong asthe:0values remain within theinterval ofconvergence and aslong asyou arereasonably sure that you have enough terms inthe series togiveyouvalues that areaccurate tothedesired number ofdecimal places. Ahigh-speed calculating machine can, forexample, addwith ease 100terms ofaseries. EXERCISE 46 Apply themethods ofthislesson tosolve theproblems which follow. UseTaylor series toterms ofthefourth order. 1.Usethedirect substitution method toﬁndapproximate values when :1:=0.1, 0.2,0.3,0.4ofaparticular solution ofthedifferential equation y’=2:+y forwhich y(0) =1.Take h=0.1. Solve theequation andcompare your results with actual values. 2.(a)Solve problem 1byusing thecreeping upmethod. Intheabsence ofa solution interms ofelementary functions orofanerror formula, how could youdetermine theaccuracy of1|/(0.4)? (b)Why could not“oneuseerror formula (46.41) todetermine theupper bound oftheerror iny(0.4), justaswedidinExercises 44,11 and45,10 toﬁndanupper bound oftheerror inthenumerical value ofy(2)? Ans. Here y’isafunction of:7:andy;soalsoisy("“")(z). Forexam- ple,11(4)=y'”=y”=1+y’=1+1+y.Hence onecannot determine themaximum value ofy("+1)(a:) inaninterval ofwidth hwithout knowing y(z). Buty(z)isthesolution weseek anddonotknow. Inthe previous problems y’wasafunction only of2:. Lesson 47 STARTING IVIETHOD--RUNGE-KUTTA FORMULAS 653 3.Start again with theequation y’=2:+yforwhich y(0) =1.Using the values ofy(0.2) found inonestepin1andintwosteps in2,apply formula (46.57) tocorrect y(0.2). (Norm. Formula (46.57) isbased onusing terms in aTaylor series toorder four.) Starting with thecorrected ﬁgure ofy(0.2), compute y(0.4) inonestep andintwosteps. Correct y(0.4) bymeans of (46.57). Compare results with those obtained in1and2andwith actual values ofthesolution. 4-.Error formulas (46.57) arebased onstopping with thefourth order term ina Taylor series. Find comparable error formulas ifonestopped with athird order term, with aﬁfthorder term. 5.Using thedirect substitution method, ﬁndapproximate values when 2:=0.1, 0.2,0.3ofaparticular solution oftheequation y’=—a:y forwhich y(0) =1. 6.Solve problem 5,using thecreeping upmethod. How could youdetermine theaccuracy ofy(0.3)? Hint. Seeanswer toproblem 2(a)above. 7.Canthedirect substitution method beused in5toﬁndapproximate values of1/(0-4). ll/(0-5). 1/(0-6). y(0-7). 1/(0-3). '''? ANSWERS 46 1.1.1103417, 1.2428000, 1.3996750, 1.5834667. Actual values: 1.1103418, 1.2428055, 1.3997176, 1.5836494. 2.(a)1.1103417, 1.2428052, 1.3997171, 1.5836486. Compute g/(0.4) bythe creeping upmethod withh=0.05. Then make useofComment 46.4—3. 3.Corrected values: y(0.2) =1.2428055, y(0.4) =1.5836493. 4.Third order term: El(1@+%)+=%l=%Ii[(1@+%)+%l~(<3»+h>l*Ea.+h)=§:11[<15++—yo.+0|- Fifth order term: 4<»~s>+s1-3|y1<-~s>+s1~5+h»|.E<w.+1»)=gla/[(5.+--ya.+0|- 5.0.99501, 0.98020, 0.95601. 6.0.99501, 0.98019, 0.95598. 7.Read Theorem 38.14. Themaximum interval ofconvergence given bythe theorem is <§.Hence, intheabsence ofadditional information, the series found in5cannot beused for >Q. LESSON 47. Starting Method—Runge-Kutta Formulas. The starting method weshall develop inthislesson forﬁnding anu- merical solution of (47-1) y’=f(=v.y) 654 NUMERICAL Mrrrnons Chapter 10 satisfying theinitial condition (47-11) 1/($0) =l/01 wasdeveloped bythetwopeople after whom itisnamed, Runge (1856- 1927) andKutta (1867-1944). Ithasanadvantage over theTaylor series method inthatitdoesnotusederivatives. Asmentioned intheprevious lesson, ifthefunction f(z,y) isvery complicated, itmay beextremely diﬂicult toobtain itssecond derivative, letalone itsthird and fourth. Because ofthisfactanditshighdegree ofaccuracy, itisinpractice prob- ably themost widely used starting method. By(46.12), (47.12) ye.+10=yo.)+1/<x.>h+ 1*+ he +1/(4;('$o) h4+____ Differentiating (47.1), weobtain (47-13> 1/'6)=$76.1/>+ %r<w.y>y'<w). Substituting (47.1) and(47.13) in(47.12), andstopping with theh”term, wehave (47-14) 1/($0 +h)=?!($o) +f(1?01!Jo)h 2 + + r<w..y.>] Ouraimwillbetorewrite theright side of(47.14) sothat itwillhave theform (47-15) 1/($0 +h) =I/($0) —|—Ahf($o»!lo) +Bhflfvo -l"Ch,yo+Dhf($o.!/0)], andthen ﬁnd values ofA,B,C,Dsothat theright side of(47.15) will actually equal theright sideof(47.14). By(38.12) and(38.13), theTaylor series expansion ofafunction oftwo variables about apoint (x0,y0), is,with :1:=so+a,y=I/0+b, (47.16) f(1'3o+a»!lo+b)=f($o,y0)+¢ +b +"' In(47.16) leta=Chandb=Dhf(x0,yo). There results (47-17) flro+Ch.1/0+Dhf(Ivo.y0)l =f(wo.1/0) +ch +1>hf<x..y.) +--- Lesson 47 Srxnrmo METHOD-—RUNGE-KUTTA FORMULAS 655 Substituting (47.17) in(47.15) andsimplifying theresult, weobtain (47-18> ye.+h)=yo.)+(A+B)hf(wo.y0) +Bh’[0 +1>f(-5.1/5) - Comparing (47.18) with (47.14), weseethat both right sides willbealike if,forexample,* (47.19) A+B=1, B=3, C=1, D--=1, from which weobtain (47.2) A=Q-, B=4, C=1, D=1. Substituting these values in(47.15), wehave (47-21) ?/(10 -l"h)=3/(930) —|—=i‘hf($0.?/0) +ihfllo —|—h,1/0-l"hf($0,Z/o)l- Formula (47.21) canbewritten inthesimpler form (47-3) 1/(110 -l"h)=1/($0) —|—2("1 -1-M2), where (47-31) "1=hf(1'»‘o,?/0), "2=hf($o -1-h,1/0—|—M1)- Iftheﬁrst four terms oftheseries (47.12) areused, then theRunge- Kutta formula becomes (47-32) 1/(we+h)=2/(110)+t(v1+4v2+vs). where (47-33) "1=h-f($0,1/0) ="1. "2=hf($o +ill»1/0+2'11). 03=hf(x0 +h,yo-1-2122-—v1). And iftheﬁrst ﬁveterms oftheseries (47.12) areused, then the Runge-Kutta formula becomes (47-34) 1l($0 +h)=2/($0) +§(w1 -1-21112+2'93+W4), where (47-35) wl=hf(93o,y0) ="1, W2=hf($0 ‘l’2(1)yo-l"21111) ="2- ws=hf($o +277'»1/0'1'2192), W4=hf($o +h.1/0-l"we)- "There areother possible choices ofA,B,C’,D,butthese arethesimplest. 656 NUMERICAL METHODS Chapter 10 Since (47.3), (47.32), and (47.34) were obtained byincluding thesecond, third, and fourth order terms respectively ofaTaylor series, these for- mulas have been called respectively thesecond, third, and fourth order forms. AsinTaylor series, twomethods areavailable forﬁnding y(x0 +0.1), y(:z:0 +0.2), 3/(:00 +0.3)bytheRunge-Kutta method, onebydirect sub- stitution informula (47.3) or(47.32) or(47.34), theother bycreeping up toy(:z:0 +0.2)andthen toy(x0 —|—0.3)bykeeping h=0.1ﬁ:z:ed andusing anyoneoftheformulas insuccession. Example 47.36. Usethefourth order form andthecreeping upproc- esstoﬁndanapproximate value when :1:=0.2oftheparticular solution of (2) y’=1’+y, forwhich 3/(0) =1. Solution. Comparing theinitial condition with (47.11), weseethat 1:0=0,y0=y(x0) =1.Hence, by(47.34), with :00=0,h=0.1, (b) 1/(0-1) =3/(0) +‘5(W1 '1'2W2'1'2W3-l"W4)~ By(47.35), with :00=0,y0=y(:c0) =1,h=0.1andf(:c,y) =:02+y, wehave 1.01=().1f((),1) =0.1(1) =0.1, wg=0.1f(0.05, 1+0.05) =0.1[(().()5)2 —|—1.05] =0.10525, w3=0.1f(0.05, 1+0.052625) =0.1(0.052 +1.052625) =0.1055125, w.=0.1f(0.1, 1+0.1055125) =0.1(0.12 +1.1055125) =0.11155125. Hence, bytheabove values, theinitial condition and(b),anapproximate value ofy(0.1) is (c) y(0.1) =1+}(0.1 +0.2105 +0.211025 —|—0.11155125) =1.1055127. With x0=0.1andh=0.1,(47.34) now becomes (d) I/(0-2) =ll(0-1) -l"'l‘(Wi -l“2W2-l"2W3'l'W4)- By(47.35), with 2:0=0.1,y0E1/(2:0) =y(0.1) =1.1055127 andf(z,y) = 2”+1/. wl=0.1f(0.1, 1.1055127) =0.1(0.12 —|—1.1055127) =0.11155127, w,=0.1f(0.15, 1.1055127 +0.0557756) =0.1(0.152 +1.1612883) =0.1183788, Lesson 47 STARTING METHOD—RUNGE-KU1TA FORMULAS 657 w3=0.1f(0.15, 1.1055127 +0.0591894) =0.1(0.152 +1.1647021) =0.1187202, w4=0.1f(0.2,1.1055127 +0.1187202) =0.1264233. Hence by(c)and(d),anapproximation ofy(0.2) is (e)y(0-2)=1.1055127 +1-(0.1115513 +0.2367576 +0.2374404 +0.1264233) =1.1055127 +0.1186954 =1.2242081. Comment 47.37. Ifwehadfound y(0.2) inonestep, formula (47.34) would have become, with h=0.2,:00=0,y(x0) =y0=1, (f) 1/(0-2) =1+t(w1+2w-2+2w-3+wi). where (g) wl=0.2f(0,1) =0.2(1) =0.2, w.=0.2f(0.1, 1+0.1)=0.2[(0.1)’ +1.1]=0.222, w;;=0.2f(0.1,1 +0.111) =0.2[(0.1)2 +1.111] =0.2242, w4=0.2f(0.2, 1+0.2242) =0.2[(0.2)2 +1.2242] =0.25284. Hence by(f)and(g) (11) y(0.2) =1+%(0.2+0.444+0.4484 +0.25284) =1+30.34524) =1.2242067. Comment 47.4. Comment onError inRunge-Kutta Method. 1.Ithasbeen proved that theerror foraninterval ofwidth h,inusing theRunge-Kutta fourth order form, is (47.41) E(x0+7.)=Ch‘, where Cisaconstant. Although theconstant Cin(47.41) isnotthesame asthecin(46.51), ifwestart with (47.41) andfollow thesteps after (46.51), wearrive atthesame conclusion (46.57). Hence by(46.57), 1::1(1):1:"l-to E($o+h)=—?/ 110+‘ '"-*y($o+h)'+ InExample 47.36, wefound, with 2:0=0,h=0.2,[see(e)] h(5) 3/[(20+5)+=y[0.1+0.11=y(0.2)=1.2242081, 658 NUMERICAL METHODS Chapter 10 and,inComment 47.37, (b) y(a:0 +h)=y(0.2) =1.2242067. Therefore by(47.42), with :00=0,h=0.2, (5)ER“ ++=E(0.1+0.1)=1*r(1.22-12081 -1.2242067) =0.0000001, which istheapproximate error iny(0.2) obtained intwosteps. Theactual error in3/(0.2) obtained intwosteps is0.0000002, seeTable 46.31. 2.Allother remarks onerrors inComment 46.4 carry over without change totheRunge-Kutta method. Comment 47.5. This method could alsobeusedasacontinuing one. The objection toitsuseforthispurpose liesinthelarge number ofcom- putations which must bemade ateach step. Weshall later give acon- tinuing method which islesstime consuming andjustasaccurate. EXERCISE 47 Insolving thefollowing problems, usethefourth order Runge-Kutta method. 1.Prove error formulas (47.42). 2.Using direct substitution, ﬁndapproximate values when 2:=0.1and0.2 ofaparticular solution oftheequation y’=2:+yforwhich y(0) =1. Take h=0.1. Compare results with actual values. 3.Solve problem 2byusing thecreeping upmethod. Intheabsence ofasolu- tioninterms ofelementary functions orofanerror formula, how many decimal place accuracy could youassume inthevalue ofy(0.2)? Compare with actual value. 4.Using thevalues ofy(0.2) found inonestepin2andintwosteps in3,apply formula (47.42) toapproximate theerror iny(0.1 +0.1). Compare with actual error. 5.Using direct substitution, ﬁndapproximate values when :1:=0.1and0.2 ofaparticular solution oftheequation y’=2:2+yzforwhich 1/(0) =1. 6.Solve problem 5,using thecreeping upmethod. 7.Using thevalues ofy(0.2) found in5and6,determine, bymeans offormula (47.42), theapproximate error iny(0.1 +0.1). ANSWERS 47 2.y(0.1) =1.1103417, y(0.2) =1.2428000. Actual values: 1.1103418, 1.2428055. 3.y(0.2) =1.2428052. Canassume ﬁvedecimal accuracy, seeComment 46.4—3. 4.E(0.1—}— 0.1)=0.0000003. Actual error =0.0000003. 5.y(0.1) =1.1114629, 1/(0.2) =1.2529908. 6.y(0.15) =1.1776078, y(0.2) =1.2530162.7.E(0.1+ 0.1) =0.0000017. Lesson 48A Fmrrs DIFFERENCES 659 LESSON 48. Finite Differences. Interpolation. Before wecandevelop acontinuing method which isjust asaccurate asRunge-Kutta butlesstime consuming, weshall need additional infor- mation. We momentarily digress, therefore, todevelop this needed information. LESSON 48A. Finite Differences. Deﬁnition 48.1. Theﬁrst difference ofafunction f(z), written as Af(x) (read “delta fof2:”)isdeﬁned as (43-11) Af(I) =f(rv+h)—f(r), where hisaﬁxed constant. Itisthedifference invalues ofthefunction fortwoneighboring values ofx,hunits apart. Thesecond difference A21‘(ac)isdeﬁned asthedifference oftheﬁrstdif- ference off(z)fortwoneighboring values ofre,hunits apart. Bydeﬁnition, therefore, (43-12) A2f(Iv) =A[Af(r)] =A[f(1 +h)—f(¢v)]=Af(r+h)—Af(I)- Similarly thethird difference A3f(a:) isdeﬁned asthedifference ofthe second difference off(z) fortwoneighboring values of2:,hunits apart. Therefore (48.13) A3f(x) =A[A2f(x)]. And ingeneral, wedeﬁne (48.14) A"f(x) =A[A""f(x)], n=1,2,---. With theaidoftheabove deﬁnitions, weshall now explain how to construct atable ofdifferences Af(:z;), A2f(:c), ---,A"f(a:). Example 43.15. If (3) f(I)=6’ and h=0.1,construct atable ofdifferences ofe’forvalues ofrvfrom 0to0.5. Solution. (See Table 48.16). Since h=0.1,weshall need values of f(z) =e‘that, beginning with :0=0,are0.1unit apart. Intheﬁrst column ofourtable, wetherefore write x=0,0.1,0.2,0.3,0.4,0.5. In thesecond column wewrite thevalues off(z) =e‘when ac=0,0.1, ---,O.5.* By(48.11) Af(:c) isthedifference inthevalues off(x) fortwo ‘Various texts areavailable that contain tables ofvalues ofe‘,0", log2:,sinx, cos2:,etc. Excellent tables ofvalues ofthese andother functions, correct totwelve andmore decimal places, have been published under thesponsorship oftheNational Bureau ofStandards, Mathematical tables project. 660 NUMERICAL Mrrruons Chapter 10 Table 48.16 1=f($) = Affl) = A2f(1) = Aafll) = A4111) =A5f(1) = e‘= Ae‘= A20” = A3e“ = Afe‘ = A512” = 0 1.00000 0.1 1.10517 0.2 1.22140 0.3 1.34986 0.4 1.491820.10517 0.11623 0.12846 0.14196 0.156900.01106 0.01223 0.01350 0.014940.00117 0.00010 0.00007 0.00127 0.00017 0.00144 0.5 1.64872 neighboring values of1:,0.1unit apart. Hence weobtain thethird column ofthetable bytaking thedifferences ofthesecond column. Forexample, by(48.11), Ae°'2=e°'3—e°‘2=1.34986 —1.22140 =0.12846. Anal- ogously A2f(:c), by(48.12), isthedifference ofAf(x) fortwoneighboring values ofrc,0.1unit apart. Therefore weobtain thefourth column by taking differences ofthethird column. Forexample, by(48.12), A2e°'2 = A(Ae°~’) =Ae°'3-Ae°'2=0.14196 -0.12846 =0.01350. Theelements oftheﬁfth column A3e"' arethedifferences ofthefourth column, etc. Comment 48.17. Ifyouwant theﬁfthdifference off(z), i.e.,A5f(:c), when :1:=a,youmust knowf(a),f(a +h),---,f(a +5h). And ingeneral ifyouwant themth difference off(z) when :1:=a,youmust know f(a), f(a+h),---,f(a+mh). Intheabove example, a=0,andwetherefore hadtoknow f(0), f(0.1), ---,f(0.5) inorder toevaluate theﬁfthdiffer- ence ofe’,i.e.,Ase‘, when :1:=O. Comment 48.18. Callyo,yl,---,y,,thevalues off(ac)corresponding tof(xo), f(:ro —|—h),---,f(:r:o +nh). ByDeﬁnition 48.11, (48-181) A!/0=2/1—1/0- By(48.12), (48.18l) andDeﬁnition 48.11, (43-132) A2?/0 =A142/ol =A0/1 -*I/0)=AZ/1*-Ayo =1/2-2/1 —y1+z/e=y2—2y1+1/6 By(48.13), (48.181) and(48182), (48-183) A31/6=/1142061 =M112—21/1+yo)=A?/2—2Ay1+Aye =3/3—?l2"'2?/2+2!/1+?/1 -*1/o= ya-31/2+3!/1 -110- Acomparison ofthecoefficients ontheright sides of(48.182) and(48.183) with therespective coeﬁicients intheexpansion of(re-—1)2and(:1:—-1)3 Lesson 4-8B POLYNOMIAL INTERPOLATION 661 shows that they arealike. This leads onetosuspect that thefourth dif- ference ofyois (48-184) A‘!/e=94—411:;+692—491+yo- You canverify that this suspicion isindeed correct. Infact itcanbe proved byinduction, using anargument similar totheonegiven inLesson 49A, that thecoefficients intheexpansion ofA"y0 arethesame asthe coefficients intheexpansion of(:1:—1)". Therefore .."<—1>"n! . 48.19 A= -i— .._, where itisunderstood that Aoyo =yo.Hence, interms off(z), (48.19) becomes (48191) me.)=ii fl$o +<11—1011. ,_0.. LESSON 48B. Polynomial Interpolation. Letusassume that the only information wehave about afunction isitsvalues when 1:=$0, :00+h,xo+2h,---,2:0+mh,andthat wewish toﬁndavalue ofthe function when 1:0=1:0+dh,d960,1,2, .This situation occurs, forexample, when weneed thevalue ofsin0.5245 andthetable wecon- sult gives values ofsin0.5236, sin0.5265, sin0.5294, etc., i.e., itgives values over intervals ofwidth h=0.0029. The usual procedure inthis case istoaddtothevalue ofsin0.5236, 29;ofthedifference invalues be- tween sin0.5236 and sin0.5265. When wedothis, weareimplicitly assuming that thegraph ofthesinefunction between these twopoints is approximated byastraight line. Wearethus ineffect using alinear interpolation toapproximate thesine function between twopoints ofits graph. Abetter method ofinterpolating thevalue ofsin0.5245 from the tabular values istoapproximate thesine function byapolynomial f(z) whose graph coincides with thesinefunction atmore than thetwopoints sin0.5236 andsin0.5265. Ifitcoincides with three values ofthesine function, wegetasecond degree polynomial toapproximate sin:0;ifit coincides with m+1values, wegetapolynomial ofdegree mtoapproxi- mate sin2:.This approximating polynomial f(z) will then give, when :1:=0.5245, amore accurate value ofsin0.5245 than willthelinear inter- polation method described above. Two questions now arise. Given aset ofvalues ofafunction f(z) at1:0,9:1,---,:v,,,. Does apolynomial exist that cantake onthese values atthegiven points, andifthere issuch a ‘The fraction n!/[k!(n —k)!]isalsofrequently written as - x=662 N111/11-11110111. Mmuons Chapter 10 polynomial, isitunique? The answer toboth questions isgiven bythe following existence anduniqueness theorem which westate without proof. Theorem 48.2. Letf(x0), f(x1), ---,f(:c,,,), bem+1distinct values of afunction f(z). Then there isoneandonly onepolynomial F(:e) ofdegree lessthan orequal tomwhich coincides with these m—|—1values off(z). Deﬁnition 48.21. The unique polynomial F(a:) ofdegree lessthan or equal tom,whose values coincide with m-1-1distinct points ofafunc- tionf(z), iscalled apolynomial interpolating function of Comment 48.22. Weshall beinterested only inapolynomial inter- polating function, although functions other than polynomials may beused forinterpolation purposes. Comment 48.23. Very fewofyouhave used nonlinear interpolation toapproximate thevalue ofafunction f(z) that isnotgiven intables, from those which are. Tousethismethod, itisnecessary toﬁndaninter- polating polynomial F(a:) which agrees with f(z) atthree ormore points and isofdegree g2. Itwould betroublesome indeed, ifeach time we wished tousepolynomial interpolation, wehadtoﬁndtheapproximating function F(:z:). Weareindebted toNewton (1642-1727) forformulas by which wecanreadily make polynomial interpolations whenever the abscissa points, forwhich thefunctional values areknown, areevenly spaced. These formulas arederived inthenext lesson. EXERCISE 48 1.Iff(z) =log1andh=0.1,construct atable ofdifferences oflog:2:forvalues of:2:from 1to1.5. 2.If_f(0) =sin(0)andh=5°,construct atable ofdifferences ofsin0 for values of0from 10°to35°. ANSWERS 48 1. f(I)= Af(1=) = A2f(1) =431(1) =A‘f(I) = A5f(1=) = loga: = Aloga: =Azlogx =A3loga: =Afloga: =A51oga: = 1.1 0.09531 1.2 0.18232 1.3 0.26236 1.4 0.33647 1.5 0.40547 1.6 0.470000.08701 0.08004 0.0741 1 0.06900 0.06453-0.00697 0.00104 —0.00022 0.00004 —0.00593 0.00082 —0.00018 —0.00511 0.00064 —0.00447 Lesson 49A 2.NI-JWTON’B (Fonwxnn) Irrranromvrron Foauom 663 f(9)= 0:sin0=Af(@)=A’f(6) =A3110) =A‘f(9) =1151(0) =Asin0 =A2sin0 =A3sin0 =A4sin0 =A5sin0 = 10° 0.17365 15° 0.25882 20° 0.34202 25° 0.42262 30° 0.50000 35° 0.57358—0.00l97 —0.00063 0.00001 0.00003 ——0.00260 -0.00062 0.00004 —0.00322 —0.00058 —0.003800.08517 0.08320 0.08060 0.07738 0.07358 LESSON 49. Newton’s Interpolation Formulas. LESSON 49A. Newton’s (Forward) Interpolation Formula. We assume thatf(:0)isafunction which isdeﬁned onaninterval (:00,1:0+mh), andthat itsvalues have been given only forthem—|—1distinct abscissa points: 2:0,:00—|—h,:00—|—2h,---,:00-1-mh,ashappens, forexample, when weconsult atable ofvalues ofthefunction orwhen these isolated, distinct values have been found experimentally, Fig.49.1. Ourobject willbeto f(xo+2h) f(Io+h) f(xo+mh) ‘f(Ie) xo xo+h xo+2h xo+mh Figure 49.1 ﬁnd apolynomial F(a:), ofdegree §m that agrees with f(z) atthese m+1points. ByTheorem 48.2, weknow that there issuch apolynomial F(:v) andthat itisunique. ByDeﬁnition 48.21, thisunique polynomial F(a:) isaninterpolating function off(z), bywhich wemean that itwill give anapproximate value ofthefunction f(z) forana:which does not coincide with thepoints mo,:80+h,---,mo+mh.Toobtain aformula forF(x), weproceed asfollows. ByDeﬁnition 48.1 Af(I8o) =f($o+h)-f($o)- Solving forf(:e0 —|—h),wehave (49-11) f(w6+h)=f(Io) -1-Af(4'3o)- Applying Atoboth sides of(49.11) gives (8) Af($o +h)=Af(¢o) +A2f(=vo)- 664 NUMERICAL Mrrrnons Chapter 10 Again byDeﬁnition 48.1, weobtain (b) Af(¢o +h)=f(wo+2h)-f($o+h)- Since theleftsides of(a)and(b)arethesame, wecanequate their right sides. Hence (<1) f($o '1"271)=/‘($0 -frh)+Af($o) +A2f($o)- Byreplacing in(c)thevalue off(:to +h)asgiven in(49.11), wehave ﬁnally (49-12) f(w6+2h)=f($0)+2Af($o) +A2f(¢o)- Applying Atoboth sides of(49.12) gives <d> Af($o+2h)=Af($o)+24%.) +A3f($0)- ByDeﬁnition 48.1, (8) Af(Iv6 +2h)=f(fvo+3h)—f($0+2h)- Hence equating theright sides of(d)and(e),weobtain (1) f(wo+3h)=f(wo+2h)+Af(wo) +2A2f(wo) +A°f(wo)- Replacing in(f)thevalue off(x0 +2h)asgiven in(49.12), wehave ﬁnally <49-13) fee+sh)=f(5'30)+3Af($o) +34960) +A3f($o)- Ifyouwillcompare theconstant coefficients intheﬁnal expressions for f(:c0 -1-h),f(a:0 +2h),f(:eo +3h)asgiven in(49.11), (49.12), and(49.13), with thecoefficients intherespective expansions of(:1:-1-1),(0:—|—1)”, (a:-1-1)3,youwilldiscover that they arethesame. This observation leads onetosuspect that thecoeﬁicients intheexpansion off(xo +nh) will bethesame asthose intheexpansion of(2:+1)". Weshall now prove byinduction that thissuspicion isindeed correct. Proof. We assume that the coefficients inthe expansion of f[a:0 +(k-—1)h] arethesame asthecoefficients intheexpansion of (:1:+1)"'1. That isweassume (a)flwo+(Iv—1)h]=f(w6)+(Iv—1)Af(we) _|_ A2f(,;o) lc—1Ic-—- -—- _+ A3f(x°) _|_..._|_A1‘1f(,;0) isavalid equation. Wemust then prove, byComment 24.1, that the coefficients intheexpansion off(:c0 —|—kh)arethesame asthecoeﬂicients Lesson 49A NEw'roN's (Foawxan) INTERPOLATION Fonmxnx 665 intheexpansion of(:1:+1)”. Applying Atoboth sides of(a),weobtain (b)Ame+(k—1)h1=Ame)+<1—1)A’f<ze) + A3160) +---+A"f(xe). Adding theleftsides of(a)and(b)there results, by(48.11), (c)f[xo +(Ic-1)h]—|—Af[xo +(Ic——1)h] =flxo+(Iv—1)hl+f(1o +kh)—flxo+(70-"1)hl =f($o +1911)- Theaddition oftheright sides of(a)and(b)gives (<1)re.)+kArc.)+ii 4%.) +<1-1)<k—2)<k-3:1)+30-no-2)Aaﬂxo)+___ (k—l)(k—2)---(lc—m)—|—m(lc—l)(Ic-2)---(lc—m+1) + m! ><Amf($o) +---+A"f<we). which simpliﬁes to (e)re.)+1Arc.)+i‘%',"—” A216.)+81%"-31 4%..) k(k—1)(k-2)---(Ic—— +1) _|_...+ m‘ mA"'f($o) +---+A"f(Io)- Since (c)isthesum oftheleftsides of(a)and(b),and (e)isthesum of their right sides, wecanequate these twotoobtain <0re.+11)=re.)+ItArc.)+4% 4216.) + A3_f(;,;o) _|_... __ _ ...1;_ M +M"11”‘ 22,1,("‘+1)Afoe)+---+A"f(Io)- The coefficients intheexpression ontheright sidearethesame asthose intheexpansion of(:1:+1)". Hence oursuspicion isproved. Wetherefore cannow write, by(f), -1<49-14) re.+nh)=re.)+nAre.)+’1’12—,—) 4%.) _ A3f(,,0) +... +n(n-—l)(n —23n-!- -(n—m+ 1)Amfcto)’ n= 0,1,2,--~,'m. 666 NUMERICAL Mrrruons Chapter 10 [Note that when n=m,thelastterm simpliﬁes toA’"f(a:0).] Foreach n=1,2,---,m,theright sideof(49.14) gives aformula forcomputing therespective ordinates f(:e0), f(x0 +h),---,f(xo -1-nh)interms of f(:c0) andthedifferences off(:c0).* Since theright sideof(49.14) ismeaningful even when nisnotzero or aninteger, wecanlet _ n(n—1)2(49-15) F(iv)—f($0)+nAf(5'50) +—2,? Af($0) + A3f(x0) _|_... +"f"_‘)8’“22”,‘'<”"’"+11A"'f(a:o). where nisanynumber between 0andm,andwhere (49.16) :1:=2:0+nh, n=2%“- Since theright sideof(49.15) isthesame astheright sideof(49.14), we know that (49.161) F(:eo +nh)=f(x0 -1-nh), n=O,1,2,---,m. Substituting thesecond equality of(49.16) in(49.15), weobtain (49.17) F6)=f<e.)+if‘,f—°’(x-1'30)+ <1-a)c-$0—h) +4?-($8<x—w..)<»-we-h)<e-e.—2h)+--- Amf(~'$0)+W(I “‘x0)($ -150 "'11) ><(x——:e0——2h)---(x-—:e0—[m—-1]h). By(49.17), weseethat F(:e) isapolynomial ofdegree §m. Wehave thusproved, by(49.l61) and(49.17), that: 1.The function F(:z:) deﬁned by(49.15) or(49.17) agrees with thefunc- tion f(z) atthem+1points forwhich x=xo,xo+h,avo+2h, .4.’$0 + mh, 2.F(:z;) isapolynomial ofdegree §m. Therefore, byTheorem 48.2 and Deﬁnition 48.21, F(:e) istheunique ‘Note theresemblance between thisformula andtheTaylor series formula (37.27) bywhich onecancompute avalue ofj(:c0+nh)interms off(:v0) andthederivatives off(z)at:1:=zoo. Lesson 49A N1:w'roN’s (Fonwxnn) Irrrsarotzrrron Fommtx 667 polynomial interpolating function ofthegiven function f(z). This means that ifwewish toﬁndanapproximate value off(z), when (49.18) as=1:0—|—nh, n=%» nreal, wecanusethefunction F(x) deﬁned byeither ofthetwoformulas (49.15) or(49.17). These formulas areknown asNewton’s (forward) inter- polation formulas. Example 49.19. Use Newton’s (forward) interpolation formulas (49.15) and(49.17) andthefollowing table ofvalues (a) e°~1=1.10517, e°'2=1.22140, e°'3=1.34986, e°~'*=1.49182, e°'5=1.64872, toﬁndanapproximate value ofe°‘14. Solution. Byuseof(49.15). Here f(z) =e’,x0=0.1andh=0.1. Since wehave been given ﬁveequally spaced values ofthefunction e’,by Comment 48.17, weshall beable toﬁndonly thefourth difference, namely A“f(:c0), informula (49.15) or(49.17). Weshall therefore have tostop with m=4.And since weseek thevalue ofe°'“, the:1:inthese formulas is .14—— .1 0.14. Therefore by(49.18), n=Li)——1—()— =0.4. Substituting the above values in(49.15), itbecomes (1))F(0.14) =f(0.1)+(0.4)Af(0.1) + l A2f(0.1) +(2-i.*l(£4_":§'1li A3,~(0_1) (0.4)(0.4 —l)(0.4 —2)(0.4 —3)4+ 4,p Af(0.1). Theneeded differences have already been calculated inTable 48.16 inthe rowbeginning with 0.1. Therefore (b)becomes (c)F(0.14) =1.10517 +(0.11623)(o.4) -(0.01228)(0.12) +(0.0o127)(o.064) -(0.00017)(o.0416) =1.15027, which istheapproximate value ofe°1‘. 668 Numsnrcxr. METHODS Chapter 10 Solution. Byuseof(49.17). Inthis formula rt—x0=0.14 - 0.1=0.04. Asbefore h=0.1, m=4.Substituting these values in (49.17), itbecomes (<1)F(0.14) =f(o.1)+96%Af(0.1) +%51;,)- (0.04)(0.04 -0.1) + (0.04)(0.04 —0.1)(0.04 ——0.2) + (o.04)(0.04 -0.1)(o.o4 -o.2)(0.04 -0.3), which isequivalent to(b)above. Comment 49.191. The actual value ofe°'“‘ toﬁve decimals is 1.150 27,thesame asthatobtained in(c)above byusing afourth degree polynomial interpolating function. Ifyou hadused linear interpolation, i.e.,obtained thevalue ofe°‘14 byadding toe°‘1two-ﬁfths ofthedif- ference invalues between e°"ande°'2, youwould have obtained 1.151 66 which iscorrect toonly twodecimal places. LESSON 49B. Newton’s (Backward) Interpolation Formula. New- tonalsogave aformula bywhich itispossible tointerpolate backward instead offorward. LetV,which isanupside down delta, called “del,” bedeﬁned by (49-2) Vf(w) =f(w)—f(w-h)- Asbefore weassume wehave been given values off(z) only forthose points whose abscissas are2:0,x0——h,:00——2h,---,:00——mh,Fig.49.21. me-mh) ("‘°'2") f[1-("I—1)hl I I0 I lﬂxo _h) lﬂxo) x0—mh x0-(m-1)h xo-2h xo-h x0 Figure 4-9.21 LetF(:c), ofdegree §m,beapolynomial interpolating function off(z). Byfollowing theprocedure outlined inLesson 49A, you willobtain, in place of(49.15), Lesson 4-9B NEw'roN’s (Bxcxwxnn) INTERPOLATION FORMULA 669 (49-22) Fe)=16.)—nvm.)+m v’r(».) .._ v3f(x0) +... +"f""1”"‘22,,‘'‘"'’"+1)(—1)"v"r(e.). where (49.28) a:=x0——nh, n=%=—:%,nreal. Wehave lefttheproof toyouasanexercise. If,in(49.22), wereplace nbyitsvalue asgiven in(49.23), weobtain thealternate form 2 (49.24) F6)=foe)+3%?!(4—4.)+ (4—a)c—4.+h) +Y%§,",—°)(e-e.)(»—e.+h)(e—e.+2h)+--- +Y},’5,5,§3—’(e-e.)(e—e.+h) ><(:c—a:0+2h)---(:c—a:0+[m—1]h). Formulas (49.22) and(49.24) areknown asNewton’s (backward) inter- polation formulas. Example 49.25. Usethevalues ofe’given in(a)ofExample 49.19, and Newton’s backward interpolation formula toﬁnd anapproximate value ofe°'4°. Solution. Asintheprevious Example 49.19, f(z) =e‘,h=0.1and since wehave been given ﬁveevenly spaced values ofe‘,themin(49.22) is four. Here, however, x0=0.5,2:=0.46andby(49.23), n= = 0.4.Substituting these values in(49.22), weobtain (a)F(0.46) =f(o.5)-o.4vf(0.5) +9-%°_'6l v'2f(0.5) _ V3]-(0_5) + v4_f(Q_5)_ The needed differences aregiven inTable 49.26. They areslightly dif- ferent inthiscasefrom those inTable 48.16. Here :00=0.5.Wemust therefore ﬁnd differences of_f(0.5) instead off(0.1), and remember Ve°‘5 =e°'°—e°"‘, Ve°'4 =e°'4—e°'3, etc. Substituting in(a)the 670 NUMERICAL METHODS Table 49.26Chapter 10 f(1)= 2:’ e‘=Vf(¢) =Ve"=V2f(1¢) = Vze‘ =V°f(1) = V4f(1) = V3e" = V4e" = 0.1 0.2 0.3 0.4 0.51.10517 1.22140 1.34986 1.49182 1.648720.11623 0.12846 0.14196 0.156900.01223 0.01350 0.014940.00127 0.00144 0.00017 ﬁgures shown intherowbeginning with 0.5ofTable 49.26, weobtain (b)F(0.46) 1.64872 -0.4(0.15690) -(0.12)(0.01494)-(0.064)(0.00144) -(0.0416)(0.00017)1.64872 -0.064651.58407, which istheapproximate value ofe°"°. Itsactual value toﬁveplaces is 1.58407 which agrees with (b)toﬁveplaces. Byalinear interpolation, youwould have obtained 1.58596. LESSON 49C. The Error inPolynomial Interpolation. When we useapolynomial F(a:) ofdegree §masaninterpolating function forf(:z:), weknow only that both agree atthose points whose abscissas arex0, 1:0+h,---,9:0-1-mh. For other values of11:,thedifference between F(x) andf(z) may bevery small orvery large depending onwhat the graph off(:e) looks likeforthese intermediate values. The following error formula (49.31), which westate without proof, is based ontheassumption that f(x) hasacontinuous derivative oforder m+1inaninterval I.'a§x§b,which contains thepoints 1:0,x0+ll; ---,x0—}-mh. Let (49-3) E(w)=f(fv)—F(Ii). where E(x) istheerror function ortheremainder function. Itisthe difference between thevalue ofthefunction f(z) and theinterpolating function F(x). Then where x0,:01=:c0—|-h, :r2=:z:0—|-2h, ---, x,,,=x0+mh arethe values of:1:forwhich f(z)andF(:0)agree, andXisanumber between the Lesson 49—-Exerci se 671 smallest andlargest values ofx0,1:1,---,mmand1:.Unfortunately the formula does nottellushow tochoose X.However, ifwecandetermine theminimum andmaximum values ofIf("‘+1)(:c)| inthegiven interval I, then If<"‘+1)(X)| must bebetween them. Therefore by(49.31), (49.32) (”Ix°)(x(7.f§)1)§'(Z—x'")l1min|r<"+1>(e)|1 s|E(w)| é (73"_ '(37_'xm) '[max If-(m+1)(x)H, min I. Formula (49.32) isalso valid if1:1=1:0—h,1:0=1:0—2h,---, xm=1:0—mh. Hence (40.32) isvalid forboth Newton’s forward and backward interpolation formulas. Example 49.33. Find themaximum error incurred inusing Newton’s forward interpolation formula inExample 49.19. Solution. InExample 49.19,h =0.1,1:0=0.1,:1: =0.14, f(1:) =e‘, m=4.Hence1:1 =1:0+h= 0.2,:c0=x0+2h =0.3,:c0 =:v0+3h= 0.3, 1:4==1:0+4h=0.4.Since m=4,wemust, in(49.31), stop withx —1:4. Hence f""+1)e“ =f(5)e" =e’.Intheinterval (0.1,0.5), e’hasthelargest value when 1:=0.5.Therefore by(49.32), (0.14-0.1)(o.14 -0.2)(o.14 -0.3) ><(0.14-o.4)(o.14 -0.5)05 5' °(9)|E(=v)| é <0.0000005. EXERCISE 49 l.UseNewton’s (forward) interpolation formulas (49.15) and(49.17) andthe following table ofvalues, e°-3=1.34986, e°-4=1.49182, e°-5=1.64872, e°~6=1.82212, e°-7=2.01375, toﬁnd anapproximate value ofe°'35. Compare with actual value andvalue obtained byalinear interpolation. 2.Use(49.15) and(49.17) andthefollowing table ofvalues, log1.1=0.09531, log1.2=0.18232, log1.3=0.26236, log1.4=0.33647, log1.5=0.40547, log1.6=0.47000, toapproximate log1.12. Compare with actual value and value obtained bya.linear interpolation. Hint. Theneeded table ofdiffer- ences isgiven inanswer toExercise 48,1. 3.Thefollowing temperatures were recorded athourly intervals: 611.14., 2°; 74.14., 12°;811.11., 17°;911.11., 20°;104.11., 22°. What wastheapproximate temperature at6:3011.11.? 4.Prove formula (49.22). Hint. Follow theprocedure used toobtain (49.15). 5.UseNewton’s (backward) interpolation formulas (49.22) and(49.24) and thetable ofvalues given in1and2toapproximate respectively (a)e°'°5, (b)log1.53. Compare with actual values andwith those obtained bya linear interpolation. 672 Numznrcxr. Mrrruons Chapter 10 ANSWERS 49 1.e°-35 =1.41906. Actual value: 1.41907. Linear interpolation: 1.42084. 2.log1.12 =0.11333. Actual value: 0.11333. Linear interpolation 0.11271. 3.Approximately 7.9°. 5.(a)e°~65 =1.91554. Actual value 1.91554. Linear interpolation 1.91794. (b)log1.53 =0.42527. Actual value 0.42527. Linear interpolation:0.42483. LESSON 50. Approximation Formulas Including Simpson’s Rule and Weddle’s Rule. Wearenow ready tocontinue ourstudy oftheproblem ofﬁnding a numerical solution of (50-1) y’=f(¢.y) satisfying theinitial condition (50-11) 1/($0) =1/0- Lety(1:) beasolution of(50.1) fulﬁlling (50.11). Then, by(50.1), (50-12) y’=flw.y(=v)l, where fisnow afunction of1:alone. Integration of(50.12) andinsertion oftheinitial condition gives I Z I (50-13) I dy=I f[r.y(w)] dw. y=ye+ff[w.y(w)] dw-y=1/O x=:o xo If1:=1:0—|—nh,(50.13) becomes, with y0replaced byitsequal y(1:0) of (50.11),zo-|-nh (50-14) y(wo+nh)=y(wo)+I fl=v.y(w)] dw-$0 Since y(1:) istheunknown solution of(50.1), thefunction f[1:,y(1:)] is alsounknown. Hence wecannot possibly hope toperform theintegration in(50.14). However, wecan useapolynomial interpolating function F(12)forf[x,y(1:)] justaswedidintheprevious lesson forf(1:). LetF(a:) besuch apolynomial interpolating function, agreeing with f[1:,y(x)] atthem+1points whose abscissas are1:0,1:0+h,1:0+2h, ---,1:0+mh. Therefore, by(49.17), with f(:c0) replaced byf[:z:0,y(1:0)] which, by(50.12), isequal toy’(1:0), (50.15) 17(5)=[1+"7‘8°A+(”_“°)g”,h",' ’”°*h)A’ +(e—a)c—4.3;no-1..-21»),.+___!3 +(1-a)c—$0—h)(e—wsﬂzmzh) ~--(e-$0—rm-11h)A..]y.(,,o,’ Lesson50 APPROXIMATION Fonmums. Smrson. Wnnnnn. 673 where forconvenience inwriting wehave placed y’(x0) outside thebrackets. Comment 50.16. Ifwestop with theA2term in(50.15), then m=2 andF(x)becomes apolynomial ofdegree lessthan orequal totwoagreeing with f[:v,y(x)] =y'(x) atthethree points whose abscissas are1:0,2:0+h, mo+2h,see1after (49.17). Ifwestop with theA3term, then m=3 andF(x)becomes apolynomial ofdegree lessthan orequal tothree that agrees with f[a:,y(:z:)] =y’(:::) atthefour points whose abscissas are I0, ''',170+3h, etc. By(50.15), anapproximation totheintegral in(50.14) is,therefore, 20-4-nh (50.2) f F(x)da: "°+""' :z:—x (:1;-—:z:)(a:—:c—h) [<16—wo)(w—10—hm—$0—21»)--+ (Z_$0 —im _ Am dz. m!h"' Tosimplify integrating theright sideof(50.2), wemake thesubstitution u=at—2:0.Therefore du=dx,u=0when at=xoandu=nhwhen x=:00+nh.Hence wecanreplace (50.2) by zo+nh (50.21) / F(a:) dx “*0 ‘ nh ("‘h) (-71) -270 =fo l1+%A+u;!h2 A2+uu swig A3+"' +“‘“'"><“"2%,;<“"P""11")Am]I/($0)it M u uz—hu2us—3hu2 -1-2h2u 3 =f(, (1+ZA+ 21112A+ 3!h3 A u‘-6hu3+11h2u2 -6h3u + 411.4 A4 us—l0h'u4 +35h2u3 —50h3u2 +24h4u 5 + 5!h5 A [us——15hu5 +85h2uA ——225h3u3l +274h'Au2 -120h5u 6 + ems A +--y'(1¢o) du- 674- NUMERICAL Mmnons. Chapter 10 Integration of(50.21) gives so-1-nh (50.22) / F(:c) dz= 1 2h2 1 3h3 2h3 l""+n%A+m(% "'1? A244+&§(_"_47_l__n3h4+n2h-1) A3 as 45 as1 25 4 +2W(5 2+3 3"h)A 1n°h° 5635n‘h° 50n3h6 265 +m(T*2""+—4—-T+12"” A 1Mn’ 5n°h’ 5,22sn‘h’ +mzE(T"T+"" "-—4-37 +%g-1‘- -601%’) A6+--1/(:00). Simpliﬁcation of(50.22) results in :04-nh (50.23) / F(a:) dz: 2° 2 a =nh[1+gA+%(%—g)A2+%(%—n2+n)A3 143“11’ +ﬁ(%_%+Tn_3" ~1n5 435113 5011.2 5 +i%(€*2" +T*T+12")A +%(T—i+l7n -—i+-—_--—60n A1n“5125 422511“ 274112 6 2 4 3 ....].'<..>.Ifn=1in(50.23), weobtain 30+)! (50.24) I F(a:)dx=h[1+.3A-15¢’+g,;A“-71§;;A‘*0 +T2'UA5—r6'"".%<rA° +--?/'($o)- Therefore, by(50.14) and (50.24), anapproximation toy(x° +h)in Lesson 50 Arraoxrmrron Fonmunxs. SIMPSON. WEDDLE. 675 terms offorward differences ofy’(x0) is (50-25) ?l(1’?o —|—h)=I/($0) -l"H, where Histheexpression ontheright of(50.24). Ifin(50.23), westopwith theA term, then byComment 50.16, thepoly- nomial interpolating function F(x) isofdegree lessthan orequal toone, agreeing with f[x,y(x)] atthetwopoints forwhich re=x0and2:0+h. With n=1andusing only terms including Aof(50.23), weobtain, with thehelp of(48.11), 20+}! (50.3) /; F(x) da:=h[1+§A]y’(:c0) =h{y’(wo) +iii/'(=vo +h)—1/’(w<>)]}~ =§[y’(w<>)+1/($0+h>1. Therefore, by(50.14) and(50.3), anapproximation ofy(:c0 —|—h)is (50.31) Z/($0+1»)=yet)+Q[y’(w<>)+1/($0+h)-]- Formula (50.31) isknown asthetrapezoidal rule forapproximating a numerical solution of(50.1) satisfying (50.11). Ifin(50.23), westop with theA2term, then byComment 50.16, F(:c) isthepolynomial interpolation function off[:c,y(:c)] ofdegree lessthan or equal totwo, agreeing with fatthethree points forwhich :1:=2:0,:00+h, :00—|—2h.With n=2,andusing only terms including A2of(50.23), we obtain, with thehelp ofComment 48.18, 30-}-2h (50.32) Fe)<1»=zhu+A+as—M11/($0) ° =2hw'<x0> -l-A1/($0) -l-%A21!'(=vo)l =2hly'<x.> +um+h>—1/($0) +&w'<x. +2h)—2u'(wo+h>+y'<w.>>1} =§W0)+4?/($0+h>+I4/(930+2h>1. Therefore by(50.14) and(50.32), anapproximation toy(x0 —|—2h)is <50-83> ye.+21»)=yew+§rm»)+4m.+h>+1/($0+2h>1. Formula (50.33) isknown asSimpson’s rule forapproximating anu- merical solution of(50.1) satisfying (50.11). 676 NUMERICAL M1-mrons Chapter 10 Ifin(50.23), westop with theA3term, then byComment 50.16, the polynomial interpolation function F(2:) isofdegree lessthan orequal to three, agreeing with f[2:,y(2:)] atthefour points forwhich 2:=2:0,2:0+h, 2:0+2h,2:0+3h.With n=4,*andusing only terms including A3of (50.23), weobtain, with thehelp ofComment 48.18, :0-I-4h (50.34) /#0 F(2:) d2: =4hl1+2A+id‘;-2)A’+M16—16+4)A“1y'(xo> =4h[y'<w0> +2Ay'<w.>> +%A’y'<x0> +§A3u'(wo)l =at/($0) +2n/e..+h>—I/($o)] +§l?/($0 +2h)"‘2?/($0 +h)-l"I/($o)l -l"§ly'($0 +3h)-'3?/($0 +2h) +31/($0 -l"h)A‘?/'(93o)l} =%‘[met+h>—1/($0+21»)+21/(Z.+am Therefore, by(50.14) and(50.34), anapproximation ofy(2:0 +4h)is (50-35) We+4h)=1/(wt) +%[2y'<x.>+h>—um+21»)+21/($0+am Prhceeding aswedidabove, wecanobtain approximating polynomials F(2:) ofstill higher degrees, agreeing with f[2:,y(2:)] atmore and more points. The only other twoapproximating formulas, however, that will interest usarethose inwhich F(2:) isofdegree ﬁveandsix;seeExercise 50,2 forthecase when F(x) isofdegree four. Foraﬁfth degree poly- nomial interpolating function, (50.23) willgive 30-}-5'! (50.36) /to F(2:)d2: =%[19y'(w<>) +151/(wt+h)+sow».+21»)+5<w'<w0+sh) +75?/($0 +4h)-1-19!/($0 +571)]- Therefore, by(50.14), anapproximation of3/(220 +5h)is (50-37) 1/($0+5h)=y(Ivo) +K, where Kistheexpression ontheright of(50.36). ‘The reason fortaking n=4instead ofn=3willbecome apparent when weuse theresulting Formula (50.35). Forthecase when n=3,seeExercise 50,1. Lesson 50 Arrnoxrmrrron FORMULAS. Smrson. Wapnm. 677 Forasixth degree interpolating function, (50.23) willgive Z0-|-Oh (50.38) f F(2:)d2: 10 3h =m[411/(we) +216?/($0 +h)+271/($0 +2h) +272y’(wo +3h)+27y’(w<> +4h) +216?/(10 —|—5h)+41y'(xo +5h)]- Hence, by(50.14), anapproximation ofy(2:0 —}—6h)is (50-39) I/(10+6h)=1/($0) +G, where Gistheexpression ontheright of(50.38). Formula (50.39) isusually replaced bythefollowing formula, (50-4) y(rvo+6h)=1/(we) +3%,[42y'(1>o) +210m. +h)+42y'<x.+21») -1-2521/($0 —|—3h)+42y'($o —|—4h)+210y'(1¢o +5h)'1"42!/(10 +6h)]- Note that itadds alittle tosome terms in(50.38) andsubtracts alittle from others, sothat itsoverall accuracy isvery close tothat given by (50.38). Itsadvantage over (50.38) liesinitsbeing reducible tothe simpler form (50-41) y(wo+6h)=2/($0) +%{y'<x0> +am.+h)-l-3/($0+21»)+em.+sh) +I/($0 -l"4h)-l"51/($0 +5h)-l"I/($0 +2h)]- This lastformula isknown asWeddle’s rule. Weobtained (50.23) byusing Newton’s forward interpolation formula (49.17). Asanexercise, start with Newton’s backward interpolation for- mula (49.24) andshow that /30+"). n 1n2 n2to n2 3 4 3 +%<%+n2+n)V3+%(%+%+%-+3n>VA 5 3 2 +%("§+2n‘+?”T”+5°T"+12:1)v‘+---]y'(xo>- For convenient reference wecollect theapproximation formulas de- veloped inthislesson. 23$ $28 A33 ﬁg ﬁgwggukwm :5+23$+GE+g<€+As+ORV‘;+QM”+cayg+ g+Si+S+3%+“ORE%+AC2HAg+cgBEm_%_§$$3+§_€+A5+°ﬁV_\Nm_N_+Ag+83%+ wQR+EEK+Q+°§_$_~_+€\§_%+Ac“:HAg+cg__a€o___Q&95%ﬁg$3+5%+Ag+8%I3+§_§_w+2%HA5+8%_d_Eo___€H02%E5 Ea+8%+Q+£5+A__5\m_W+ﬂogHGa+cg®___MH°MMQH@w N2*+oaim+Aoavxrw+Ac":HS+963ABE_Gwmo@@~ﬁH%w_ﬂmCgsgéémAss\____§__~__m_Essg8§_§§€§_§_§£Riggs QSEV A808 828 Agegay Lesson 50 APPROXIMATION Fonmmxs. SIMPSON. WEDDLE. 679 Wealso collect below forconvenience, theerror term associated with each oftheabove formulas.* Interpolating Polynomial Error ha (50.7) a+b2:(trapezoidal rule) —--Ey”'(X) h"-(50.71) a+bx+02:2(Simpson's rule) —%y(5'(X) , 28(50.72) Third degree polynomial g6h5g/(5) (X) . . 215(50.73) Fifthdegreepolynomial -iﬁﬁ h’y">(X) 717<1) 9719 (9)(50.74) Weddle’s rule -—-1?)y(X1)—E y(X2) Inalltheabove error formulas, Xisavalue of2:intheinterval (:00,2:0+nh), corresponding tothe:00—|—nhintherespective formulas (50.6) to(50.64). Comment onApproximation Formulas (50.6) to(50.64). We assume inthediscussion which follows that y(z) isasolution of (50-8) y’=.f(w,y) satisfying theinitial condition (50-81) Z/($0) =yo- Hence when 2:=2:0,y’by(50.8), hasthevalue (=1) y’(wo) =f[rvo,1/(rvo)l- Since 3/(2:0) istheinitial condition given in(50.81), wecancalculate 3/(2:0) of(a). Again by(50.8), when 2:=2:0+h, (b) 1/($0 +h)=flfvo+h,y(1o+h)]- Todetermine 3/(:00 +h)of(b),wemust therefore know thevalue of y(x0 —|—h).But wecannot know 3/(210 +h)unless weknow y(z). And since y(z) isthevery solution weseek, wecannot hope toevaluate y'(2:0 —|—h)byuseof(b). Ittherefore follows that wecannot, forexample, useformula (50.6) unless wehave other means available bywhich wecan estimate avalue ofthey'(:c0 +h)term whiclrappears init. ‘William Edmund Milne, Numerical Calculus, Princeton University Press, Princeton, N.J., 1949; Numerical Solutions ofDiﬁeremial Equations, John Wiley &Sons, New York, 1953. 680 Nuurznrcxr. Mmnons Chapter 10 Alltheremaining formulas (50.61) to(50.64) have asimilar drawback. Formula (50.61), forexample, canbeused only after wehave been able toobtain, inaddition to3/(:00), estimated values ofy(2c0 +h)and y(z0 +2h). With them, wecanthen, by(50.8), estimate values ofthe terms I/($0 +h)=flxo +h!1/($0 +h)l and 3/($0 +2h)=fl$0 ‘l’2h,1/(10 -l"271)] that appear intheformula. Thelastformula (50.64) requires sixpre- liminary estimates inaddition tog/(x0), before itcanbeused, namely estimates ofy(2:0 +h),---,y(2c0 +6h). By(50.8), wecanthen estimate values ofy’(x0 +h),---,3/(:00 +6h). With theexception of(50.62), there isanother unusual feature about allthese formulas. Consider forexample formula (50.61). After the needed twopreliminary values ofy’(2;0 +h)andy’(:c0 —|—2h)have been estimated byﬁnding approximate values ofy(2:0 —|—h)andy(2:0 +2h)by other means, all(50.61) will doisgive avalue ofy(2:0 +2h)allover again. Why then istheformula necessary atall? Ifwehave approximated y(2:0 +2h)bysome other method, why calculate y(2:0 +2h)once more? Itturns outthatformula (50.61) isactually acorrector formula, i.e.,re- peated application oftheformula willimprove anestimated approxima- tionofy(2:0+2h)computed byalessaccurate formula than itself (as determined bytheir respective error terms). Weshall clarify thispoint at thetime weusetheformula. Itistherefore called appropriately acorrector formula. Ifyouwillcarefully examine alltheother formulas inthislist, youwillﬁndthat with theexception of(50.62), allarecorrector formulas. Allthey willdoiscorrect thelastestimated approximation ofy(x0 +nh) [needed toapproximate y'(2:0 +nh)intheformula] obtained byother less accurate methods. Formula (50.62) ontheother hand isacontinuing formula. With it,wecanevaluate y(2:0 +4h)provided wehave obtained estimated values Of1l(¢'10) andI/(Io+h),y'(Io+2h).1/(re+3h)- Itisevident, therefore, thatwecannot start touseanyoftheformulas (50.6) to(50.64), whether tocorrect ortocontinue, unless wehave acer- tain number ofpreliminary estimates. Itisforthisreason that thefor- mulas developed inprevious lessons, since they donotrequire preliminary estimates fortheir use,have been called appropriately starting formulas. There are,therefore, asmentioned intheintroduction tothischapter, three types offormulas innumerical methods. 1.Starting formulas. 2.Continuing formulas. 3.Corrector formulas. Wehave already developed various starting formulas. Inthenext lesson, weshall describe asimple continuing andcorrector combination formula. Lesson 50—Exereise 681 EXERCISE 50 1.In(50.23), take n=3andstop with theA3term. Thepolynomial in- terpolating function F(2:) istherefore ofdegree lessthan orequal tothree, agreeing withf[2:,y(2:)] of(50.1) atfourpoints. Prove that zo-f-3h / F01)11¢=%[y'(1>o) +31/(Io +h)+31/($0 +2h)+1/'(¢o+371)]-=0 This formula isknown asthethree-eighths rule. 2.In(50.23), take n=4andstop with theA4term. Thepolynomial inter- polating function F(zz)istherefore ofdegree lessthan orequal tofour, agree- ingwithf[2:,y(2:)] of(50.1) atﬁvepoints. Prove that 20-{-4h <w.s2> /Fe)e=%‘[71/(1o)+ 32?/(Io+h)+12?/(Io +21»)’° +szree+sh)+71/(10+4101- 3.Prove (50.36) bytaking n=5andstopping with theA5term in(50.23). 4-.Prove (50.38) bytaking n=6andstopping with theA6term in(50.23). 5.Prove (50.5). Hint. Start with Newton’s backward interpolation formula (49.24) andfollow theprocedure used inthetexttoarrive at(50.23). 6.Prove that, ifn=1,(50.5) reduces to 20+» (50.83) f F(2:)d2: ”°=hu+iv+av’+%v“+#2-iv‘+rev“+--'l1/'(10)-Hence anapproximation toy(2:0+nh)of(50.14), with n=1,interms of backward diﬁerences, is (50-84) 1/(Io+h)=1/(Io) +It where kistheexpression ontheright (50.83). Themethod which makes use offormula (50.84) isknown asAdams’ method ofapproximating anu- merical solution ofy’=f(z,y) satisfying y(2:0) =yo.Tocalculate V"'y'(2:0), you must know m—|—1evenly spaced values ofy'(2:), namely y’(2:0), y'(2:0 ~—h),----,y'(2:0 —mh), seeLesson 49B. These values canbe obtained from thegiven differential equation y’=f(z,y) only after corre- sponding values ofy(2:0), ---,y(2:0 -mh)have been found bystarting formulas. Hence (50.84) isacontinuing formula. Bymeans ofAdams’ method, ﬁndanapproximate value, when 2:=0.5, oftheparticular solution ofthedifferential equation y’=2:2+yforwhich y(0) =1.Take h=0.1andstop with theV4term in(50.84). Hint. By (50.84), with 2:0=0.4andh=0.1, (50.85) y(0.5) =y(0.4) +0.l[1/(0.4) +§Vy'(0.4) +-155-V21/(0.4) +g-V31/(0.4) +5-Z-#V‘Ay'(0.4)]. Tocalculate V4y’(0.4), youmust know y'(0.4), 1/(0.3), y'(0.2), y'(0.1), y'(0). From thegiven equation, when 2:=0,y=1,g/(0) =0+1=1.Use Table 46.31, creeping upprocess column, toﬁndtheother needed values ofy'.* Then construct atable ofdifferences ofy’(0.4). Head theﬁrst column y’=2:2+yandenter thevalues ofy'(0), y’(0.1), y’(0.2), 3/(0.3), ‘The actual values will, ofcourse, giveamore accurate result fory(0.5). However, weassume these actual values areunknown. 682 Nnurznrcxr. Mmnons Chapter 10 y’(0.4). Head thesecond column Vy’andenter initthedifferences ofthe ﬁrstcolumn—rememberVy’ (0.1) =y'(0.1) —y'(0.0),Vy'(0.2) =1/(0.2) — y’(0.1). Head thethird column V2;/’andenter initdifferences ofthesecond column. Head thefourth column V3;/’andenter initdifferences ofthethird column. Head thelastcolumn V4y’andenter initdifferences ofthefourth column. Thelastrowshould then contain allthedifferences needed inthe above formula (50.85). 7.In(50.1), letf(z,y) beafunction only of2:sothat y’=f(z) forwhich 1/(2:0) =yo.Hence, by(50.14), 20-1-nh (50-9) y(ro+ nh)=1/(10)+[2 f(z)dr- LetF(2:) betheinterpolating function forf(z), agreeing with f(z)atthe n—|—1points whose abscissas are2:0,2:0-1-h,2:0+2h,---,2:0+nh, / \\\\ \ // re.+e-1>h1= \\\\\ f(xo+h)= /// F[x0+(n “Dh] \\ F(x0+h) ,/ f(x +2h)= f(xo+nh)= r':i§°i= \‘’/ Fe§+2h) F("°*"") 0 x0 x0+h x0+2h x0+(n-1)h x0+nh Figure 50.91 Fig.50.91. Letthegraph ofF(2:) bethestraight linejoining theendpoints ofthese ordinates. Then anapproximation ofy(x0+nh)of(50.9) is :0-Q-nh (50-92) y(1o+nh)=1/(wo) +‘/I F(w)11¢, where F(z) isthefunction whose graph consists ofthese straight lines. By(50.6) and(50.7), with y’replaced byitsequal f(z), wehave (50-93) 1/(ro+h)=y(wo) +2lf(Io) +f(wo+h)]+E, where E=—h3f”(X )/12,andXisavalue of2:intheinterval (2:0,2:0 +h). The term g[f(2:0) +f(a:0+h)]may alsobelooked atasthearea ofa trapezoid ofwidth handheights f(2:0), f(a:0+h),Fig. 50.91; hence the name trapezoidal rule. (a)Byadding theareas ofeach trapezoid inFig.50.91, show that =,,+,.n (50-94) ft F(1)11¢=g{f(¢o) +2f(1o+h)+2f(ro +2h)+--- ° +2.7110+e-1>h1+fee+nh)}-Hence anapproximation to(50.9) is (50-95) y(wo+nh)=y(wo)+H, where Histheright sideof(50.94). Lesson 50—Exereise 683 (b)Following theprocedure outlined inExercise 44,11, show that the upper bound oftheerror informula (50.95) duetousing F(z) toap- proximate f(2:)in(50.9), is 3 (50.96) |E|s where Misthemaximum value ofIf’’(2:)|intheinterval (2:0,2:0+nh). 8.(a)Useformula (50.95), with n=4,h=4,toobtain anapproximate value, when 2:=2,oftheparticular solution oftheequation y’=1/2: forwhich y(1) =0.Show howthisvalue ofy(2) canbeused toap- proximate log2.Hint. Here :0=1,y(2:0) =0,nh=1.Thesolution ofy’=1/2:forwhich 1/(1) =0isy=log2:.Therefore y(2) =log2. (b)Use(50.96) toﬁndanupper bound totheerror intheresult obtained in(a). Compare with actual error. Hint. n=4,h=1,f(2:) =1/2:, M=max. |f”(2:)| =max. I2/2:3] =2ininterval (1,2). (c)What isthelargest value ofhwhich canbeused toinsure thattheerror Einthecomputation oflog2islessthan 0.0005? Intohowmany parts would itbenecessary todivide theinterval (1,2)? 9.In(50.1), letf(z,y) beafunction only of2:,sothat y’=f(z) forwhich y(2:0) =yo.Hence, by(50.14), 2:0-i-nh (50-97) y(==o+nh)=£I1(1o)+/ f(z)dw- LetF(2:) betheinterpolating function forf(z), agreeing withf(z)atthe n+1points whose abscissas are2:,2:0—|—h,2:0—|—2h,---,2:0+nh,where niseven. Letthegraph ofF(2:) consist oftheparabolic arcconnecting the three ordinates f(2:0), j(2:0+h),f(2:0+2h),solid curve inFig.50.91, plus theparabolic arcconnecting thethree ordinates f(2:0+2h),f(2:0+3h), f(2:0+4h),etc.Then anapproximation toy(a:0+nh)of(50.97) is zo-1-nh (50-98) y(1o+nh)=I/(10)+/0 1"(1)dr,10 where F(2:) isthefunction whose graph consists ofthese parabolic arcs. By(50.61) and(50.71), with y’replaced byitsequal f(z), wehave (50-99) I/(wo+2h)=y(wo) +g[f(1o) +4f(=vo +h)+f(I¢o+2h)]+E, where E=—h5f (4)(X)/90,andXisavalue of2:intheinterval (2:0,2:0 +2h). The term g[f(2:0) +4f(2:0 +h)—|—f(a:0—|—2h)], ithasbeen proved, seea calculus text, isthearea under aparabolic arcjoining theends ofthree ordinates f(1o), f(1o+h),f(1o+2h)-(a)Byadding theareas under each parabolic arcinFig.50.91, show that zo-I-nh (50-991) /Fe)e-gfree)+4f(1o+h)+2f(I0+2h)=0 +4f(10+sh)+2/ee+41-)+--- +wee+e—2)h1 +4f[lo+e—1)h1+1ee +nh)}. 684- NUMERICAL METHODS Chapter 10 Hence anapproximation to(50.97) is (50-992) 1/($0+nh)=y(zo) +It where kistheright sideof(50.991). Formula (50.992) isknown as Simpson’s rule. (b)Following theprocedure outlined inExercise 44,11 [note that here the interval is(2:0,2:0+2h)instead of(:0,2:0—|—h)],show that theupper bound oftheerror informula (50.992) duetousing F(2:) toapproximate f(z)in(50.97), is he (50.993) §E6nM, where Misthemaximum value of|f<47(a:)| intheinterval (2:o,2:0 +nh). 10.Useformulas (50.992) and(50.993), withn=4,h=1toanswer questions (a),(b),(c)ofproblem 8.Hint. Here M=max. |f(4)(2:)| =max. I24/2:5| = 24ininterval (1,2). ANSWERS 50 6.y(0.5) =1.6961610. 8.(a)y(2) =0.69702. Actual value: log2 =0.69315. (b) <0.01042. Actual error =0.00387. (c)h=0.05totwodecimal places. Therefore weneed todivide theinterval (1,2) intotwenty parts. 10.(a)y(0.2) =0.69325. (b) <0.00052. Actual error =0.00010. (c)h<0.24. Since nmust beeven, weneed todivide theinterval (1,2) intosixparts. Compare with 8(0). LESSON 51. Milne’s Method ofFinding anApproximate Numerical Solution ofy’=f(x,y). The method weareabout todescribe forﬁnding anumerical solution ofthedifferential equation y’=f(2:,y) satisfying theinitial condition y(2:0) =y0isprobably theonemost widely used today. Itissimple in form andhasarelatively high degree ofaccuracy. Themethod uses con- tinuing formula (50.62) toestimate orpredict avalue ofy(:e0 —|—4h)and then employs Simpson’s formula (50.61) tocorrect it.Forconvenience werewrite these formulas below, using asubscript pfortheformula we shall useasapredictor orestimator andasubscript cfortheoneweshall useasacorrector. <51-1) nee+4h)=ye.) +5;‘We.+1-)-re.+21-)+2y'e.+sr->1, (51.11) y,(x0 +4h) =1/($0 +2h) +§‘[y'eg+ 2h)+4-re.+sh)+y./ee+4h)]- When these twoformulas areused incombination, they areknown col- lectively asMilne’s method. Inorder touse (51.1) wemust know Lesson 51 Mu.NE’s METHOD Arrtmn 'roy’=f(z,y) 685 y’(2:0 +h),y’(2:0 +2h), y’(2c0 +3h). Wecan determine these values from thegiven differential equation y’=f[x,y(2:)] only after weknow the corresponding values ofy(x0 +h),y(x0 +2h), y(:c0 +3h). Hence we must usestarting formulas inorder toobtain these needed preliminary estimates. Formula (51.1) will then predict orestimate avalue of 3/(x0 +4h); formula (51.11) willcorrect thisestimate. The new value of y(2:0 +4h)thus found caninitsturn beused asanew estimated value and (51.11) used over again tocorrect it.Ithasbeen proved that, ifthe original estimate isnottoofaraway from thetruevalue andifhissuffi- ciently small, therepeated useof(51.11) willgive asequence ofvalues of y(2:0 +4h)which willeventually converge. Example 51.12. Use Milne’s method toﬁnd anapproximate value when so=0.5oftheparticular solution ofthedifferential equation (5) y’==1’+y, forwhich y(0) =1. Solution. InExample 46.21, wefound byTaylor series methods, (b) y(0.1)=1.1055125, y’(0.1) =1.1155125; 3/(0.2)=1.2242077, 1/(0.2) =1.2642077; y(0.3)=1.3595755, 1/(0.3) =1.4495755. By(51.1) and(51.11) with x0=0,h=0.1, e)5.6-4)=5(0)+72570.1) —7/(0.2)+2540-3)]. 5.6.4)=yo-2)+[1/(0-2)+4y'<0.:>-)+5./(0.4)). Substituting thevalues of(b)andtheinitial condition y(0) =1,inthe ﬁrst equation of(c),weobtain (d) y,,(0.4) =1+ [2(1.1155125) —1.2642077 +2(1.4495755)] =1.5154624. By(2)and(<1), y,/(0.4) =0.16+1.5154624 =1.6754624. Hence thesecond equation of(c)becomes (e) y,(0.4) =1.2242077 + [1.2642077 +4(1.4495755) +1.6754624] =1.2242077 +0.2912657 =1.5154734. 686 Numanrcn. M1-zrnons Chapter 10 Wenowmake useof(51.11) once more toseewhether itwillcorrect the value in(e). Using thevalue in(e)asanew estimate, wehave (f) y,/(0.4) =0.42+1.5154734 =1.6754734. Ourcorrector formula (c)becomes, with thehelpof(b)and(f), (g) y,(0.4) =1.2242077 + [L2642077 +4(1.4495755) +1.6754734] =1.5154738. You canverify that athird application of(51.11), using thevalue in(g) asanew estimate, willnotchange thisvalue. Since ourcorrector will make nomore corrections, weaccept thisvalue of1/(0.4). From here onwecontinue tomake useofourpredictor andcorrector formulas, using theﬁrsttoestimate anapproximation, thesecond tocor- rect thisestimate; thecorrector formula being used repeatedly until no further correction results. By(51.1), with h=0.1andmo=0.1, <11) mo-5)=y<0.1>+ [2y'(0-2) —1/'<0.s>+2y'<0-4)]. By(a)and(s) (i) 1/(0.4) =0.16 +1.5154738 =1.6754738. Therefore by(b)and(i),(h)becomes (j) y,(0.5) =1.1055125 + [2(1.2642077) -—(1.4495755) +2(1.6754738)] =1.6961508, 1"which isanestimated value ofy(0.-5). Using thisvalue in(a),weobtain (k) y,/(0.5) =0.25 +1.6961508 =1.9461508. By(51.11), with h=0.1,xo=0.1, <1) y.<o.s>=y<o.s>+Qmos)+41/(0.4)+y./<o.s>1. By(b),(i),and(k),(1)becomes (m) y,(0.5) =1.3595755 + [1.4495755 +4(1.6754738) +1.9461508] =1.6961629. Using thevalue 3/(0.5) =1.6961629 asanew estimate, weﬁndby(a), (n) y,/(0.5) =0.25 +1.6961629 =1.9461629, Lesson 51 M1Lma’s Marnon APPLIED T0y’=f(z,y) 687 andby(51.11) (6) y,(0.5) =1.3596755 +%[1.449675s +4(1.67547ss) +1.9461629] =1.6961633. You canverify that further use.of(51.11) willnotchange thevalue in (0). Wetherefore accept thisestimated value ofy(0.5). Comment 51.2. Inthismanner, alternately using predictor formula (51.1) andcorrector formula (51.11) asmany times asneeded, youcan obtain y(0.6), y(0.7), y(0.8), etc. Itwillbefound desirable toconstruct a table inwhich torecord allrelevant calculations asthey arefound. Its form isgiven inTable 51.22 below. The letter Dwhich appears initis deﬁned as (51.21) D(:c0 —|—4h)=y¢(a:0 +4h)—-y,,(x0 —|—4h), where y,,istheestimate computed by(51.1) andy,istheﬁrst correction ofthisy,,computed by(51.11). Itsimportant purpose willbeexplained later. Wehave leftittoyouasanexercise tocomplete thelinebeginning with 0.6. Table 51.22 I y,of 1;,of y, y, Dof Actual (51.1) (51.11) P c (51.21) Value 0.4 1.5154624 1.5154734 1.6754624 1.6754734 0.0000110 1.5154738 1.6754738 1.5154741 0.5 1.6961508 1.6961629 1.9461508 1.9461629 0.0000121 1.6961633 1.9461633 1.6961638 0.6 1.9063564 Comment 51.23. Formula (51.1) isacontinuing formula. However, itisrarely, ifever, usedforthispurpose since, asweshall show below, itis considerably lessaccurate than corrector formula (51.11). By(50.72), the error term E,of(51.1) is (61.24) E,=;=3h“y‘5’(X,). By(50.71) theerror term E,of(51.11) is (61.25) E,=——§15h5y‘5’(X2). Ifwemay assume that hissufficiently small sothat thevariation invalue ofy(5)(X1) and1/(5)(X2)inaninterval ofwidth 4h[X1isavalue of:1:in 688 NUMERICAL METHODS Chapter 10 theinterval (mo,:50+4h); X2isavalue of:1:intheinterval (xo+2h, xo+4h)]isnegligible, then weseefrom error formulas (51.24) and(51.25) that thecorrector formula ateach single stepisapproximately 28times asaccurate asthepredictor formula. Comment 51.3. Comment onError inMilne’s Method. 1.Call E,,(:v0 +4h)theerror incomputing y,,(a:0 —|—4h)byusing for- mula (51.1); call E,(x0 +4h)theerror incomputing y¢(a:0 —|—4h)by using formula (51.11) tocorrect this predicted value ofy,_,(:z:o +4h). LetY(xo +4h)betheactual value ofy(xo -1-4h). Then (51.31) Ec(x0 +4h)=Y(:c0 +4h)——y,,(2:o +4h), E,,(xo -1-4h)=Y(:c0 +4h)—y,,,(a:o -1-4h). Subtracting theﬁrst equation from thesecond, wehave (51.32) yc(:z:0 -1-4h)—-y,_,(xo +4h)=E,,(:co —|—4h)—E,(:c0 —|—4h). Ifwemay assume that hissufficiently small sothat thevariation between y(5)(X1) oferror formula (51.24) andy‘5)(X2) oferror formula (51.25) is negligible, then wecommit asmall error byusing 1/(5)(X)astheir approxi- mate common value. Hence subtracting (51.25) from (51.24) and re- placing y(5)(X1)andy‘5’(X2) byy‘°)(X), weobtain (51.33) E,,(:2:0 —|—4h)——E',,(:c0 +4h)=§-§h5y‘5)(X). By(51.21), theleftside of(51.32) isD(:z:0 -1-4h). Hence by(51.32), (51.33), and(51.25), wehave (51.34) D(x0+4h)=H-h5y5(X) =—-29E,(:c0 +4h). Therefore by(51.34), (51.35) E44,,+4h)=- - Formula (51.35) tells usthat theapproximate formula error intheﬁrst corrected value ofy,(x0 +4h)is-11; thedifference between theﬁrst corrector andpredictor values ofy(:c0 +4h). 2.The column headed DinTable 51.22 thus serves avery useful pur- pose. Dividing itby29willgive anestimate oftheformula error inthe ﬁrst corrected value ofy,over onehstep. Iftherefore wewant < 0.000005, then ID/29| must be<0.000005 orID]must be<0.000145. Aslong asID]remains lessthan thisﬁgure, weassume that theerror in ourﬁrst corrected estimate y¢,over each hstep, islessthan 0.000005. 3.AslongasD/29 islessthan thedesired accuracy, wemay continue tousethesame hinterval. AssoonasD/29 becomes greater than desired accuracy, wemust reduce h.Conversely, ifD/29 isconsiderably smaller Lesson 5l—Exe1-cise 669 than desired accuracy, wecansafely increase h.The method ofreducing andincreasing hinthecourse ofanextended computation isdiscussed in thenext Lesson 52. 4.The corrector formula may beused repeatedly ateach step, until there isnodifference between twosuccessive values of1/(:50 +4h). How- ever, asthedifference between predictor andﬁrstcorrector values in- creases, you will ﬁnd that you must usethecorrector more and more times ateach step. When thishappens, even though D/29 may stillbe lessthan desired accuracy, itwillusually befound besttoreduce h. 5.The column headed Dserves another useful purpose. Itcontrols in some measure arithmetical accuracy ateach step. Whenever anentry in Dshows asudden change from adeﬁnite behavior pattern, thepreceding andcurrent calculations should bechecked. 6.Ifpredictor andcorrector values agree tokdecimal places after being properly rounded off,then Icdecimal accuracy isassumed. EXERCISE 51 1.Complete line0.6inTable 51.22. 2.Replace (b)ofExample 51.12, bytheactual values ofy(0.1), y(0.2), y(0.3) as given inTable 46.31. Following themethod ofthislesson, compute y,,(0.4), y,(0.4), andD(0.4). Then use(51.35) tocorrect y,(0.4). Addthiscorrected value ofy(0.4) toyour table. Now proceed tocompute y,,(0.5), y,(0.5), and D(0.5). Use(51.35) tocorrect 1/,(0.5). Addthiscorrected value ofy(0.5) to your table. Calculate y,,(0.6), y,(0.6), andD(0.6). Correct y,(0.6). Compare these values with those obtained previously andwith actual values given in Table 51.22. 3.Using thesixpreliminary results obtained in2,compute y.,(0.6) bymeans of Weddle’s rule(50.64). Intheabsence ofasolution, howmany decimal place accuracy could youassume iny(0.6)? Hint. See3after “3.Cumulative Errors, ”Lesson 52A. 4.Find numerical approximations when :1:=0.4,0.5,0.6oftheparticular solu- tionofthedifferential equation y’=a:-1-yforwhich y(0) =1.Take h= 0.1. Follow themethod employed inExample 51.12, using ﬁrstthepredictor formula, then thecorrector formula asoften asisnecessary until nofurther correction results. Forpreliminary values, takey(0.1) =1.1103418, y(0.2) = 1.2428055, 1/(0.3) =1.3997176. Solve theequation andcompare your results with actual values. 5.Using thesixpreliminary results obtained in4,compute y.,(0.6) bymeans of Weddle’s rule(50.64). Intheabsence ofasolution, howmany decimal place accuracy could youassume iny(0.6)? SeeHint in3. 6.Find numerical approximations when :0=1.4,1.5oftheparticular solution ofthedifferential equation y’=myforwhich y(1) =1.Take h=0.1. Follow themethod used inExample 51.12, employing ﬁrstthepredictor and then thecorrector formula asoften asnecessary. Forpreliminary values take1/(1.1) =1.11071, y(1.2) =1.24608, y(1.3) =1.41199. Solve theequa- tionandcompare results with actual values. 7.Using theresults obtained inproblem 6,compute y(1.5)bymeans ofcorrector formula (50.63). Here 2:0=1,y(:c0) =1,h=0.1.Intheabsence ofasolu- tion, howmany decimal place accuracy could youassume? SeeHint in3. 690 NUMERICAL Mnrnons Chapter 10 1. 2.1'11»(0-6) 1/p(0-4) y(0.4) =1.5154733, y,,(0.5)1>(0.5) =0.0000126, y(0.5)ANSWERS 51 1.9063415, y,(0.6) 1.5154627, y.(0.4)1.9063561, D(0.6)1.5154742, 1>(0.4)1.6961512, y.(0.5)1.6961634, y,(0.6) y,(0.6) =1.9063561, D(0.6) =0.0000137, y(0.6)0.0000146.0.0000115.1.6961633,1.9063424,1.9063556. 3.y.,,(0.6) =1.9063563. Can assume sixdecimal place accuracy ifrounded offtosixdecimals. y.(0.4) =1.5336497, y.(0.5) =1.7974429, y.(0.6) =2.0442334. Actualvalues: 1.5836494, 1.7974425, 2.0442376. y.,,(0.6) 2.0442377. Canassume sixdecimal place accuracy. y.(1.4) 1.61609, y.(1.5) =1.36326. Actual values: 1.61607, 1.36325.y(1.5) =1.86825. Canassume fourdecimal place accuracy ifrounded offto fourdecimals.4-. 5. 6. 7. LESSON 52. General Comments. Selecting h.Reducing h. Summary and anExample. LESSON 52A. Comment onErrors. 1.Formula Errors. With each approximating formula wehave also given acompanion error term which measures themagnitude oftheerror. Since these error formulas cannot usually beused, wehave alsosuggested practical means bywhich themagnitude oferror canbeestimated. Unless anumerical method hasassociated with itauseful error formula, the method isoflittle value. 2.Rounding oﬂ"Errors. Inapractical problem, thenumber ofsteps needed willusually beknown, also theaccuracy desired. Itwillthus be possible toestimate thenumber ofdecimal places that should beused at thestart inorder tooffset rounding offerrors under themost unfavorable circumstances. Forexample, suppose you want four decimal accuracy andexpect touseeight steps. Ifyouround offtosixdecimals, then the maximum possible absolute value oftheerror dueonly torounding off, i.e.,duetodropping theseventh andlater decimals, is,after eight steps, 0.0000005 ><8=0.000004. Itmust beremembered, however, that the useofaformula mayconsiderably magnify, ateach step, theeffect ofthe rounding offerror. 3.Cumulative Errors. Ateach step ofabuilding upprocess, twoerrors occur. One, because westart offwith aninherited error, andtwo, because weareusing anapproximation formula. Since each succeeding step de- pends ontheprevious estimate, itwillbeunusual indeed ifweobtain increasing accuracy asweproceed. Itmay happen inarare case that one error may beoffset byasucceeding one. Intheusual case, itwillnot happen, andaccuracy willdecrease ateach step. Although formulas exist which willgive theupper limit oferror atthe endofeach step dueboth toformula andcumulative errors, they arenot Lesson 52B Cnoosmo runS121: orh691 easy touse. Wegive below three practical suggestions forestimating accuracy, ones which have been found adequate inmost cases. 1.Start with many more decimals than youneed. 2.Make allcalculations over again with anhequal toonehalf itspre- vious value. Ifthenew ﬁnal result agrees with theprevious onetoIc decimal places, after being properly rounded off,then lcdecimal place accuracy isassumed. 3.Apply corrector formula (50.63) after ﬁvepreliminary estimates have been obtained orWeddle’s corrector (50.64) after sixsteps. Ifthere- sultobtained bythese corrector formulas agrees with thelastestimate used intheformula tokdecimal places, after being properly rounded off,then lcdecimal place accuracy isassumed. Weddle’s rule, inpar- ticular, issimple instructure and isextremely good atdiscovering errors. Ifthere islittle agreement between thelastestimated value and itscorrected value, then either anarithmetical error hasbeen made orhistoolarge. Aﬁnal word ofcaution. Inmost problems, thepractical approach to errors asoutlined above, willwithin reasonable certainty, assure youofa result which iscorrect tokdecimal places. However, only aformula which gives theupper bound oftheerror duetorounding off,formula and cumulative errors, cangive, with certainty, themagnitude oftheerror in anumerical computation. LESSON 52B. Choosing theSize ofh.You may have been saying toyourself, “How does oneknow what size htoselect atthestart?” Weused h=0.1inourexamples, butwhat made uspick 0.1instead of 0.2or0.05 or0.3? Ifitispossible touseanerror formula foragiven method, then itisalsopossible todetermine hsothat theerror duetothe approximating formula willremain within thedesired limits. Frequently aknowledge oftheproblem plus practical experience will determine a starting value ofh. Intheabsence ofauseful error formula orpractical information, then allyoucandoistostart with anhwhich seems reasonable. Sayyoudecide tostart with h==0.3and tousetheRunge-Kutta method togetyour ﬁrst approximations. Calculate y(0.3) inonestep andthen intwosteps, i.e.,calculate 1/(0.3) andy(0.15 +0.15). Formula (47.42) willthen give youanapproximate value ofthemagnitude oftheerror. Itstates that theapproximate error in3/(0.15 +0.15) isone-ﬁfteenth thedifference between y(0.15 +0.15) and 1/(0.3). Ifone-ﬁfteenth this difference is greater than thedesired error, youmust reduce h;ifitisreasonably less than thedesired error, youmay retain h;ifitisvery much lessthan the desired error, youmay increase h. 692 NUMERICAL Mnrnons Chapter 10 Letusassume h=0.3issatisfactory. Starting with thevalue ofy(0.3) found intwosteps, wethen proceed toﬁndy(0.6) andy(0.9). The initial condition plus these three preliminary estimates, y(0.3), y(0.6) andy(0.9), willenable ustoswitch toMilne’s method. From here on,webegin to watch thesizeofthecolumn Dinourtable. By(51.21), itisthediffer- ence between thepredictor value computed byusing (51.1) andtheﬁrst corrected value computed byusing (51.11). Dividing Dby29willkeep usposted astotheapproximate magnitude oftheerror intheﬁrst yc ﬁgure inourtable. When itbecomes larger than thedesired error, we must reduce h.This brings upthequestion ofhow toreduce hinthe course ofanextended computation; alsohow toincrease h. LESSON 52C. Reducing and Increasing h.Asweproceed step by step, anhwhich issatisfactory inearly stages may become toolarge in later ones. Suppose, forexample, wearesatisﬁed that thevalues of y(0.1), y(0.2), 3/(0.3) found byastarting method have thedesired accuracy. Weswitch totheMilne method anddetermine that thevalues ofy(0.4), y(0.5), 1/(0.6), y(0.7) stillhave thedesired accuracy. However toobtain y(0.8), weﬁnditnecessary toreduce hto0.05. The next value wemust ﬁnd is,therefore, y(0.75). TouseMilne’s predictor formula (51.1) with h=0.05, :60=0.55, wemust know 3/(0.55), 1/(0.6), 1/(0.65), y(0.7). We already know y(0.6) andy(0.7). How doweﬁnd1/(0.55) and3/(0.65) without thenecessity ofstarting from thebeginning allover again with h=0.05? One way ofobtaining g/(0.55) and1/(0.65) isbyuseofNewton’s back- ward interpolation formula (49.22). Since weknow eight evenly spaced values ofy(:c), 0.1unit apart, namely y(0), y(0.1), ---,1/(0.6), y(0.7), we canuseterms inthis formula toV7y(0.7). By(49.23), with :60=0.7 :6=0.65 and h=0.1,weﬁnd n=(7-—0.65)/0.1 =1}.With n=3 3'1 yo lys lyz 1 1 '°:'I\'J§-'<=:x0—3h x°—2h x0—h x0- —3h —2h —h — Figure 52.1 Lesson 52C Ranucmo ANDINCREASING h693 andtheneeded values intheformula obtained byconstructing atable of differences ofy(0.7), wecan determine y(0.65). Similarly, bytaking x=0.55sothatn=£3wecanapproximate 3/(0.55). Asecond andperhaps easier waytoobtain these values isbyuseofafor- mula which hasapproximately thesame order ofaccuracy asSimpson’s rule. Letthethird degree polynomial F(:c) =a—|—bx+cxz+119:3bean interpolating function fory(ac). Since F(x) isathird degree polynomial, we may assume byTheorem 48.2, that itagrees with y(:c)atfour points whose abscissa values arehunits apart. Callthese fourabscissas xo,:60—h, xo—2h,:60—3h,andtheir respective ordinates yo,yl,yz,y3,Fig. 52.1. Forconvenience incalculation, wetake theorigin at:60,sothat thecoordi- nates ofthepoints atwhich F(x) and y(z) agree are(0,yo), (—h,y1), (—2h,y2), (——3h,y3). Since each ofthese points satisﬁes theequation (52.11) F(a:) =a+bx+cm’+data, wehave (52.12) yo=a, y1= a— hb+h2c—h3d, yg=a—2hb—|—4h2c —8h3d, ya=a—3hb+9h2c —27h3d. Solving (52.12), fora,b,c,d,weobtain (52.13) a=yo, b=11?/0 —18111 +9?/2—2113, 6h c_ 2110-51/1+4!/2-1/3 “ 21.2 ’ d=3/0-33/1+3?/2-1/3_ 6h3 Substituting (52.13) in(52.11) willgive theequation oftheinterpolating function F(x)fory(z). Therefore when a:=——h/2, F(—h/2) willgivethe approximate value ofy(—h/2). Making thesubstitution (52.13) in (52.11) andreplacing acby—h/2, wehave (52.14) F(_ =yo__M>“18%; 91/2*2% +2!/0 —51/1+492 -1/3 ___!/0 "'31/1-l-31/2-'1/3 8 48 =5y6+15y1 ——5y4+1r3,16 which isanapproximation ofy(—h/2). Hence interms ofouroriginal 694- NUMERICAL METHODS Chapter 10 abscissas, seeFig.52.1, weobtain theapproximation formula, (52.15) y(:60— = 1’5[51/($6) +151/(16 ~h)-5y(w6 —2h)+1/(we—3h)]- Comment 52.16. Todouble hisasimple matter. Ifyoudecide that thehyou have been using canbesafely doubled, allyou need doisto take every other preceding estimate. Forexample, ifyouhave been using h=0.1andﬁndh=0.2willbesatisfactory, youneed useonlythepre- vious estimates, y(a:o +0.2), y(a:0 +0.4), etc. LESSON 52D. Summary and anIllustrative Example. Wehave given various starting, continuing, predictor, and corrector formulas. Which ones youshould choose inaparticular problem willbedetermined bythedegree ofaccuracy desired, therelative difficulty ofthemethod, and theamount oflabor involved. Ifyou have many occasions touse numerical methods, experience willbeyour best guide. Itistheusual practice tostart with theRunge-Kutta method. After theneeded preliminary estimates have been obtained, itisthen customary toswitch totheMilne method. Insolving problem 52.2 below, weshall assume that weknow nothing about itssolution. Hence weshall have to depend foradetermination oftheapproximate accuracy ofourresult on thesuggestions made inLesson 52A. Example 52.2. Find anapproximate value when 2:=0.6ofthepar- ticular solution ofthedifferential equation (9) 9'=$2+y. forwhich y(0) =1.Assume wewish theerror tobelessthan 0.00005. Solution. Tobring more methods into thediscussion, weshall start with Taylor series instead ofwith theusual Runge-Kutta formulas, then switch toRunge-Kutta andendﬁnally with Milne. First wemust choose anh.Wedecide totryh=0.1,andproceed to calculate y(0.1) intwo steps andinonestep inorder toseehow much agreement there isinthetworesults. Using ourbasic equation (46.12) with h=0.05 and:60equal successively to0and0.05, weobtain (b)y(0+0.05)=1/(0-0'5) =2/(0)+y’(I<'1I)(0-05) (4) +1-Lg(0.o5)’ +yff”(0.05)“ +yT§°) (0.05)*, y(0.05+0.05)=1/(0.1)=y(0.05) +1/(0.05)(0.05) + (0.05)” + (0.05)=* + (0.05)*. Lesson 52D Smmnnr ANDANInnusrmrrvn Exrmrtn 695 Thederivatives of(a)toorder fourare (9)9’=av’+y, y"=25+9'. 9”’=2+9", 9"’=9"’- Hence when :6=0andy(0)=1,weﬁndfrom (c) (d) 1/(0) =1, 1/"((1) =1. I/’”(0) =3, 1/("(0) =3- Substituting these values intheﬁrstequation of(b),weobtain (e) y(0.05) =1+0.05 —|—§(0.05)2 +%(0.05)3 +§(0.05)4 =1.05 +0.00125 +0.0000625 +0.0000008 =1.0513133. With5=0.05,y(0.05) =1.0513133, we66.1from(5) (1) 1/(0.05) =1.0533133, 1/"(0.05) =1.1533133, 1/"(0.05) =3.1533133, y“’(0.05) =3.1533133. Substituting (e)and(f)inthesecond equation of(b),weobtain (g)y(0.05 +0.05) =1.0513133 +1.0533133(0.05) +s(1.1533133)(0.05)” +4(3.1533133)(0.05)* +,1.,(3.1533133)(0.05)* =1.0513133 +0.05269067 +0.00144227 +0.00006570 +0.00000082 =1.1055128. InExample 46.2, wefound thevalue ofy(0.1) inonestep. By(f)of thatexample, y(0.1) =1.1055125. Wenownotethatthevalue ofy(0.1) obtained inonestep agrees, when rounded off,tosixdecimal places with thevalue ofy(0.1) obtained intwosteps. Wenote also by(46.57), that (h) E(0.l) =-}-§[y(0.5 +0.5) -—y(0.l)] =4}-§(0.0000003) =0.00000032. Since thiserror issuﬁiciently smaller than thedesired one,andbecause y(0.l) and1/(0.05 +0.05) agree tosixdecimals, weaccept h=0.1asa proper starting value. Ournext task istodetermine how many decimal places tocarry. Since h=0.1andwewant y(O.6), there willbe,ifhdoes notneed tobere- duced, atotal ofsixsteps. Iftherefore wecarry seven decimal places, dropping theeighth, theabsolute value oftheerror dueonly torounding off,inthemost unfavorable circumstances, islessthan 6(0.00000005) = 0.0000003, which does notaffect thesixth place. Since wewant ourerror tobe<0.00005, wedecide that seven decimals willgive usasufﬁcient margin ofsafety. Wearenow ready tocalculate y(0.2). With xo=0.1 696 Nounnrcn. Mrrrnoos Chapter 10 andh=0.1,ourbasic equation (46.12) becomes <0no.2)=no.1)+y'<0-1)<9-1)+ <9-1)’ +”i—"'§?'1) (0.1)‘*+y-L311) (0.1)*+..._ Although thevalue of1/(0.1) =1.1055128 found intwosteps with h= 0.05 ismore accurate than 3/(0.1) =1.1055125 found inonestep with h=0.1,weshall usethelastﬁgure because oftheadoption ofa0.1 interval. Weshall then beable tousethose error formulas that arebased onh=0.1. By(c),with 2:=0.1,y=1.1055125. y'(O.1) =1.1155125, y”(O.1) =1.3155125, y"’(0.1) =3.3155125, y‘*>(0.1) =3.3155125. Substituting theabove values in(i),weobtain (k) 3/(0.2) =1.1055125 +(0.1)(1.1155125) —|— (1.3155125) +9%(3.3155125) +930-1 (3.3155125) =1.2242077. Before proceeding toﬁnd3/(0.3) weshall check theaccuracy ofy(0.2) = 1.2242077 asgiven in(k). By(46.57), (l) E(0.1 +0.1) =T15[y(0.1 +0.1) —y(0.2)]. By(k) (m) y(0.1 +0.1) =1.2242077. By(f)ofExample 46.2 (n) y(0.2) =1.2242000. Hence E(0.1 +0.1) =-f1;(1.2242077 ——1.224200) =0.0000005, which istheapproximate error in3/(0.2) of(k)found intwosteps. Since thiserror isstillmuch lessthan thedesired oneof0.00005, wecontinue touseh=0.1. Toﬁnd y(0.3), weswitch totheRunge-Kutta method. By(47.34), with :60=0.2,h=0.1,y(0.2) =1.2242077, (o) y(0.3) =1.2242077 -1-3?,-(wl —|—2w2—|—2103+w4). By(47.35), With h=0.1, :60=0.2, yo=3/(2:0) =jl/(0.2) =1.2242077, Lesson 52D SUMMARY ANDANIttosrnxrrvn Exxmrta 697 f(z)!/) =$2+ya (p) w,=0.1f(0.2, 1.2242077) =0.1(0.22 +1.2242077) =0.12642077, w2=0.1f(0.25, 1.2242077 +0.0632104) =0.1(0.252 +1.2374131) =0.1349913, 1.03=0.1f(0.25, 1.2242077 +0.0674959) =0.1(0.252 +1.2917036) =0.1354204, w4=0.1f(0.3, 1.2242077 +0.1354204) =0.1(0.32 +1.3596281) =0.1449628. Substituting (p)in(o),wehave (<1)1/(0-3) =1.2242077 +%(0.1264208 -1-0.2699836 +0.2708408 +0.1449628) =1.2242077 +0.1353680 =1.3595757. Wearenow ready toproceed with Milne’s method. Forconvenience, we collect theresults thusfarobtained. (52.21) y(0) = 1/(0-1) = 1/(0.2) = y(0.3) =1.0000000, 1.1055125 1.2242077 1.3595757y'(0) =1.0000000; y’(0.1) =1.1155125; 3/(0.2) =1.2642077 [by(a)and(k)]; y'(0.3) =1.4495757 [by(a)and(q)]. The Milne formulas are,by(51.1) and(51.11), (52-22) 1/p(1>o -1-4h)=I/($0) +$1296. +1)—re.+21>+2711.+361. (52.23)7.6.+4h>=1/<.~».+2h) +§6'0.+2h)+4.6.+310+9.-<4.+4101. Using (52.22) andthevalues in(52.21), wehave with :60=0,h=0.1, <r>y.<94>=1/<0)+12310-1) —.102)+23103)] =1.0+ [2(1.1155l25) —1.2642077 +2(1.4495757)] =1.5154625. 698 NUMERICAL M11/r1-1o1>s Chapter 10 By(a)and(r), (.) y,,'(0.4) =1.6754625. By(52.23) with5.,=0,1.=0.1, (t)1/.(0-4) =y(0.2)+%[1/(0.2) +4;/(0.3) +y,,’(0.4) 1.2242077 +0;[1.2642077 +4(1.4495757) +1.6754625] 1.5154735. And by(a)and(t), (u) y/(0.4) =1.6754735. Using thevalue in(u)asanew estimated value inSimpson’s formula (52.23), wehave, with thehelp of(52.21), (v) yc(0.4) =1.2242077 + =1.5154738 Athird application ofSimpson’s formula willnotchange thevalue in (v)-By(9)9119 (v). (w) y/(0.4) =1.6754738.0;[1.2642077 +4(1.4495757) +1.6754735] InTable 52.24, westart torecord theMilne values, keeping acareful watch onthecolumn headed D,which by(51.21), isthedifference between Table 52.24- yof "(52.22)y.of (52.23)l/P, ye,D= ﬁrsty,~—y,ID/29‘ < 0.4 1.5154625 0.5 1.6961508 0.55 1.7972722 0.6 1.90636021.5154735 1.5154738 1.6961631 1.6961635 1.7972764 1.7972765 1.90635731.6754625 1.9461508 2.0997722 2.26636021.6754735 1.6754738 1.9461631 1.9461635 2.0997764 2.0997765 2.26635730.0000110 0.0000004 0.0000123 0.0000005 0.0000042 0.0000002 —0.0000029 0.0000001 thepredictor value and theﬁrst corrector value. And, by(51.35), the error intheﬁrst corrector value foronestep isapproximately -D/29. Since, for1/(0.4), [D/29| islessthan thedesired error of0.00005, wecon- tinue with h 0.1. Lesson 52D SUMMARY ANDANILLUSTRATIVE EXAMPLE 699 By(52.22) with h=0.1,:60=0.1, (X)9.0-5)=1/(0.1)+12910.2) -1/<03)+2.10-4)] =1.1055125 + [2(l.2642077) ——1.4495757 +2(1.6754738)] =1.6961508. By(a)and(x) (aa) y,,’(0.5) =1.9461508. Therefore by(52.23) with h=0.1,:60=0.1, (bb) y.(0-5) =1/(0-3) +9§—1ly’(0-3) +41/(0-4) +U./(0-5)] =1.3595757 + [1.4495757 +4(1.6754738) +1.9461508] =1.6961631. Asecond application ofSimpson’s corrector willchange thevalue in(bb) to (6.) y.(0.5) =1.6961635. Athird application of(52.22) willnotchange thevalue in(cc). With this value ofy.(0.5), weﬁnd from (a),y.'(0.5) =1.9461635. Although |D/29| isstillmuch lessthan desired accuracy, andwemay safely continue touseh=0.1,weshall calculate 3/(0.6) byreducing hto0.05 inorder to demonstrate how touseformula (52.15). With hnow equal to0.05, wecannot usethepredictor formula (52.22) toﬁnd y,,(0.55) unless weknow four preceding values ofy,0.05 unit apart, namely y(0.5), y(0.45), y(0.4), andy(0.35). Wealready know y(0.5) andy(0.4). Toﬁnd 3/(0.45) and 1/(0.35), wemake useofformula (52.15). With h=0.1,theformula becomes (dd)y(0.45) =3/(0.5—0.05) =T1;[5y(0.5) +15y(0.4) -51/(0.3) +y(0.2)] =T1,[5(1.6961635) +15(1.5154733) —5(1.3595757) +1.2242077] =1.6024534, y(0.35) =y(0.4-0.05) =115l5y(0-4) +15!/(0-3) —52/(0-2) +y(0.1)l =g[5(1.5154733) +15(1.3595757) -5(1.2242077) +1.1055125] =1.4347174. 700 NUMERICAL Marnons Chapter 10 By(a)and(dd), (ee) y'(0.45) =(0.45)2 +1.6024534 =1.8049534, Wenow have theneeded preliminary values tousethepredictor formula (52.22) toestimate y,,(0.55). With h=0.05, :60=0.35, itbecomes (ff) y,,,(0.55) =y,,(0.35 +0.20) =1/(0.35)+ 12310.4) —310.45)+2.105)] =1.4347174 + [2(1.6754738) ——1.8049534 +2(1.9461635)] =1.7972721. By(a)and(ff) (gg) y,'(0.55) =(0.55)” +1.7972721 =2.0997721. Thecorrector formula (52.23), with h=0.05, :60=0.35, now becomes (hh)9.(0-55)=1/.(0.35 +0.20) =1/(0.45)+1./(0.45) +4./<05)+y.'<9.s5>1 =1.6024534 + [1.8049534 -1-4(1.9461635) +2.0997721] =1.7972764. Asecond application ofthecorrector formula changes thevalue in(hh) to (ii) y.(0.55) =1.7972765. With thisvalue ofy.(0.55), weﬁndfrom (a),y¢’(0.55) =2.0997765. Returning tothepredictor formula (52.22) with h=0.05, 2:0=0.4, (ii)u.(0-6)=y,,(0.4+0.2) =.614)+ 124/(9.45) —1/<05)+2./<0.55>1 =1.5154733 +OT?[2(1.3049534) —1.9461635 +2(2.0997765)] :1.9063602. By(9)and(ii) (kk) y,/(0.6) =(0.6)2 +1.9063602 =2.2663602. Lesson 52—Exercise 701 By(52.23) with h=0.05, xo=0.4, (ll) y,(0.6) =3/(0.4 +0.2) =y(0.5)+%[1/(0.5) +4./(0.55) +y,,'(0.6)] =1.6961635 + [1.9461635 +4(2.0997765) +2.2663602] =1.9063573. Asecond application offormula (52.23) does notchange thevalue in(ll). Hence, by(a), (mm) y/(0.6) =2.2663573. Asaﬁnal check ontheoverall accuracy ofourresult, wemake useof Weddle’s_corrector formula (50.64). With h=0.1,xo=0,itbecomes (weusethesubscript wforWeddle) (I111) 1/..(0-6) =1/(0)+ [y'(0) +5y'(0-1) +3/(0.2) +6y'(0-3) +1/(0-4) +5y’(0-5) +y'(0.6)] =1+0.0311+5(1.1155125) +1.2642077 +6(1.4495757) -1-1.6754738 +5(1.9461635) +2.2663573] ==1.9063562, which isacloser approximation tothetrue value ofy(0.6) than istheone in(ll). Since thedifference between y.(0.6) andy,,,(0.6) isnotsigniﬁcant, weassume their common value 1.90636 rounded offtoﬁvedecimals is accurate toﬁveplaces anditserror istherefore <0.000005. [The actual value of1/(0.6) is1.9063564 sothat theerror inthevalue of1/(0.6) in(nn) is0.0000002. Rounded offtoﬁve decimal places, theactual value of y(0.6) is1.90636, aﬁgure which agrees with ourﬁnal result rounded off toﬁvedecimal places.] EXERCISE 52 1.InExample 52.2, wefound 1/(0.1) =1.1055125 and y(0.05 +0.05) = 1.1055128. Useformula (46.57) tocorrect 1/(0.1). Using thiscorrected ﬁgure of y(0.1), compute byseries methods y(0.2) andy(0.1 +0.1), i.e.,compute the value ofy(0.2) inonestepandintwosteps. Thevalue of1/(0.2) =1.2242000, computed inonestepcanbefound in(n)ofthisexample. Useformula (46.57), with 2:0=0,h=0.2,tocorrect y(0.2). Starting with thiscorrected value ofy(0.2), switch totheRunge-Kutta method, fourth order form, andcom- pute y(0.2 —|—0.1)andy[(0.2 +0.05) +0.05], i.e.,compute y(0.3) inonestep andintwosteps. Apply formula (47.42) tocorrect y(0.3). Replace (52.21) bythese newcorrected ﬁgures justobtained. UseMilne’s method toﬁnd y,.(0.4), y,(0.4), D(0.4). Apply formula (51.35) tocorrect y,(0.4) andaddthis 702 NUMERICAL M1-rrnoos Chapter 10 corrected ﬁgure toyour table. Compute y,,(0.5), y,(0.5), D(0.5). Correct y,(0.5) by(51.35). Dothesame fory,(0.6). Finally apply Weddle’s rule (50.64) tocompute y,,(0.6). How many decimal place accuracy doyounow have? Compare allresults with previous ﬁgures andwith actual values. 2.Usethemethod ofthislesson toﬁndanapproximate value when 2:=0.6of theparticular solution ofy’=:1:+yforwhich y(0) =1.Assume youknow nothing about thesolution andthatyouwishtheerror tobelessthan 0.00005. UseTaylor series fortheﬁrsttwoapproximations, Runge-Kutta forthethird approximation. Then switch toMilne’s method. Inthecourse ofyour com- putations reduce hbyone-half even though itmay notbeessential. Use formula (50.63) tocheck theaccuracy of1/(0.5) andWeddle’s rule(50.64) tocheck theaccuracy ofy(0.6). Intheabsence ofasolution, how many decimal place accuracy could youassume in1/(0.5), iny(0.6)? See3after "3.Cumulative Errors,” Lesson 52A. Solve theequation andcompare your results with actual value of3/(0.5) andy(0.6). 3.Follow theinstructions in2toﬁndanapproximate value when 1:=1.6, oftheparticular solution ofy’=xyforwhich y(1) =1. ANSWERS 52 1.y(0.1) =1.1055123, y(0.2) =1.2242035, y(0.3) =1.3595771,y,,(0.4) =1.5154628, y,,(0.5) =1.6961512, y,,(0.6) =1.9063429, 1/¢(0.4) =1.5154745, y.(0.5) =1.6961646, y.(0.6) =1.9063565, D(0.4) =0.0000117, D(0.5) =0.0000134, D(0.6) =0.0000136, y(0.4) =1.5154741, y(0.5) =1.6961641, y(0.6) =1.9063560, y.,(0.6) =1.9063566. Canassume ﬁvedecimal place accuracy. Actual value: 1|/(0.6) =1.9063564. 2.y=2e‘—:1:—1.Actual values: 1/(0.5) =1.7974425, y(0.6) =2.0442376. 3.y=e(”"1)/2. Actual value: 3/(1.6) =2.1814723. LESSON 53. Numerical Methods Applied toaSystem ofTwo First Order Equations. Weconsider asystem oftwoﬁrst order equations d if =f(t)z7y)! d if=9(tw.9). forwhich (53-11) $00) =$0. 1/(10) =1/0- InLesson 39,Example 39.17, weshowed how toﬁndaseries solution of asystem such as(53.1). Hence wecanusethismethod toobtain starting approximations, provided thederivatives canbeobtained without exces- sive diﬁiculty. Ifthey cannot, wecanusethefollowing Runge-Kutta fourth order formulas. (53-12) $(1o'1'h)=$00) 'l'6611 +2112'1'2113'1'"4). Illfio'l'h)=I/(10) 'l'zi=(w1 'l'21112'1'21113'1'104). Lesson 53 SYSTEM orTwo Frnsr Onnnn EQUATIONS 703 where (53-13) P1=hf(1o.1>0,3/0). W1=h9(10,$0,?/0). 1/'2=hf(10 'l'271.150 'l'2111.110 'l'21111). W2=h9(10 'l'211.160 'l'11711.1/0 'l'21111). 113=hf(10 'l'211.$0'l'1112,yo'l'21112). W3=h§(io 'l'th»$0'l'2112.1/0 'l'21112). '14=hf(10 'l'71.110 'l''13,!/0 'l'1113). W4=h9(10 'l'11,110 'l'03.1/0 'l'we)- The Milne predictor and corrector formulas forthesystem (53.1) are (53.14) :z:,,(t0 +4h)=a:(to) +%1'[2x’(to —[-h)—:1/(to +2h)+2a:’(to +3h)], 9.0.+41>=90.)+%12./(1. +1»)—1/(1.+21»)+270.+3h>1. (53.15) :t,,(l0 —]-4h) =wo.+2h)+§1w'0. +2h)+4w'<1.+3h)+9./(1.+4101. 9.0.+41> =ya.+210+§lw’(t.+2h)+4./<1.+sh)+9./(1.+4101. Allcomments and formulas inregard toerrors, made inconnection with thenumerical solution oftheequation y’=f(z,y), apply toeach function making upthesolution ofasystem. The ﬁfth degree formula (50.63) orWeddle’s formula (50.64) canbeused tocheck theoverall accuracy ofthevalues of:v(t)andofy(t). Example 53.16. Find approximate values when :6=0.4andy=0.4 ofaparticular solution ofthesystem (9) w’(t)=19. 2/(t)=92/. forwhich 16(0) =1andy(0) =1. Solution. Tobring intothediscussion alltheabove formulas, webe- ginbyusing Taylor series.* The series solution of(a)hasalready been ‘By(f)ofExample 39.17, theinterval ofconvergence is|t|<0.0625. We,therefore, cannot useTaylor series toﬁnd:6(0.1), y(0.1), z(O.2), y(0.2) unl wecanestablish a larger interval ofconvergence. Thechances arethat theseries solution does actually have such alarger interval since thetheorem gives only aminimum interval. But unless wecanprove both series converge fort=0.1,t=0.2,wewould need touse Runge-Kutta orsome other starting method inplace ofTaylor series. Wehave used Taylor series only forillustrative purposes. 704- NUMERICAL Mmnons Chapter 10 found inExample 39.17. Itis[see(k)] <1») x0>=1+5+5+5+--- 23s '2 3 4 y<0=1+¢+‘§+‘§+;i4+---. Hence, by(b),with t=0.1and0.2respectively, weobtain (0)a:(0.1)=1+%+@%l::+§%+---=1.0053, 2 3 4 y(0.1)=1+0.1+@§1l-+(%1)-+32%)-+---=1.1054, :c(0.2)=1+(0%2+£95;E+£9182£+---=1.0229, y(0.2)=1+(0.2)+@§+@+K%+...=1.2231, Remark. Wehave used direct substitution toﬁnd:z:(O.2) and3/(0.2). Itwould have been more accurate, buthave involved much more labor, ifwehadused thecreeping upmethod. Toﬁnd:c(0.3) and3/(0.3), weswitch totheRunge-Kutta method. With to=0.2, h=0.1, x(0.2) =1.0229, y(O.2) =1.2231, (53.12) becomes (d) :c(0.3) =1.0229 +%(v1 +202-1-203+04), 1/(0.3) =1.2231 +%(w1 +2w;+2w3+w4). With h=0.1, to=0.2, 2:0=1J(t0) =x(0.2) =1.0229, yo=1/(to) = 1/(0.2) =1.2231, f(t,z,y) =tyandg(t,1:,y) =xy,(53.13) becomes (e) 1);=0.lf(0.2, 1.0229, 1.2231) =0.1(0.2)(l.223l) =0.02446, wl=0.lg(0.2, 1.0229, 1.2231) =0.1(l.0229)(l.223l) =0.1251, 0,=0.1f(0.25, 1.0229 +0.0122,1.2231 +0.0020) =0.1(0.25)(1.2s57) =0.0321, w,=0.1g(0.25, 1.0229 +0.0122,1.2231 +0.0020) =0.1(1.0351)(1.2ss7) =0.1331, 03=0.1f(0.25, 1.0229 +0.0101, 1.2231 +0.0000) =0.1(0.25)(1.2s97) =0.0322, w3=0.1g(0.25, 1.0229 +0.0101, 1.2231 +0.0000) =0.1(1.0390)(1.2s97) =0.1340, Lesson 53 Srswsm orTwo Fmsr ORDER Eqozmons 705 114=0.1f(0.3, 1.0229 +0.0322,1.2231 +0.1340) =.0.1(0.3)(1.3571) -=0.0407, w.,=0.1g(0.3, 1.0229 +0.0322,1.2231 +0.1340) =0.1(1.0551)(1.3571) =0.1432. Hence (d)becomes (f):c(0.3) =1.0229 +H0-0245 +2(0.0321) +2(0.0322) +0.0407] =1.0552, 1/(0.3) =1.2231 +%[0.125l +2(0.1331) +2(0.1340) +0.1432] =1.3568. Since wenow have theneeded number ofpreliminary estimates, weswitch toMilne’s method toﬁndx(0.4) and1/(0.4). Before wecanuseformulas (53.14) and (53.15), however, wemust know a:'(0.l), :c’(0.2), :v’(0.3) and y’(0.1), 3/(0.2), 1/(0.3). Weobtain these values byuseof(a)and the values of:c(0.1), :z:(0.2), :c(0.3), y(0.1), g/(0.2), 1/(0.3) asfound in(0)and (f)above. Therefore (2) x'(0.1) =0.1y(0.1) =(0.1)(1.1054) =0.1105, a:'(0.2) =0.21/(0.2) =(0.2)(l.2231) =0.2446, a:’(0.3) =0.3y(0.3) =(0.3)(l.3568) =0.4070, y'(0.1) =a:(0.1)y(0.1) =(1.0053)(1.l054) =1.1113, y'(0.2) =:::(0.2)y(0.2) =(1.0229)(1.223l) =1.2511, 1/(0.3) =x(0.3)y(0.3) =(1.0s52)(1.350s) =1.4317. Hence with to=0,h=0.1,and with theinitial conditions :c(0) =1, y(0) =1,wecanwrite (53.14) as (11):c,,(0.4) =1+%;5[2@'(0.1) -no.2) +2:c'(0.3)] =1+ [2(0.1l05) —0.2446 +2(0.4070)] =1.1054, y.,<0-4)=1+12140.1) —1'02)+21/<0-311 =1+%[2(l.11l3) —1.2511 -1-2(1.4317)] =1.5113. By(a)and(h),weobtain (i) :0,/(0.4) ==0.4;/(0.4) =0.4(l.5113) =0.6045, y,/(0.4) =:1:(0.4)y(0.4) =(1.l054)(l.5113) =1.6706. 706 NUMERICAL Mmnons Chapter 10 The corrector formulas (53.15) aretherefore (1) 340.4) =:c(0.2)+%[:c’(0.2) +4x’(0.3) +2,/(0.4)] =1.0229 + [0.2446 +4(0.4070) +0.6045] =1.1055, 3.01.4)=yam)+040.2)+4y'<0-3)+y.'<o.4>1 =1.2231 +9§[1.251l —|—4(1.4317) —]-1.6706] =1.5114. EXERCISE 53 Using thevalues of:c(0.4) andy(0.4), given in(j)ofExample 53.16, asnew predictor values, usecorrector formulas (53.15) toseeifthey willcorrect these results. Ifthey do,repeat theprocess until twosuccessive values of:c(0.4) andy(0.4) agree. Then compute :v(0.5) andy(0.5) bymeans ofthepredictor formulas andtherepeated useofthecorrector formulas. With ﬁvevalues of x(t)known inaddition totheinitial condition, usecorrector formula (50.63) tocheck theaccuracy of:v(0.5). Dothesame fory(0.5). How many deci- malplace accuracy canyouassume inz(0.5), y(0.5)? Hint. See3after “3.Cumulative Errors,” Lesson 52A. Find approximate values when :1:=0.1,0.2,---,0.5,y=0.1,0.2,---, 0.5oftheparticular solution oftheﬁrstorder linear system f(t)=1—1/.1/(t)=—41+ y, forwhich 2(0) =1,y(0) =1.Take h=0.1.UseTaylor series tothefourth order, direct substitution method, tocalculate x(0.1), y(0.1), x(0.2), y(0.2); Runge-Kutta method tocalculate :c(0.3), y(0.3); Milne’s method tocalculate :v(0.4), y(0.4), :c(0.5), y(0.5). Finally apply corrector formula (50.63) to compute a:(0.5), y-(0.5). How many place accuracy canyouassume in:c(0.5), y(0.5)? Now solve thesystem andcompare results. Inproblem 2,compute a:(0.2), y(0.2) bythecreeping upmethod. Then apply (46.57) tocorrect :c(0.2), y(0.2). With these corrected values of:1:(0.2), y(0.2), useRunge-Kutta method tocompute a:[(0.2—|- 0.05) +0.05], z(0.2+ 0.1). Do thesame for1/(0.3). Correct a:(0.3), y(0.3) bymeans of(47.42). Compute :c,,(0.4), z¢(0.4), D(0.4). Correct a:,(0.4) bymeans of(51.35). Dothesame for y,,(0.4). Compute z,,(0.5), :1:,(0.5), D(0.5). Correct :c,_,(0.5). Dothesame for y,(0.5). Finally apply corrector formula (50.63) tocompute a:(0.5), y(0.5). How many decimal place accuracy canyouassume inyour results? Compare with results obtained in2andwith actual solution. ANSWERS 53 :c(0.4) =1.1055, y(0.4) =l.51l4;:c(0.5) =1.1776, 1/(0.5) =1.6938. Corrector (50.63): :c(0.5) =1.1776, y(0.5) =1.6939. Canassume fourand three decimal place accuracy. Lesson 54- 2.Saconn Onnnn EQUATION 707 1(1)values: 1.01009, 1.00940, 1.17020,1.33240, 1.57490. y(t)values: 0.03234,0.31740, —0.11807, —-0.65395, —1.33021. Corrector values (50.63): a:(0.5) = 1.57504, y(0.5) =—1.33050. Can assume three and two decimal place accuracy respectively, ifweround offtothisnumber ofdecimals. Solution: :1:=}(e3‘+ 30”‘), 1/=§(-e3‘+ 3e"‘). Actual values: a:(0.5) =1.5753203, y(0.5) =—1.3310485. 3.:::(0.2) = :z:(0.3) = :z:(0.4) =1.00953, y(0.2)1.17051, y(0.3) 1.33274, 1/(0.4) x(0.5) =1.57525, y(0.5) Corrector values (50.63)=0.31705;=-0.11355;=-0.6-5450;=-1.33091. x(0.5) =1.57530, y(0.5) =—1.33101. LESSON 54. Numerical Solution ofaSecond Order Differential Equation. Intheproof ofTheorem 62.22, weshow how asecond order differential equation y"=f(a:,y,y’) canbereduced toasystem oftwo ﬁrst order equations. Anumerical solution ofthisequation cantherefore befound bythemethod ofLesson 53.However, itisalsopossible toﬁndanumerical solution ofsuch anequation without thenecessity ofreducing ittoa system. Weillustrate themethod byanexample. Example 54.1. Find anapproximate value when :1:=0.4ofapar- ticular solution oftheequation (a) 2/”=2w+21/—1/’, forwhich y(0) =1,y'(0) =1. Solution. Inorder tobring into thediscussion Taylor series, Runge- Kutta andMilne methods, weshall ﬁndy(0.l) andy(0.2) byTaylor series method, y(0.3) byRunge-Kutta’s method andy(0.4) byMilne’s method. ByTheorem 37.2, if (54-11) 11030 -l"h)=3/050) -1-I/'(¢'=0)h then+yr/5:30) hz+yrréﬁvo) h3+y(4;('$o) h4+_._, (54.12) 5'0».+h>=1/<x.)+ y"<».>h+ 1.’+#,"—°>1*+--- From (a),andbydifferentiation of(a),weobtain yr! S2x+2y ___yr’ ylll =2+2y: __yr!’ y(4) =2311/ _yin. Bytheinitial conditions, 2:==0,y=1,y’=1.Substituting these values in(b),there results (0)y”(0)=1,y”'(0) =2+2—1=3,y“’(0) =2-3=-1. 708 NUMERICAL METHODS Chapter 10 Substituting theinitial conditions and (c)in(54.11) itbecomes, with x0 =0; h2 h3 h4 ((1) ll(h)=1+h+§+§—ﬁ-l-"'. Asintheﬁrst order case, twomethods areavailable toustoﬁnd y(0.1) andy(0.2). Bydirect substitution, weobtain from (d), (0)3/(0.1) =1+0.1+§(0.1)” +§(0.1)3 -,1;(0.1)* =1.105490, y(0.2)=1+0.2+%(0.2)* +.§(0.2)3 -.,1.,<0.2)* =1.223933. The creeping upmethod involves much more arithmetic, butinsures greater accuracy. Byourbasic equations (54.11) and(54.12), with h=0.1 andxoequal successively 0.0,0.1,weﬁnd (f)1/(0+0.1)=y(0-1) I/”(0) 2=1/(0)+1/'(0)(0-1) +*2-,— (0-1) In (4) +%§%MW+#§%MV+~» y’(0+0.1)=1/(0.1) ' ' =7@+w@o0+%QoN<4) i+3/i3s()_)(0_1)3+..., y(0.1+0.1)=y(0.2) ' =y(0.1) +1/’(0.1)(0.1) +%‘-ii) (0.1)2 + (Q_1)3 + (0_1)4= +..., 7/(0.1+0.1)=1/(0.2) I ' =ron+wono0+$§#o0’ _|_ (0_1)3+..._ Substituting theinitial conditions and (c)intheﬁrst twolines of(f),we obtain (g)y(0.1)=1+(1)(0.1) +%(0.1)" +%(0.1)3 -2440.1)‘ =1.105490, 1/(0.1) =1+(1)(0.1) +§(0.1)’ -+(0.1)3 =1.114833. Therefore with 2:=0.1,1/(0.1) =1.105496, y’(0.1) =1.114833, weﬁnd Lesson 54 Snconn ORDER Eqmmou 709 by(b), (11) 3/"(0.1) =2(0.1) +20.105490) —1.114333 =1.290159, y'”(0.1) =2+2(1.114333) -1.290159 =2.933507, y“1(0.1) =2(1.290159) —2.933507 =—0.341139. Substituting (g)and (h)inthethird andfourth lines of(f),there results (1)y(0.2)=1.105490 +(1.114333)(0.1) + (o.1)1 . 0.+W <01)“-—%%1§9 (0.1)1=1.223943, y’(0.2) =1.114333 +(1.290159)(o.1) +E27 <01)’ -gél <01)"=1.259000. Wenow switch totheRunge-Kutta method toﬁnd_y(0.3). The fourth order form forthesecond order equation y"=f(:c,y,y’) forwhich y(:c0) = 1/013/($0) =1/1is (54-13) 1/0110 +71)=1/(5150) +3011 -1-2112'l'2113'1"114), 1/($0 +h)=3/($0) +30111 —|—21112-1-21113+1114), where, with yo’=y’(a:0), (54.14) v1=hyo’, 1111=hf(x0,1l0,!/0'), 112=hlil/0' -l"21111), "12=hf($o +2h,1/0-l"‘E111,1/0'-l"111111), 113=Ill!/0' 'l'21112), 1113=hf($9 +271,1/0-1’2112,1/0'+21112), 114=710/0' —|—1113), 1114=hf(93o +h,1/0+113,1/0'-1-1113)- Therefore with 2:0=0.2,h=0.1,y(.7:o) =y(0.2) =1.223948, y’(0.2) = 1.259060, (54.13) becomes (1)y(0.3)=y(0.2+0.1)=1.223943 +30».+25,+2»,+5.), y’(0.3) =y'(0.2+0.1)=1.259000 +%(w.+215,+21»,+15.). With h=0.1,yo’=3/(xo) =1/(0.2) =1.259060, yo=y(:1:0) ==1/(0.2) = 710 NUMERICAL Mmnons Chapter 10 1.223948, andf(x,y,y’) =21+2y-—y’,(54.14) becomes (k) 111= w1= U2: W2: 113 1113 114 11140.1(1.259060) =0.125906, 0.1f(0.2, 1.223943, 1.259000) 0.1[2(0.2) +2(1.223943) -1.259000] =0.153330, 0.1(1.259060 —|—0.0794418) =0.1338502, 0.1f(0.25, 1.223948 +0.062953, 1.259060 +0.079418) 0.1(0.5 —|—2.573802 —-1.338502) =0.173530, 0.1(1.259060 +0.086765) =0.1345825, 0.1f(0.25, 1.223943 +0.000925, 1.259000 +0.030705)0.1(0.5 +2.531740 -1.345325) =0.173592, 0.1(1.259060 -1-0.173592) =0.1432652, 0.1f(0.3, 1.223948 +0.1345825, 1.259060 —|—0.173592) 0.1(O.6 +2.717061 -—1.432652) =0.188441. Substituting (k)in(j),there results (1) y(0.3) =1.223948 +§(0.125906 +0.267700 -1-0.269165 —|—0.143265) =1.223948 +Q-(0806036) =1.358287, 1/(0.3) =1.259000 +&(0.153330 +0.347000 +0.347134 +0.133441)=1.259000 +3(1.041521)= 1.432047. Wesummarize theresults thus farobtained (I11) 3(0) 3/(0.1) 1/(0-2) 3/(0.3)1.000000, 1/(0) 1.105490, y’(0.1) 1.223943, 1/(0.2) 1.353237, 1/(0.3)1.000000 ; 1.114833; 1.259060 ; 1.432647. The Milne predictor formulas forthesecond order equation y”= f(w.9,9’) forwhich y(wo)=1/0,1/(Io) =91are (54-15) (9)1/p'(ro +4h)=y'(¢1o) +%127%.+h)—1/"<40+21»)+21/"<4.+31>]. (b)4.14.+41>=1/<4.+2h) +§[9740+21»)+49740+sh)+y./<40+41)]. The corrector formulas are (54-16) (9)1/1/(1130 +4h)=1/(10+2h) +§1y"<wo+21»)+41/"<40+sh)+4./'<w.+41)]. Lesson 54- Ssconn ORDER EQUATION 711 (b) 3/40110 -1*4h)=1/0110 -1-211) +§1y'<==o +21»)+41/<4.+sh)+1/.'(4o+41)]. Asintheﬁrst order case, formulas (54.16) canbeused again andagain until twosuccessive values ofy¢(:1:o +4h)agree. By(54.15) and(54.16), with xo=0,h=0.1, <11)4./<0-4)=1/<0)+12y"<o.1) —y"<0_2)+2y"<o3)1. 9.19.4)=70.2)+1970.2)+41/'<9-3)+y.'<0.4)1. 5.79.4)=1/<02)+11/"<0-2)+41/'<93)+1/."<0-4)], 4.0.4)=1/<9-2)+11/<0-2)+4479-3)+1.70-4)]. Before wecanusethese formulas, wemust know thevalue ofy"when :1:=0.1,0.2,0.3. By(a)and(m), these needed values are (0) y"(0.1) =2(0.1)+2(1.105490) -1.114333 =1.290159, 3/"(0.2) =2(0.2)+2(1.223943) -1.259000 =1.533330, 1/'(0.3) =2(0.3)+20.353237) -1.432047 =1.333927. Substituting in(n),theinitial conditions and thevalues in(m)and (0), weobtain (p) y,’(0.4) =1+ [2(1.296159) ——1.588836 +2(1.883927)] =1.636178, y,,(0.4) =1.223948 -1- [1.259060 —|—4(1.432647) +1.636178] =1.511476. Tousethethird formula in(n),weneed toknow y,,"(0.4). With :1:=0.4 andy,,(0.4), y,’(0.4) having thevalues in(p),weobtain from (a), (q) y,,"(0.4) =2(0.4) -1-2(1.511476) —1.636178 =2.186774. Hence thelasttwoformulas in(n)become (1)y,,’(0.4) =1.259000 +%[1533330 +4(1.333927) +2.130774] =1.030104, y.(0.4) =1.223943 +951[1.259000 +40.432047) +1.030104] =1.511473. 712 NUMERICAL METHODS Chapter 10 EXERCISE 54 Using thevalue ofy(0.4), given in(r)ofExample 54.1, asanewpredictor value, apply corrector formulas (54.16) repeatedly until twosuccessive values ofy(0.4) agree. Then ﬁndy(0.5) bymeans ofthepredictor formulas andre- peated useofthecorrector formulas. Dothesame fory(0.6). Finally apply Weddle’s rule(50.64) tocompute y..,(0.6). How many decimal place accuracy canyouassume iny(0.6)? See3after “3.Cumulative Errors, ”Lesson 52A. Solve (a)ofExample 54.1andcompare your results with actual values. Apply themethod ofthislesson toﬁndanapproximate value when 2:=0.5 ofaparticular solution oftheequation y"=as-1-2y+y’forwhich y(0) =1, 1/(0) =1.Apply corrector formula (50.63) toevaluate 3/(0.5). How many decimal place accuracy canyouassume? Solve theequation andcompare your results with actual values. TheAdams method, seeExercise 50,6, canalsobeusedasacontinuing formula forﬁnding anumerical solution ofasecond order equation. Theneeded formulas are (54-2) (11)1/(I0+11)=211(wo) +h[y'(1o) +1Vy'(wo) +152721/'(wo) +%V3y'(wo)]- (b)y'(10+h)=1/(wo) +hla/"(1:0) +1Vy"(wo) +r‘2V’y”(ro) +%V“z/”(wo)l- Usethese formulas toﬁndapproximate values ofy(0.4) and1/(0.4) ofthe particular solution oftheequation y"=2:1:—|—2y-y’forwhich 1/(0) ==1, y’(0) =1.Hint. With 2:0=0.3,h=0.1,youmust know y'(0), 1/(0.1), 1/(0.2), y'(0.3); y"(0), y"(0.1), y”(0.2), 3/"(0.3) inorder tosetuptheneeded tables ofdifferences. Youwillﬁndthese values inExample 54.1. UseMilne’s method toﬁndanapproximate value ofy(0.4) oftheparticular solution oftheequation y'”=y”+my’+2yforwhich y(0) =1/(0) = y"(0) =1.Thenecessary Milne’s formulas are 1/."<o.4) =1/70)+121/"0.1) —1/"<9.2)+21/"<93)1. 7./(0.4)=1/(9.2)+11/'<9.2)+41/"<0-3) +1/."<0-4)]. 7.0-4)=70.2)+11/<02)+41/<9-3)+1./<0-4)1. Obtain theneeded preliminary values bymeans ofTaylor series, direct substitution. ANSWERS 54 y(0.4) =1.511473, jl/(0.5) =1.686538, 1/(0.6) =1.886643. y.,,(0.6) =1.886640. Canassume ﬁvedecimal place accuracy. Solution: y=-§e"—-Q-e'2‘ —x—Q. Actual value: y(0.6) =1.886666. y=gag’ —-3+2.Actual value: y(0.5) =2.038711. y(0.4) =1.51145, 1/(0.4) =1.03012. 1/(0.4) =1.520. Lesson 55 PERTURBATION Msmon. Fmsr ORDER EQUATION. 713 LESSON 55. Perturbation Method. First Order Equation. Inphysical problems, wefrequently encounter adifferential equation, asforexample, thedifferential equation (55-1) y’+y’=0,1/(1)=1, which hasbeen disturbed byasmall effect, sothat (55.1) hastobemodi- ﬁedtoread (55-2) y’+y’=er,11(1)=1, where eissmall. Itthen becomes necessary todetermine byhow much thesolution of(55.1) hasbeen altered because ofthepresence ofthedis- turbing function ex.Werefer tothis change inthesolution asaper- turbation. Aprecise perturbation theory isextremely diﬂicult. Inthislesson, we shall aimtogiveonlyarough outline ofamethod bywhich thisproblem canbehandled. Call y0(x) asolution of(55.1) satisfying y(l) =1,and denote thesolution of(55.2) by (55-3) y(¢)=yo(w)+20(1) where p(a:) istheperturbation. Wenext expand y(z) inaseries inpowers ofe,sothat (55-4) 1/(w)=yo(w)+¢y1(w) +£21/2(w) +ea;/3(5) +--- Comparing (55.3) with (55.4), weseethat (55-5) P(¢)=¢y1(w) +621/2(5) +53!/a($) +---- The ﬁrst term ey1(a:) iscalled theﬁrst order perturbation; thesecond term e2y2(:c) iscalled thesecond order perturbation, etc. Substituting (55.4) in(55.2), weobtain (55-6) 2/0'+em’+521/2'+¢3y3'+--- +(yo+@111+62112+£31/3+---)2=ﬁv- Carrying outtheindicated multiplication, then collecting coeﬁicients of likepowers ofe,wehave (55-7) (yd+yo’)+(y1’+2y<>y1)¢ +(1/2'+2?/0?/2 +Z/12)E2 —|—(''''">63+'''==£113- Next weequate likepowers ofe.There results (55-3) yo’+yo2=0, yl’+2?/0?/1 =w, 1/2'-l"Zyoi/2 -l"Z/12=0» 714- NUMERICAL M1-rrnons Chapter 10 Bysolving each equation of(55.8) insuccession, wecanthus determine thefunctions y1(x), 1/2(1), ---in(55.4). Each ofthese functions, how- ever, must satisfy aninitial condition. Since theinitial condition asso- ciated with theoriginal equation (55.1) isy(1) =1,and since yoisa solution of(55.1) sothat yo(1) =1,thisinitial condition willbesatisﬁed if,in(55.4), weassume (55-81) 3/0(1) =11 3/1(1) =0» 1/2(1) =0»‘‘‘- Weillustrate thedetails oftheabove method bysolving Example 55.9 below. Inpractice theﬁrstandsecond order perturbation terms of(55.5) areusually sufficient. Example 55.9. Find theﬁrst andsecond order perturbation terms in thesolution of (a) y’—|—yz=0,forwhich y(1) =1, duetothepresence ofadisturbing function ex,where eissmall. Solution. Because ofthedisturbing function ex,(a)must bemodiﬁed toread I/I+1/2=ex: forwhich y(1) =1.Following theprocedure outlined above welet,see (55.4) and(55.81), (C) y(w)=yo(w)+¢y1(w) +£21/2(1) +---, withinitialconditions (<1) 2/o(1)=1,y1(1)=0, 112(1)=0, Substituting (c)in(b),weobtain, see(55.7), (e)(yo'+2/02)+(1/1'+Zyoi/i)¢ +(1/2'+290112 +2/1’)? +---=ex. Equating coeﬁicients oflikepowers of5°,e,62,weobtain from (e)the system ofequations (f)I/0''1"U02=0, 2/1'—|—2?/0?/1 =1'1, 1/2'+2!/092 +U12=0- Asolution oftheﬁrst equation of(f),satisfying theinitial condition 1/0(1) ==1Of(<1),is (g) yo= Lesson 56 Psnrunnxrron Mmnon. Snconn ORDER Eqmvrrons. 715 Substituting (g)inthesecond equation of(f),weobtain (11) 3/1'-l"'22/1 =$- Asolution of(h)satisfying y,(1) =0of(d)is . 1 1(1) 3/1=Z(112 "-5;)‘ Substituting (g)and(i)inthethird equation of(f),weobtain . ,2 1 1(1) y-.»+;y-.>=—E(w‘~2+;)- Asolution of(j)satisfying y2(1) =0of(d),is 12:5 2:: 1 2 <1‘) 1/*=-n(?"§";)"§1?-»"' Substituting (g),(i),(k)in(c),weobtain 1e 1 é’ 2132 <1)1/=;+z(”2*n)“%("’”“*14“*n+;)' Thesolution of(a)satisfying y(1)=1,i.e.,itssolution ifthere were no disturbing function expresent, is1/ac. Because ofthedisturbing function ex,theﬁrstandsecond order perturbation terms are,respectively, the second andthird terms in(l). EXERCISE 55 1.Find theﬁrstandsecond order perturbation terms inthesolution ofy’+ 3/2=0,forwhich y(1) =1,duetothepresence ofadisturbing function e:c2, where eissmall. ANSWERS 55 lF.td_¢3 1_ dd___t’21_921s2s_.ll'Sorer.-51-; ,secon orer. E 2: :c—;§+; LESSON 56. Perturbation Method. Second Order Equation. InLesson 55,weoutlined amethod ofdetermining theperturbation of asolution ofaﬁrstorder differential equation duetoasmall disturbance. Weshall now apply thismethod todetermine theperturbation ofasolu- tionofasecond order equation. Consider thedifferential equation (56.01) 1/”+y=0, 716 Ntmmucan METHODS Chapter 10 forwhich (56.02) 1/(0) =0, y’(0) ='1. Note that (56.01) isthedifferential equation ofsimple harmonic motion. Letthedisturbing function be—2e(y’)2, where eissmall. Therefore (56.01) becomes (56-1) 1/"+y=-2¢(y')2 forwhich (56.11) y(0) =0, y’(0) =1. Wewish toﬁndtheﬁrstandsecond order perturbation terms ofasolution of(56.01) satisfying (56.02) resulting from thepresence ofadisturbing function —2e(y’)2. Cally°(x) asolution of(56.1) satisfying (56.11). Let (56-12) y=yo+E2/1+ @2212+---; therefore yr! =you +eyln +622/21! +____ Inorder that(56.12) maysatisfy theinitial conditions (56.02), weassume (56-13) 3/0(0) =0, 2/1(0) =0- 1/2(0) =0, - yo’(0) =1,y1’(0) =0, 1/2’(0) =0, Substituting (56.12) in(56.1), weobtain, using only terms toe2, (56-14) yo”+@111”+ezuz”+yo+eyi+£21/2 =-2¢l(yo')’ +¢’(2/1')’ +¢‘(z/2')’ +2¢yo’y1’ +2521/0'1/2' +2¢31/1'!l2'l- Collecting coeﬂicients oflikepowers of6,wehave (55-15) (yo”+1/0)+(2/1'.’+1/1)¢-1-(1/2"+1/2)62 =—2(i/o')2¢ —4yo’yi'¢’- Equating coefficients oflikepowers ofe°,e,ea,weobtain from (56.15), (56-15) yo"-l"yo=0,1/1"+I/1=—2(?/o')2i 92”+I/2=-4!/0'1/15 Asolution oftheﬁrst equation of(56.16) satisfying 1/0(0) =0,yo'(0) =1is (56.2) yo=sin2:, yo’=cos:0. Substituting (56.2) inthesecond equation of(56.16), weobtain Lesson 56 Psnrunnzmon Mrrnon. Srzconn Onnsn EQUATIONS. 717 (56.21) yl”+y,=-2cos2:0. Ageneral solution of(56.21) is 56.22 1=clsinx +c2cosx —4sinzx —cosza: 3' v yl’=clcosx —C2SiIl$ —ﬁsinxcosx. Aparticular solution of(56.22) satisfying y,(0) =0,y1’(0) =0is (56.23) yl=§cos 2:-—§sin2x—§cos2x, yl’=——§sin::: -—-gsinxcosx +ﬁsinxcosx =—§sin:z: —ﬁsinxcosx. Substituting inthethird equation of(56.16), thevalues ofyo’andy1'as given in(56.2) and(56.23), weobtain (56.24) yz”+yg=—4(——§ sinrvcosac—§sin:0cos”2:) =-Esinxcosx —|—139-sinxcoszx =-5-sinxcosx +139-sina: —135-sin3x. Thecomplementary function of(56.24) is (56.25) ye=clsinx+c2cosac. Aparticular solution (56.251) ofyg”+yz=§sinaccosx,isyg=—--§sin2:cosx, ofI/2”+yg=13¢sin:0, isyz=—§-ac cosac, ofyz”+I/2=15‘-sins ac,byExample 21.32, is1/2=-}sin3:2:-—-2:2:cos:0. Hence ageneral solution of(56.24), by(56.25) and(56.251), is (56.26) yo=clsinx—|—C2cos:1:——Q-sin :1:cos:2:—§xcos:1:——ﬁsin 32:, yo’=c1cosx—c2sinx+-8-(sing a:-—cos’x) +§(:csinax—cosx)—-1;cos3:0. By(56.13), 1/2(0) =0,y2'(O) =0.Inserting these values in(56.26), weﬁnd (56.27) 0=c2, c1=§--l—f;+§=-'1‘-Q’-. Therefore by(56.26) and(56.27), aparticular solution of(56.24) satisfying 1/2(0) =0,y,'(0) =0is (56.28) yo=-Y-}sinx —-3-sinxcosa: —-§:vcos2: —-1-sin 32:. 718 NUMERICAL Marnons Chapter 10 Substituting, in(56.12), thevalues ofyo,yl,andyaasfound in(56.2), (56.23), and(56.28), weobtain (56.29) y=sinx +e(§cosx ——§sin2:z: —§cos2 az) +e2(§-ﬁsina: -—tgsinxcosx ——§a:cosa: —-}sin3x). Thesolution of(56.01) satisfying (56.11), i.e.,itssolution ifthere were no disturbing function present, issinas.Because ofthedisturbance function ——2e(y’)2, theﬁrstandsecond order perturbation terms are,respectively, thesecond andthird terms in(56.29). EXERCISE 56 1.Find theﬁrstandsecond order perturbation terms inthesolution ofy"— y=0forwhich 1/(0) =0,1/'(0) =2duetothepresence ofadisturbing function ey’,where eissmall. ANSWERS 56 I —l 1.First order: e<% —-—-—xe2); second order: ea[E8-:(-1+2:+x2)-1-£5(1+2:—12)]- Chapter 11 Existence and Uniqueness Theorem fortheFirst Order Differential Equation y’=f(x,y). Picard’s Method. Envelopes. Clairaut Equation. Introductory Remarks. Aswehave repeatedly emphasized, differ- ential equations whose solutions canbeexpressed explicitly orimplicitly interms ofelementary functions arerelatively fewinnumber. Even a ﬁrst order differential equation (57-1) 2/’=f(w,y) willusually nothave anelementary solution. Inthese cases, itisdesirable tohave theorems which willanswer thefollowing questions forus. 1.Does (57.1) have a1-parameter family ofsolutions? SeeExamples 4.21 and4.22 fordifferential equations which have nosolutions; Exam- ple4.2foronewhich hasonly onesolution; (4.652) orExample 5.3 foronewhich hastwo1-parameter family ofsolutions. 2.If(57.1) hasa1-parameter family ofsolutions, isitageneral solution aswedeﬁned this tenn inDeﬁnition 4.7, i.e.,does itcontain every particular solution? Seetheﬁrst example inLesson 4Cfora1-para- meter family which does notcontain every particular solution. 3.Isthere aparticular solution of(57.1) valid onsome interval andsatis- fying agiven initial condition 3/(mo) =yo? 4.Isaparticular solution satisfying aninitial condition y(:co) =yo unique? See(b)ofExample 5.3where thepoint (0,1)liesonaninﬁnite number ofparticular solutions. Fortunately there aretheorems which willgiveus,under appropriate hypotheses, theanswers tothese questions. Atheorem which answers questions 1,2,and3,i.e.,onewhich tellsuswhether asolution exists is 719 720 EXISTENCE Tnsonsu: y’=f(z,y). Pronto. CLAIRAUT. Chapter ll called anexistence theorem. Atheorem which answers question 4,i.e., onewhich tells uswhether asolution isunique iscalled auniqueness theorem. Itshould beemphasized thatanexistence anduniqueness theorem only guarantees orassures theexistence anduniqueness ofasolution. Itwill nottellyouwhether thesolution canorcannot beexpressed interms of elementary functions, orhelp youtoﬁnd thesolution. Forexample, the particular solution ofthedifferential equation (57.11) y’=-re’, forwhich 3/(mo) =yois I (57.12) y(z)=yo+/e“"ada:. =0 Ithasbeen proved that thisintegral cannot beexpressed interms ofele- mentary functions. However, theexistence and uniqueness theorem, which weshall state later, willnothelp youtodiscover thisfact. Allthe theorem willtellyou, isthat since thefunction e_“2 andthepoint (:v0,y0) satisfy itshypotheses, anunique particular solution satisfying (57.11) and theinitial condition exists. There areseveral ways ofproving theexistence anduniqueness theorem fortheﬁrstorder differential equation y’=f(z,y), satisfying thecondi- tion y(:co) =yo.The oneweshall useisdependent onamethod which is known asPicard’s approximation method, named after theFrench mathematician, Charles Emile Picard (1856-1941). Hence before wecan prove thetheorem, weshall ﬁrst need toexplain Picard’s method. LESSON 57. Picard’s Method ofSuccessive Approximations. Weassume forthemoment that aunique particular solution ofthe differential equation (57-2) 1/’=f(w,i/), satisfying theinitial condition (57-21) 1/($0) =Zlo- exists. Lety(z) betherequired particular solution. Then by(57.2), y’=f[:z:,y(:z:)], where f[x,y(:c)] isnow afunction only of2:.Integrating thisequation between thelimits :00and :1:and noting, by(57.21), that when :1:=1:0,y=yo,weobtain (57-22) /dil=Ifl¢,y(1)]d1, y(<v)=3/o+If[1,!/(l)]d1- Lesson 57 PIc.utr>’s M1-zrnoo 721 Inemploying numerical methods toapproximate theparticular solution y(z) of(57.22), weused apolynomial interpolating function inplace ofthe integrand, f[x,y(a:)]. InPicard’s method wealsousetheidea ofanapproxi- mation butofatotally different kind. Inthismethod, weobtain asequence offunctions y°(x), y1(:z:), ---,y,,(x), each ofwhich satisﬁes theinitial con- dition (57.21). The existence anduniqueness Theorem 58.5 which follows will then give theconditions that f(z,y) of(57.2) must fulﬁll inorder that aninterval about :00exist, onwhich, asn—>oo,this sequence of functions approach theparticular solution y(z)of(57.22). Furthermore, each function inthesequence isanapproximation oftheparticular solu- tion y(z): alater one, ingeneral, being abetter approximation than a preceding one. Hence thename successive approximations. Weillustrate themethod bymeans ofexamples. Weconcentrate forthe moment only onthemechanics ofthemethod without considering thesize oftheinterval about moforwhich thesequence offunctions thus obtained converges totheparticular solution y(z) of(57.22). (For themeaning of the convergence ofasequence offunctions, seeDeﬁnition 58.1 and Example 58.13.) Theﬁrstapproximation ofasolution of(57.2) satisfying (57.21) iscalled yo(:v). The function y0(:v) may, asweshall show later, beanyarbitrary continuous function deﬁned inaneighborhood ofxo.Intheabsence of additional information, itisusually taken tobetheconstant function (57-23) 3/0(5) =yo, where yoistheinitial value given in(57.21). Itisevident that thisap- proximation tothesolution isnotavery satisfactory one. Itistheequa- tion ofastraight lineparallel tothexaxisandyounits from it. The subsequent members ofthesequence ofapproximating solutions of (57.2) satisfying (57.21) arecalled y1(:v), y2(:c), ---,y,,(:c), ---,andare obtained inthefollowing manner. I (57-24) yi(w)=yo+Iflw,yo(w)l <11, 1/2(w)=yo+Lfl1=,1/1(1)] drv, Z ya(w)=yo+Loflm/2(w)l dw, -.-.-.--o.o....... 3 3/"($11) =I/0 fix)!/n—1(x)l div: where :00andyoaregiven in(57.21). (Remember f[:c,y°(:c)] means replace 1/inf(fv,1/) byy0($);fl$,!/1($)l means TBP19-66 1/inf(x,y) byy1(1=), 816-) 722 EXISTENCE Tm-zonsmz y’=f(z,y). PICABD. CLAIRAUT. Chapter ll Comment 57.241. Fordifferent permissible starting approximations y0(x), different sequences y0(:t:), y1(x), ---,y,,(a:) willresult. However, each willhave theproperty that foran:1:inaninterval about xo, limy..(w)=1/(w), where y(t) isthesolution of(57.2) satisfying (57.21). The rapidity with which thesequence ofapproximations willconverge tothesolution y(z) willdepend onhow closely thestarting solution yo(:z:) approximates the actual solution y(z); thecloser theapproximation, thequicker thecon- vergence. Example 57.25. Find theﬁrstfour Picard approximations if (a) 1/’=wy, andy(0) =1. Solution. Comparing (a)and theinitial condition with (57.2) and (57.21), weseethat f(z,y) =xy,:00=0,yo=1.By(57.23), ourﬁrst approximation istherefore 3/0(1) =1.The succeeding approximations, by(57.24), are I I 2 <1») no)-1+[]r<w.@/.)d-=1+A-d.-=1+%. m(m)=1+[0f(m/1) dw ‘ 2 2 4:1: a: 2:-1+/Q :z:(1+72—)dx-1—|——§-—|—-8-» ” ’ Z2 $4. ll3(517)=1+_/ f(1’?i1/2)df'5 =1-Ff 1?(1 -l-§-l"-8-)d1%o 0 2 4 6 =-+%+%+s-Comment 57.251. Ifwesolve (a)bythemethod ofseparation of variables, weobtain theparticular solution.y =e’”/2whose series expan- sion is 2 4 6:1: x x1-l"?-1"?-FE-l"-'-. Note that each succeeding function inthesequence yo,yl,yg,ya,---isa closer approximation totheactual solution than istheprevious one. Example 57.26. Find theﬁrst three Picard approximations if (a) y’=rv’—1/ andy(1) =2. Lesson 57 Picxrtrfs Msrnon 723 Solution. Comparing (a)and theinitial condition with (57.2) and (57.21), weseethat f(z,y) =$2—y,xo=1,yo=2.By(57.23) our ﬁrstapproximation istherefore y0(x) =2.Thesucceeding approximations are,by(57.24), I (b) 111(1) =2-l-I1 f(%-1/o)d1 =2+[1(:c2—2)da:==€:—2:z:+-13l, I 1/2(2)=2+1f(w,1/i) dw " s =2-1-/1(a:2—%+2.'c—1—;)dx, __a:3 as‘ 2 I1 53 -s*n+“ -?”+n' Picard’s Method Applied toaSystem ofTwo First Order Equa- tions. Picard’s method ofsuccessive approximations canalsobeapplied toasystem ofﬁrstorder equations. Weillustrate themethod forthepair ofﬁrstorder equations 1% =f1('5ﬂ1,?/)1 % =f2(tsx)i'/)1 satisfying theinitial conditions. (57-31) 27(¢o) =210, y(t0) =yo- The ﬁrst approximations ofthesolution ofthesystem (57.3) satisfying (57.31) arecalled x0(t) andy0(t). Intheabsence ofadditional information, they areusually taken tobetheconstant functions (57-32) 20(1) ==$0, 1/0(1) =I/01 where xoandyoaregiven in(57.31). Thesubsequent approximations are t (57-33) 1-151(1) =$0-1"/I f1l1)230(1)>?/0(1)] dt)lo 3/1(1) =yo+/tofzlt-1150(1))?/0(1)] dt- t 2- =1:0+jt5of1Itv1:1(t);f/l(t)ldti 724 EXISTENCE Tmzonrzmz y’=f(z,y). Pronto. Cmrnaur. Chapter 11 t 3/2(1) =1/o+];of2l1i$1(1),?/1(1)] 611- I 7|"1611(1) =$0 f1[t:xn—l(():yn—1(t)]dt) t y..(1)=yo+ﬁofzltw-._i(1),y.._1(i)]dt- Picard’s method thus yields two sequences offunctions, 2:0,2:1,---, xn,---,and yo,1/1,---,y,,,---,each satisfying theappropriate initial condition in(57.31). Theexistence Theorem 62.12 which follows, willthen give theconditions which f,andf2of(57.3) must fulﬁll inorder that an interval Iabout toexist, onwhich, asn—>oo,theﬁrstsequence approach alimiting function :z:(t)andthesecond sequence approach alimiting func- tion y(t). This pair offunctions is,onI,theunique solution of(57.3) satisfying (57.31). Fordifferent permissible starting approximations, dif- ferent sequences willresult buteach willhave theproperty that onI, lim:c,,(t) =1(1) and limy,,(t) =y(t).ﬂ—f@ ﬂ—O@ Therapidity with which each sequence willconverge toitslimiting func- tion will depend onhow close thestarting approximations aretothe actual solutions. Example 57.34. Find theﬁrstthree Picard approximations iftheﬁrst order system is d d(a) £=t+:v, a%=t-—:z:, and10(0) =1,1/(O) =——1. Solution. Comparing (a)andtheinitial conditions with (57.3) and (57.31), weseethatf1(l,:v,y) =t+x,f2(l,x,y) =t—-1:,to=0,xo=1, yo=—1.By(57.32) ourﬁrstapproximations area:o(l) =1,yo(t) =—1. By(57.33), thesucceeding approximations are t I (11) 1-211(1) =1+/(1) f1(1,$o)I!0) dt=1+1‘) (1-l-1)d1 ,2 t l 3/1(1) ="'1+_/0 f2(1,$o-2/0)dt= -1-l‘/0 (1-' 1)dt ,2 =-1 +E—l. Lesson 57 PIc1tRn’s M1-zrnon 725 t t t2 2-112(1) =1+10 f1(1,$1,?/1)d1= 1+)’ (t+1 +5-l-1)!”0 ,3 =1+t+1’+'37! t 1/2(1) =""1+_/I) f2(1-$171,311)!” c —-—1+/‘(t—1—ﬁ-—t)dt_ 0 2 t3 --—l-—l——§,- Example 57.35. Find theﬁrst four Picard approximations iftheﬁrst order system is d d (3') E? =tyr % =xi‘/1 and:v(0) =1,1/(0) =1. Solution. Comparing (a)and theinitial conditions with (57.3) and we see 5119-1-f1(1i$-y) =if/rf2(trx:y) =xi/1 10=01x0 =1:I/0=1- By(57.32), ourﬁrstapproximations are:c0(t) =1,y0(t) =1.By(57.33), thesucceeding approximations are t t 2 (13) 1-211(1) ==1+/A f1(1»$o,yo) d1=1-l"/(1) tdi=1-l-‘5- t t 1/1(1) = 1 f2(t1x01y0) dt= 1 dt= 1'l"t- : t 2.:c2(t)=1+fo f1(t,a:1,y,)dt =1+L (1+t2)dt ,2 t3 =‘+5+5’I I 2 y2(1)=1+/f2(1,w1.y1)dt= 1+/(1+t+t§+§)dt0 0 ‘ 268’ t 3 _ 21“i‘1‘-23a(1)—-1+ 01-l-1-l-§+"6'+§ dt _ ,2 ,3 ,4 ,5 to —1"l"§-l'"§-1'?-l'3T)*l-T8» 726 EXISTENCE THEOREM! y’=f(z,y). PICARD. CLAIRAUT. Chapter 11 ‘ 4 5 0 1 113(1)=1+/;(1+1+¢”+¢”+%+‘;+fT{,+.,§q)-it :2:3l‘17:5 1“171’ as =1+‘+§+§+z+Ta+n+m+m' EXERCISE 57 Find theﬁrst IcPicard approximations after y0(:e), oftheparticular solution ofeach ofthefollowing equations 1-5,where Itisthenumber shown alongside each equation. $“tf‘§'°!°£"'@@@\\\ Q\§\1-2/-1/(0)=1,k=12+1/,1/(1)1+1/2,1/(0)1+M1,11(1)e‘+y,11(0) Paws? Papa*aaa Find theﬁrstkPicard approximations after x0(t), 3/0(1), oftheparticular solution ofeach ofthefollowing systems, where lcisthenumber shown alongside each system. dz6.E da:7.it dz:8.7‘ 9.dt2=t+y,g=t——:r,:1:(0)=2,y(0) =1,k=3. =t+y’,%-=1-¢,o(o)=o,y(o) =1,1t=s. =11% =2:—e',a:(0) =1,1/(0) =-1,1 =3. Picard’s method ofsuccessive approximations canalsobeapplied toasystem ofﬁrstorder equations greater than two. Forthesystem ofthree ﬁrstorder equations: dz:/dt =f1(t,x,y,z), dy/dt =f2(t,x,y,z), dz/dt =f3(t,a:,;i/,2) for which :z:(to) =2:0,y(t@) =yo,z(t0) =20,thesuccessive approximations are (57-4) 0-10(1) =Io, 110(1) =yo, 20(1) =Zo- rt 1-11(1) =I0+J‘f1(1,wo,1/0,10) <11.0 l y1(1) =110+itf2(1,1o,1lo/40) 111, fl 21(1) =20+J‘fa(¢,1o,i/o,1o)d1-U fl 2-112(1) =Ito-l-J‘ f1(1,f¢1,!l1i1,)d1.0 rl 112(1) =yo-l-J‘ f2(1-11,y1,Z1)d1,0 Lesson 57—Exe1-cise 727 r-I 12(3) =lo-i-J‘ f3(¢,11,7J1,11)d¢-0 ff n-r»(¢)=ro+J‘f1(t,=v,._1,y.._1,z1.-1) dt,0 rl 1/»(¢) =y0+J‘ f2(trx'\—l!yﬂ—lazn—-1) dt: 0 rl Zn“) =Z0+J‘ f3(t;xn—1;yn—1;zn—1) dt- 0 Use(57.4) toﬁndthePicard approximations 2:3,y3,23oftheparticular solution ofthesystem da: d dz Zt'=y2r %=x+z1 a=z'_y; forwhich :c(0) =1,y(0) =0,2(0) =1. ANSWERS 57 3 4 5 1_ =1.._ "’_£_ _x__L. 1“ ”+” 3+12 120 4917 25” ‘5 2-”==@+§”"%+%+%+€a"I2 I5 $8 Z11 =5+%+m+m" 7 5g0:3 2:5 2:64'-1l3=§6+$+§Z1+§+E+'éZ'3.ya 3 5.y4=4e’-—%—:c2—3a:—-4. 6.¢3=2+z—§:”—§:3—}z4—,16¢5—-I;-5:6, y3=1—4t—§t2+§t3+-ﬁt‘——§-}t5il—-11§t°—-1-ht’. :’t‘:7 :31‘2° 7-“3-‘+§+§+%§'”="‘+§-§+m' :2t‘s‘ :3:5 :' ,8-33—1+'5+§+Z§ry3—l+E'+Z6+ﬁ—8~ 43 4 5 2 3 4 2 3 9-13=1+§‘-+%+§,»/3 =2¢+‘;-%+‘§,z=».=1+¢—‘;-‘§- 728 Exrsrrzncn THEOREMZ y’=f(z,y). PICARD. CLAIRAUT. Chapter 11 LESSON 58. AnExistence and Uniqueness Theorem forthe First Order Differential Equation y’=f(x,y) Satisfying y(x0) =yo. Foraclearer understanding oftheproof oftheexistence anduniqueness Theorem 58.5 which follows, itwillbenecessary toknow themeaning of theconvergence ofasequence offunctions, themeaning oftheuniform convergence ofasequence offunctions, themeaning ofaLipschitz condi- tion, andtobeacquainted with certain theorems from analysis. Hence, before beginning theproof, weshall brieﬂy discuss these topics andlistthe needed theorems. LESSON 58A. Convergence and Uniform Convergence ofaSe- quence ofFunctions. Deﬁnition ofaContinuous Function. Deﬁnition 58.1. Asequence offunctions f1(x)1.f2(x);"'>f1l(x)1"'1 each deﬁned onacommon setSissaid toconverge toa.function f(z) onS,ifforeach :1:inSandforeach ﬁxed e>0,nomatter how small, there isanNsuch that (58.12) |f,,(x) -—f(z)} <e,when n>N. Inwords thedeﬁnition says thefollowing. Pick any:1:youwish inthis setSandchoose anypositive number eassmall asyoulike. Forthis:1: calculate f1(a:), f2(x), f3(a:), ---andf(z). Ifthesequence offunctions f1(x), f2(:z:), ---,f,,(a:), ---converges tof(z), then according tothedeﬁni- tion youmust eventually reach afunction fN+1(:c) inthesequence, such that |f1v+1(:c) —-f(:r)| islessthan this chosen positive e.Further every function inthesequence after fN+1(:v) must alsodiffer from f(ac)inabsolute value byanamount lessthan e. Example 58.13. Show that thesequence offunctions 1 1 1 (5)f1(fv)='1—_-,3’ f2(%)=f;'%* fa(1)=§§"", 1fn(x) ,--., converges tothefunction f(z) =0onI:0<:0§1. Solution. Let:1:beany number intheinterval I:0<2:§1,and letebeany positive number. Here f,,(ac) of(58.12) is1/(1—|—nx)and f(z) =0.Hence byDeﬁnition 58.1, wemust show that there exists an Nsuch that I1—_i_17L;—0i<e, whenn>N. Lesson 58A Convsnonncn. Umronu Convsnoancn. CONTINUITY. 729 Theinequality (b)isequivalent to 1 1 11(c) IE-<1+m:, ;——1<n:c, n>;(E--—l)- If,therefore, 11 @ ”=hh-M’ then (b)willhold when n>N.[NOTE. If1/e—1isnegative, take N=0.]Forexample, ifwechoose zc=Qand e=0.02, then by(d), N=2(50 -—1)=98.Itshould therefore follow that thevalue ofeach 1 1 function after the98th, namely, f99(§) = ,f1o0(:v) =1—l‘—_-%, --- should belessthan e=0.02. You caneasily verify that each such func- tionisindeed lessthan 0.02; remember, with aﬁxed numerator, thevalue ofafraction decreases asthedenominator increases. And ifwechoose 2:=1/100,then by(d),N=100(50 —1)=4900. Wewould therefore need togotothe490lth function inthesequence before coming totheﬁrst onewhose value differs from zero bylessthan e=0.02. And ifwepick a:=1/1,000,000, then by(d),N =1,000,000(50 —1)=49,000,000, i.e., weshall need togotothe49,000,00lst function inthesequence before coming totheﬁrst onewhose value differs from zero bylessthan 0.02. Itisevident from (d)andtheabove examples, that foreach ﬁxed posi- tivee,andforeach different ac,adifferent Nwillberequired inthesequence toinsure thevalidity of(58.12), i.e.,Ndepends onboth thevalues of xande.This dependency ofNonboth 2:andeiscommon toagreat many sequences. Ontheother hand, there arecertain sequences where Ndoes notdepend on:0butonly onthevalue of6,i.e.,foraﬁxed e>0,nomatter how small, itmay bepossible toﬁndanNsuch that (58.12) willbevalid forevery 2:inIwhen n>N.Inother words, wedonot,inthiscase, need tohunt foradifferent Nforeach different :c.WeﬁndanNonce andfor allwhich willhold forevery rcinI.Inthisevent, wesaythesequence of functions (58.11) converges uniformly onItothefunction f(z). Deﬁnition 58.14. Asequence offunctions (58.11), each deﬁned ona common setSissaid toconverge uniformly onStoafunction f(z), if foraﬁxed positive e,nomatter how small, there isanNsuch that (58.15) |f,,(:c) —f(:c)| <ewhen n>N, forevery :1:inS. Weemphasize once more, that here foraﬁxed e>0,wepick anNonly once. Foreach a:inS,theabsolute value ofthedifference ofeach ofthe functions f1v+1(:c), fN+2(x), ---andf(z) willbelessthan e. 730 Exrsrsncs Tmzonsmz y’=f(z,y). Promo. CLAIRAUT. Chapter ll Example 58.16. Show that thesequence offunctions <a>r.<x>=,—i—,;» r2<x>=%5» r.<x>=,—,§5~--. fn($)= ,..., converges uniformly tothefunction f(z) =0onI:1§re. Solution. Note thatthissequence offunctions isthesame asthatof Example 58.13, butthat theinterval isdifferent. Here wemust show, by Deﬁnition 58.14, that foragiven e>0, (b) ,1—;_1-E-0,<¢, whenn>N, forevery reinI:1§:1:<oo.By(c)oftheprevious example, 11(C) Tl>E —-I)' Foraﬁxed e>0andforanxinI:1§ac,theexpression ontheright of (c)willhave itslargest value when sv=1.And when x=1,(c)becomes 1n>2—1.Iftherefore (<1) 1v=E-1], then (b)willhold foreach 1:inIwhen n>N.Forexample, ife=0.02, then by(d),N=49. You canverify that each functional value inthe 1 1 sequence after the49th, namely, f50(1) =ﬁg) ,f51(1) =F51, --- islessthan 0.02. Hence forav>1,each functional value inthesequence after the49th issurely lessthan 0.02. Ourgiven sequence offunctions, therefore, converges uniformly onI:1§:c<wtothefunction ﬂw)=0. Deﬁnition 58.17. Afunction f(x) iscontinuous atapoint x=a iff(a) exists andif (58.171) limf(:c) =f('a).1—Hl Deﬁnition 58.172. Afunction f(x) iscontinuous on, orin,an interval Iifitiscontinuous atevery point ofI. Deﬁnition 58.18. Afunction f(x,y) iscontinuous atapoint (a,b), iff(a,b) exists and (58.18l) limbf(:c,y) =f(a,b).z.|/--m, Lesson 58B LIPSCHITZ Connmon. Tnaonsms rnom ANALYSIS. 731 Deﬁnition 58.19. Afunction f(x,y) iscontinuous on, orin,a region Sifitiscontinuous atevery point ofS. LESSON 58B. Lipschitz Condition. Theorems from Analysis. (The theorems from analysis which weshall need have been stated without proof. Their proofs canbefound inadvanced calculus text books.) Deﬁnition 58.2. Iff(z,y) isafunction ofasandyinaregion Ssuch that, forevery twopiints (x,y) and(z,y) inS, where Nisapositive constant, then f(:e,y) issaid tosatisfy aLipschitz condition inS(seeFig. 58.57). Theorem 58.22. Law ofthe mean. SeeFig. 58.57. Iff(z,y) isa function ofreandythathasacontinuous partial derivative with respect toy inaregion S,thenforeach 2:there exists anumber Ysuch that f(1,y) —f(I.?7) 658.23 —-—————-—-—— =— Y, <> y_y 6,f(w,> where (z,y) and(:c,y) areanytwopoints inSandYliesbetween yandy. 8 Remark. Thenotation 5/_f(x,Y)means thevalue ofthepartial deriv- ative ofthefunction f(:c,y) with respect toywhen y==Y. Comment 58.24. Iff(:c,y) hasacontinuous partial derivative with respect toyinaregion_S,_and ifthispartial derivative asafunction ofthe twovariables :c,yisbounded inS,then f(:c,y) satisﬁes aLipschitz condi- tion. The proof proceeds asfollows. Since 6f(:c,y)/By isbounded inS, there exists aconstant Nsuch that <a> I5’;raw]sN, forevery point (z,y) inS.Let(:c,y) and (z,y) beany twopoints inS. Then byTheorem 58.22 there exists anumber Ybetween yand17,suchthat Since (:e,Y)isapoint inS,wecansubstitute theinequality (a)fortheright sideof(b)andthus obtain (58.21). Theorem 58.25. Iff(z) isaRiemann-integrable function onI:a§ :0§b,then (58.26) bfa)dxlgb|f(t)|dt. 732 EXISTENCE T1-woman: y’=f(z,y). Prcann. CLAIR-AUT. Chapter ll Theorem 58.3. Iff(z) isacontinuous function ontheinterval I: a§:2:§bandif (58.31) F(a:) =f f(t)dt, a§:c§b, then F(:t) iscontinuous onI,and (58.32) F'(:z:) =f(:c), a<rt<b. Theorem 58.4. Ifasequence ofcontinuous functions f1(x),f2(x), ---, f,,(:c), ---.each deﬁned onacommon interval I,converges uniformly on I,toafunction f(av),thenf(z) iscontinuous onI. Comment 58.41. The preceding theorem isnottrue if,onI,the sequence offunctions merely converges tof(z), butnotuniformly. For instance, thesequence ofcontinuous functions ofExample 58.13 converges onI:0§:0§1,butnotuniformly. The sequence converges onO< x§1tothefunction f(z) =0.But when :0=0,each function inthe sequence hasthevalue one. Hence ontheinterval I:0§:1:§1,the limiting function ofthesequence is <a> /<1»)=3;3?1' Thefunction f(z) istherefore discontinuous onI. Theorem 58.42. Ifasequence ofcontinuous functions fl(re),f2(x), ---, f,,(:e), ---,each deﬁned onacommon interval I,converges uniformly on Itoafunction f(z), then I I I (58.43) limff,,(x) da:=/llimf,,(:v) dx=/f(z)dz, where theinterval (:c0,:c) iscontained inI. Deﬁnition 58.44. Aseries offunctions f.(w)+f2(w) +---+f..(w) +---. each deﬁned onacommon interval I,issaidtoconverge uniformly onI toafunction f(z), ifthesequence ofpartial sums F1(:e), F2(a:), ---,F,,(:c), where F»(w) =f1(@) +'~~+f..(w). converges uniformly onItof(:c). Theorem 58.45. Ifeachfunction f1(:e), f2(x), ---,f,,(:c) isdeﬁned and bounded onacommon interval I,i.e.,if éMir 1r2!"'!nr"'r Lesson 58C Pnoor orExrsrrncs Tnaomm: y’=f(z,y) 733 andiftheinﬁnite series ofpositive terms (58.47) M1+M2+---+M,,+--- converges, then theseries (58-43) f1(w) +f2(iI) +~--+f»(w) +-~- converges uniformly onItoafunction f(t). Comment 58.481. Bytheabove theorem, ifeach |f,(a:)| éM,-,and EM, converges, then foragiven e>0,there isapositive Nsuch that lf1($) +f2(r) +'''+f»(I) —-f(I)| <6,when TI>N. forevery :0inI. Example 58.49. Show that theseries offunctions, °° 1o) ZiIr.<x>=z§§,_ﬁ. 1.-ogrél. converges uniformly onI. Solution. Take M,=1/2*’. Therefore I'M“F =M.+M.+M.+---=§+§+§+--- By(a),andforeach 1:such that 0§:c§1, Q 1 1 1 <°> ,=Z,f"(")='ﬁ'i+FF._@+2"3??+"'1 1 1 §§+§+§§+"', Thelastseries ontheright of(c)isageometric series which converges to one. Hence theﬁrstseries offunctions, which isthegiven series (a),by Theorem 58.45, converges uniformly onItoafunction f(z). LESSON 58C. Proof ofthe Existence and Uniqueness Theorem fortheFirst Order Differential Equation y’=f(x,y). InTheorem 38.14, wegave asuﬂicient condition fortheexistence ofapower series solution ofy’=f(z,y) satisfying aninitial condition y(:c0) =yo. Com- pare itwith thetheorem wenow state, which gives asufficient condition fortheexistence and uniqueness ofasolution ofy’=f(:e,y) forwhich 1/(W0) =1/0- 734- EXISTENCE Tnsossmz y’=f(z,y). PICARD. CLAIRAUT. Chapter 11 Theorem 58.5. SeeFig. 58.57. Letf(:c,y) beabounded, continuous function ofavandyinaregion Softhexyplane andlet(a:0,yo) beapoint ofS.InS,letthefunction fsatisfy theI/ipschitz condition (58.21), namely (58-51) |f(@,y) —f(w,17)| §NI?!-9|, forevery twopoints (z,y) and(ac,y) inS. Then aninterval (58.52) Io:It—-:co|<h,‘h>0, exists onwhich there isoneandonly onecontinuous function y(z), with a continuous derivative onI0,satisfying thediﬁerential equation (53-53) y’==f(r.y) andtheinitial condition (58-54) 3/($0) =yo- Remark. The above theorem isaspecial case ofthemore general Theorem 62.12 onsystems ofﬁrstorder equations. Proof. Theproof ofthetheorem willbebased onPicard’s methods of successive approximations. By(57.23) and (57.24), these approximating functions are (58-55) m(m)=vo, , I/i(1'7) =yo+/; fit’!/0(5)] dt, 1/2(w)=yo+Lfltz/1(1)] dl. .---.-----.------00 Z l/n($) =l/0'l"/;ofltrl/n-l(t)l dtr where f(x,y) isthecontinuous function of(58.53), yoistheconstant of (58.54) andtheinterval (:c0,x) iscontained inS.Because theproof is long, wehave divided itintofourparts. Norm. Weshall useI:Ix—xol §htodenote theclosed interval (58.52). A.First weshall show how todetermine theinterval I0of(58.52). B.Second weshall prove that onIeach ofthefunctions y0(a:), y1(:c), ---,y,,(x) of(58.55) iscontinuous anditsgraph liesinarectangle R contained inthegiven region S. C.Third weshall prove that thissequence offunctions yo,y1,---,y,,of (58.55) converges uniformly onItoafunction y(z) which, onI0,isa solution of(58.53) satisfying (58.54). Wethus establish theexistence ofaparticular solution y(z). Lesson 58C Pnoor orEXISTENCE Tmaonnmz y’=f(z,y) 735 D.Finally weshall prove that thissolution y(z) is,onI0,theunique par- ticular solution of(58.53) satisfying (58.54). A. Determining aninterval I.Since thegiven function f(z,y) is,by hypothesis, bounded inS,there exists apositive constant Msuch that (53-56) If(1.1/)| <M, forevery point (z,y) inS.The point (x0,y0) of(58.54) isbyhypothesis a point ofaregion S.(See Fig. 58.57 andkeep referring toit.)Hence we my) |/em-r<x.;v>| ly-5/I I:|x—xo|§h /W Xv X +8 $3 Figure 58.57 canﬁnd apositive number hsuch that therectangle ofdimensions [re—:e0|§h,Iy—3/0|<Mhwhich hasthispoint (:e°,y°) atitscenter, andwhere Mistheconstant in(58.56), liesentirely inS.Callthisrectangle R.Therefore allpoints inR(since they arealsoinS)whose recoordinates areintheinterval (58.58) I:a:0—h§a:§a:o+h, |a:—x0| gh, satisfy’ (58.51) and (58.56). This interval I,without itsendpoints, isthe interval I0of(58.52) referred tointhetheorem. Proof ofB.Each function y0(a:), y,(:z:), ---,y,,(x) of(58.55) is,onthe interval Iof(58.68), continuous anditsgraph liesinR.Consider theﬁrst 736 EXISTENCE THEOREM! y’=f(z,y). PICARD. CLAIRAUT. Chapter 11 function yo(x) =yo. Itsgraph isastraight lineparallel tothe:1:axis andyounits distant from it.Hence forall1:inI,thegraph ofy0(a:) =yo liesinR(keep referring toFig.58.57). Tocomplete theproof, weusethe inductive method ofreasoning described inLesson 24A. Weassume the graph ofthefunction y,,_1(z) liesinRandmust then show that thegraph ofthefunction y,,(:e) isalsoinR. Hence weassume that forallreinIof(58.58), thegraph ofy,,_1(:z:) lies inR.This means weassume that foreach 1:inI,y,,_1(:e) willgive avalue ofywhich isinR.Therefore [:c,y,,_1(:v)] isapoint ofR.Hence by(58.56) (=1) |flw,y»_1($)l| <M- Bythelastequation of(58.55), by(58.26), (a)and(58.58), intheorder listed, weobtain I 1 (bl l?/»(1¢) —yol= toM/n-1(i)ld¢ § z0l.flt1yn--1(t)lldti < =M:z:-—x0|§Mh. Equation (b)says that foreach :0inI,thedistance from y,,(x) toyo< Mh(seeFig.58.57). Therefore thegraph ofy,,(:v) liesinR. Wehave thus proved thatforeach reinIof(58.58), [:e,y;,(:c)], k=0, 1,2,---,n,---isapoint ofRcontained inS.Hence, bythehypothesis ofthetheorem, each integrand f[x,y;,(:c)], k=0,1,2,---,n,---of (58.55) is,inR,andtherefore onI,acontinuous function. Ittherefore follows byTheorem 58.3, that each function y1(:e), --~,y,,(x), ---of (58.55) isalsoacontinuous function onI.And since y0(a:) ==yo,acon- stant, ittooisacontinuous function onI. Proof ofC. There exists aparticular solution of(68.53) satisfying (68.64). Bythesecond equation in(58.55) andTheorem 58.25, (0) ly1(rv) —yo(1=)| = f[t,yo(t)1dl é Ifltz/o(l)]| dl- InB,weproved thatforeach 1:inIof(58.58), thepoint [x,y0(x)] isinR. Therefore by(58.56) and(58.58), (c)becomes (<1) ll/1($) '"I/o($)l <M =Ml?‘ '"$0l§Mk- Subtracting thesecond equation of(58.55) from thethird, Weobtain (6) I112—y1|=_/to{fl¢,y1(l)] -fit!/o(i)l} dll § ‘/gt lfltrl/1(t)] _fit/J0(t)ll dt‘ ' Lesson 58C Pnoor orExrsrmrcs Tnnonsmz y’=f(z,y) 737 Foreach :2:inI,both points [:e,y1(x)] and [x,y0(a:)] are, byB,inR. Hence by(58.51), wecanwrite (e)as (fl ll/2—I!1|§N I/1(3) —1/o(i)| ‘ill' Replacing theintegrand in(f)byitsvalue MIt—2:0]asgiven in(d),we obtain with thehelp of(58.58), z |x—:c|2 h2(E) ly2_3l1l<MN_/:8 lt_5°0ldt =MN'#)—§MN§'U Repeating theabove procedure, weﬁnd, with thehelp of(58.55), (58.51), (g),and(58.58), intheorder listed, <11)|?/a~ya§0|ru.y2<¢>1 -ru.y.<»>1|d¢| §Nf |y2(l) —y1(i)| ‘ill <MN2 —-_”° t=MN”’_ 2L“.<MN 3, And ingeneral, itcanbeshown bythesame type ofinductive argument weused inB,that forevery :1:inIandevery n, <1) |y..<w>—1/.._.<w>| <MN""‘ n—1L"_M(Nh)"_ <MN n!_N n! By(i) . M(1) ll/1—I/0|<—N g(Nh)2’ N2!Ml1:’ ll/2*yil< M(Nh)" |!/1»"1/n-1| <F'7' Thesum ofyoandthepositive terms ontheright sideof(j)is MNh (Nh)2 (Nh)"y0+N‘[i'+"T+"'+T+"'j|Y 738 EXISTENCE THEOREMZ y’=f(z,y). Prcann. CLAIRAUT. Chapter ll _ M aseries which by(37.42), converges toyo—|——(eN" —1).Hence, by’ N Theorem 58.45, theseries (kl 2/0+(U1—1/0)+(1/2—Z/1)+''~+(I4/n"I4/n_1)» which isthesum ofyoandthefunctions ontheleftsideof(j),converges uniformly onItoafunction y(z). Butthesum (k)isy,,.Wehave thus proved that ontheinterval Iof(58.58), (1) 1/..(r)=y(z) uniformly, i.e., thesequence ofcontinuous functions yo(x), y1(x), --~, y,,(r), deﬁned in(58.55), converges uniformly onItoafunction y(ac). By Theorem 58.4, therefore, thisfunction y(z) iscontinuous onI.Moreover, itfollows byDeﬁnition 58.14 since thegraph ofy,,(:e) isinR,that the graph ofy(z) must alsobeinR. Since forevery a:inI,[x,y(x)] and[x,y,,(ac)] arepoints ofR,wehave, by(58.51), (m) lflrm/»(r)] —f[w.y(w)l| éNll/n($) —y(r)|- By(l),thesequence y,,(z) ——>y(z)uniformly onI.Therefore, byDeﬁni- tion 58.14, there exists anindex Psuch that |y,,(x) —y(a:)| <e/N, when n>Pforevery atinI.Hence, by(m), |f[r,y»(w)] —f[r,y(r)]| <E.whenn>P, forevery zeinI.Therefore, byDeﬁnition 58.14, thesequence ofcon- tinuous functions f[:e,y0(:e)], f[:2:,y1(x)], ---,f[x,y,,(x)], ---,converges uni- formly onItof[a:,y(:v)], i.e., 7lliif;flI,yn(1)] =f[@v,y($)l uniformly. Therefore, byTheorem 58.4, f[a:,y(x)] isacontinuous function onI. Hence by(1),thelastequation of(58.55), Theorem 58.42, andtheequa- tionimmediately above, intheorder listed, (I1) 1/(I)=lim1/11(1) =1/0+limffltl/n-1(t)] dl =1/0+! limflly» 1(l)]dl,_ :0n—w0 1 =1/o+fzoflm/(t)] dt- Lesson 58C Pnoor orEXISTENCE Tnsorusu: y’=f(z,y) 739 Weshowed above thatf[a:,y(:::)] isacontinuous function onI:Ix—Ivo|éh. Ittherefore follows by(n)andTheorem 58.3, that y'(w) =flw,y(w)l, Io-'Ir~-'l3o|<h- By(n),weseealsothat=0 1/(re)=3/0+Lflm/(Old! =U0+0=yo-O Wehave thus proved theexistence onI0ofaparticular solution of (58.53) satisfying (58.54). Itisthelimiting function y(z) of(1).Wemust stillprove that thisparticular solution y(z) isunique. Proof ofD. Thefunction y(z) of(I)istheunique particular solution of (68.63) satisfying (58.64). Assume g(2:) isanother particular solution of (58.53) satisfying (58.54). Therefore weassume (0) Q'(w)=f[w,9(r)l, forwhich g(:eo) =yo.Byintegrating (0)andinserting theinitial condi- tions g(:eo) =yo,weobtain Z 0(w)=yo+/;of[ta(t)]dt~ Subtracting (n)from theequation immediately above, wehave I (P) |9(w)-1/(w)I =/;{f[l,9(t)l -—f[l,y(i)l} dil éf|f[t,9(i)l —fli,y(l)l| dll' Weproved inB,that forevery xinIof(58.58), thegraph ofeach function yo,yl,---,y,,of(58.55) liesinarectangle Rcontained inS.Weproved inCthatthissequence offunctions converges uniformly onI,toacontinu- ousfunction y(z) whose graph also liesinR.Similarly, itcanbeshown that g(x) isacontinuous function onIwhose graph liesinarectangle contained inS.Therefore foreach reinIof(58.58), thepoints [x,y(x)] and[x,g(:e)] areinS.Wemay therefore apply (58.51) tothelastexpression ontheright sideof(p). Wethus obtain (q) l.<J(w)—y(t)!éNA|(y(t)—y(i))|dil- Forevery a:inI,thegraphs ofg(x) andy(z) areinbounded rectangles contained inS.Call Dthemaximum value of|g(:e) —y(x)| for:1:inI. 740 EXISTENCE Tursonsm: y’=f(z,y). Promo. CLAIRAUT. Cha terll P Then by(q) (I) |9(w)—y($)l §DN dl=DN|$ —wol- Substituting (r)fortheintegrand in(q),weobtain ___ 2 <5)we)-yousmv’ »-ssdtl=1>N”'iT""'- Again substituting (s)fortheintegrand in(q),weobtain ‘__ 2 _ a (o|g<w>-yen §DN3/lei =1>N“E‘-#- .,,2. 3. Continuing inthismanner, weﬁnd, with thehelp of(58.58), _ fl ﬂhﬂ Nhﬂ |g<x>-1/e>| sDN"'1n,”°'§DZ,=n(n} - Inthelastterm of(u),Disaconstant and (Nh)"/n! is,by(37.42), the general term oftheseries expansion ofem‘which converges forallNh. Itsnthterm, therefore, approaches zero asn—>oo.Hence D(Nh)"/nl approaches zeroasnapproaches oo.Since |g(x) ——y(z)] canbemade lessthan anynumber nomatter howsmall, itfollows that (v) 9(1)-y(w)=0, Q(x)=y(w)- Wehave thus proved ﬁnally that theparticular solution y(z) of(1)is unique. Theassumed second solution g(2:) isthesame asy(z). Comment 58,59. Itisnow evident why inPicard’s method theﬁrst starting approximation y0(:c) need notbetheliney0(a:) =yo,where yois theinitial condition given in(58.54). Itcanbeanycontinuous function through thepoint (x°,yo) whose graph isinR.Each such starting function willdetermine asequence offunctions whose limiting function isapar- ticular solution ofthegiven equation satisfying thegiven initial condition, valid ontheinterval Iof(58.52). Bythetheorem there canbeonlyone such particular solution. Comment 58.6. Let2:0beaﬁxed value of2:.Then each point (:z:0,c) ofS,where cisanarbitrary constant, determines aunique solution y(z) of(58.53). Hence thesolution y(z) isafunction notonly of2:butalsoof theordinate c,i.e.,y=Y(:e,c). There istherefore a1-parameter family of solutions of»(58.53) cutting across each line:0=2:0. Comment 58.61. Theorem 58.5 gives only asuﬁicimt condition for theexistence anduniqueness ofaparticular solution ofadifferential equa- tion y’=f(:e,y) forwhich y(:c,)) =yo. Itisnotanecessary condition. Lesson 58C Pnoor orExxsrsncs Tneonnm: y’=f(z,y) 741 Thesuﬁicient condition means that ifaregion Scontains only points which fulﬁll (58.51) and(58.56), then theconclusion ofthetheorem must follow, i.e.,each point ofSliesononeandonly oneparticular solution of(58.53). That thecondition isnotnecessary implies itsconclusion may stillbetrue forpoints inSwhich donotfulﬁll (58.51) and (58.56), i.e.,these points may stilllieononeandonly oneparticular solution. Comment 58.62. Theorem 58.5 serves another useful purpose. By Deﬁnition 5.4,anordinary point ofaﬁrst order equation y=f(:c,y) lies ononeandonly oneintegral curve. Hence every point (a:°,yo) ofaregion S that fulﬁlls (58.51) and (58.56) must beanordinary point since, bythe theorem, each such point liesononeandonly oneintegral curve. Assume that Scontains only ordinary points and that weareable towrite a 1-parameter family ofsolutions ofadifferential equation explicitly orim- plicitly interms ofelementary functions. Ifforeach point (:to,y0) ofS, the1-parameter family yields aparticular solution through it—and bythe theorem there isonly onesuch solution——then thisfamily, byDeﬁnition 4.7, must beageneral solution since itcontains every particular solution ofthe differential equation. Inthisspecial case, therefore, thetheorem makes usaware whether ourn-parameter family ofsolutions isageneral oneor not. Example 58.63. Obtain four Picard approximations if (a) y’=1+2/2 andy(0) =0.Find aninterval forwhich thesequence ofPicard approxi- mations willconverge totheactual solution. Solution. Following themethod outlined inLesson 57B, weﬁnd o)me=a, m@=LM=% ’ a y2(w>=/0 <1+w2>dw=w+%, +Q»iQ»-’ $32 xs 7 y3(:c)= 01+ :z:+§ dx=:c—|-§ —x. Here f(z,y) ofTheorem 58.5 is1+y2and =2y.Therefore inany 1'1 bounded region S,nomatter how large, f(x,y) iscontinuous andsatisﬁes (58.56). Itspartial derivative isalsocontinuous andbounded inS.Hence, byComment 58.24, f(x,y) satisﬁes (58.51). Letustake forStheregion -5<y<5(seeFig.58.64). Then theMof(58.56) isfound asfollows. <o |mwH=H+yW<1+%=16=M. 742 EXISTENCE THEOREMZ y’=f(z,y). Prcano. CLAIRAUT. Chapter ll Wemust now choose h,according toAofthetheorem, sothat therec- tangle Rofdimensions (here 1:0=Oand yo=0) §h,|y|<26h which hasthepoint (0,0) atitscenter liesinS.Remember Sisnow bounded bythelines y=5and y=-5. Iftherefore wechoose h= I:|x|‘0.19 y=5 ii S .=-_'—_'€'-' E X it-W S 2%“ R y=-5 Y Figure 58.64-wig‘ 0.19, then therectangle Rofdimensions Ixl§0.19, |y|<4.94 which has (0,0) atitscenter willlieinS.Theinterval Iof(58.58) andtherectangle Rareshown inFig.58.64. Hence thesequence offunctions yo,yl,---,y,, willconverge ontheinterval -0.19 <at<0.19, toafunction y(z)which, onthisinterval, istheactual solution of(a)satisfying y(0) =0. Comment 58.65. Theactual solution of(a)satisfying y(0) =0is (d) y=tanac. ItsMaclaurin series is 3 5 1 9 (e) mnx=e+3‘,-+?,3’,;+137%+%5+---. which converges for|x|<1r/2. Acomparison of(e)with y3(:v) of(b) shows thattheﬁrstthree terms ofeach series arethesame. Ifwehad obtained additional Picard approximations, subsequent functions inthe sequence would contain more andmore terms inagreement with (e)and would approach theactual solution tanzc.Note, however, how much smaller theinterval I,obtained from thetheorem, isthan theinterval (—1r/2, 1r/2) forwhich theseries (e)isvalid. Intheabsence ofasolution, wewould know bythetheorem only, thatfor <0.19, thesequence of approximating functions y0(x), y1(a:), ---,y,,(:c) converges toafunction y(z)which istheparticular solution of(a)satisfying y(0) =0.Wewould thus know, forexample, that y,.(0.l) isanapproximation totan(0.1). However, wecould not know from the theorem whether y0(0.5), Lesson 58—Exercise 743 y1(O.5), ---,y,,(0.5) converges totheactual solution tan(0.5) unless we could ﬁndalarger h.Thetheorem therefore does notalways giveusamaxi- mum interval ofconvergence. EXERCISE 58 1.Show that thesequence offunctions 2:,2:2,2:3,---,2:"converges onI: 0§1<1,tothefunction f(z) =0,butnotuniformly. What isthelimiting function ofthissequence ifIistheinterval 0§2:§1?Isthelimiting function continuous? 22.Show thatthesequence offunctions -3-— 1i ,---,i converges1+a:1+2:4: 1+m: onI:0<2:<wtothefunction f(z) =1,butnotuniformly. What isthe limiting function ofthissequence ifIistheinterval 0§2:<0°?Isthis limiting function continuous? 3.Show that thesequence offunctions 1/2, 1/22:, 1/32:, ---,1/nz converges uniformly onI:1§2:<=11tothecontinuous function f(z) =0. 4.Show thattheseries x2 + 2:4 _ 2:2 _,_ mu _ 2:4 +___ 1+1? 1+a:4 1+1? 1+1“ 1—|-2:4 ’ whose sequence ofpartial sums is $2 $4 (x2)n 7‘'‘7 ) 1+12’1+¢4 1+(z2)" converges tothefunction /<1)=0,I1!<1. =5» =1; =1, >1. 5.Incomment 58.59, westated that theﬁrstPicard approximation need not bethefunction y0(:c) =yo,where yoisthegiven initial value. InExample 58.63, ourﬁrstapproximation wasy0(:t) =0.Show that y0(:t) =:cwould have been abetter ﬁrstapproximation. Solve theproblem starting with this newﬁrstapproximation. 6.Solve theproblem inExercise 57,1, starting with theapproximation y0(:r) = 1—z.See5above. Isitabetter orpoorer approximation? 7.Find aninterval ofconvergence foreach sequence ofPicard approximations obtained inExercise 57,1, Exercise 57,2, Exercise 57,5. ANSWERS 58 1.f(z) =0,0§:2:<1;f(z) =1,a:=1.Limiting function isdiscontinuous. 2.f(z) =1,0<2:<01>;f(:t) =0,2:=0.Limiting function isdiscontinuous. 6.Better sapprolrimatign. Actual solution isy=2e-1 +rt—-1=1—:2:+ 2__£L_L...” s+12so ' 744 EXIST]-INCE Tnaonamz y’=f(z,y). Proxan. CLAIRAUT. Chapter 11 7.(1)If(:c,y)| §III—I—IyI<M. Any values of2:andyarepermissible for region S.Choose hsothatrectangle ofdimensions I:cI§h,Iy-1|<Mh which has(0,1) atitscenter liesinS.Theinterval is <h. (2)If(x,y)I §Iz2I+IyI<M.Any values of2:andyarepermissible for region S.Choose hsothatrectangle ofdimensions Ix-—1|§h,Iy-—3|< Mhwhich has(1,3) atitscenter liesinS.Theinterval isIx—1I<h. LESSON 59. The Ordinary and Singular Points ofaFirst Order Differential Equation y'=f(x,y). InLesson 5,weintroduced theconcept ofanordinary point andofa singular point ofaﬁrstorder equation (59-1) v’=f(w.v)- Forconvenience werepeat these Deﬁnitions 5.4and5.41. Deﬁnition 59.11. Anordinary point ofaﬁrstorder differential equa- tion (59.1) isapoint intheplane which liesononeandonly onemember ofa1-parameter family ofsolutions, i.e.,itliesononeandonly oneintegral curve. Deﬁnition 59.12. Asingular point oftheﬁrst order differential equa- tion (59.1) isapoint intheplane which meets thefollowing tworequire- ments. 1.Itliesonnone ormore than oneintegral curve of(59.1). 2.Ifacircle ofarbitrarily small radius isdrawn about thispoint, there is atleast oneordinary point initsinterior. Comment 59.13. ByDeﬁnition 59.11 andComment 58.61, allpoints (z,y) ofaregion S,inwhich f(z,y) satisﬁes (58.51) and(58.56), areordinary points of(59.1). They lieononeandonly oneintegral curve ofitsfamily ofsolutions. Iftheregion Scontains points which donotfulﬁll (58.51) or(58.56). then byDeﬁnition 59.12 andComment 58.61, they maybesingu- lar. Hence inhunting forsingular points, weneed only examine those which failtofulﬁll (58.51) or(58.56). There iscertainly noneed tolook forthem among those which do. Example 59.2. Determine whether thedifferential equation (e) y’=21 hassingular points. Solution. Here f(z,y) ofTheorem 58.5 is2xand é)f(:c,y)6y =0. Therefore inany bounded region S,f(a:,y) iscontinuous and satisﬁes (58.56). Itspartial derivative with respect toyisalso continuous and bounded inS.Therefore byComment 58.24, f(x,y) satisﬁes (58.51). Hence, byComment 59.13, each point ofSisanordinary point. There arenosingular points. Lesson 59 Onnmxnr ANDSINGULAR POINTS ory’=f(z,y) 745 Remark. The solution of(a)is ll=$2+ca which isafamily ofparabolas with itsvertices ontheyaxis. Foreach point (:r0,yo) intheplane, (b)willgive anunique particular solution of(a) satisfying thepoint (:eo,yo). Hence, byComment 58.62, wenow alsoknow that (b)isageneral solution of(a). Example 59.21. Determine whether thedifferential equation (a) 1/’=—\/1—v’, —1§v§1. hassingular points. Solution. Since \/1—y2isdeﬁned only forvalues ofyforwhich —1§y§1,weconﬁne ourattention toaregion which iscontained between these lines. Here f(z,y) ofTheorem 58.5 is—\/1 —y?and 6f(:e,y)/6y =y/\/1 —-yz. Let ustake forStheregion deﬁned by —-1<y<1, <A,where Aisapositive number, Fig. 59.22. Let Y y=-1 x: _A '(x01y0) x=A (0.0) X S ;v=—1 Figure 59.22 (:co,yo) beapoint ofS.Since itisaninterior point ofS,itcanbesur- rounded byarectangle contained inS.Inthisrectangle, f(x,y) iscon- tinuous andsatisﬁes (58.56). Itspartial derivative with respect toyis alsocontinuous andbounded. Therefore, byComment 58.24, f(z,y) satis- ﬁes(58.51). Hence each point inthisrectangle, byComment 59.13, isan ordinary point. Since each point ofScanbesimilarly surrounded bya rectangle, allpoints ofSareordinary points. Ifweenlarge Stoinclude thelines y=1andy=—1,then forthese values ofy,(58.51) isnotsatisﬁed. [In(58.51), take y=1.Then as y—>1,I(—\/1 —yz+0)/(y —1)I-—>oo.] Hence thepoints onthese lines may, byComment 59.13, besingular points of(a). That they are infactsingular points canbeseen from thesolutions of(a). These are (b) y=cos(as—I-e) 74-6 Exrsrsncrz Tnaoaauz y’=f(z,y). PICARD. CLAIRAUT. Chapter 11 where cistheusual arbitrary constant, andthefunctions (c) y=:l;1. Each point ontheliney=1,i.e.,each point (:c0,1) liesontheintegral curves y=cos(2:—mo)andy=1;each point ontheliney=—1,i.e., each point (:e0,——1), liesonthecurves y=cos(:1:+1r—:00), and y= -1. Forexample, thepoint (0,1) liesontheintegral curves y=cos:1: andy=1,thepoint (—2,1) liesontheintegral curves y=cos(:1:—|—2) andy=1,thepoint (3,—1)liesontheintegral curves y=cos(re+1r—3) andy=——1. Hence, byDeﬁnition 59.12, thepoints onthelines y=1 and_y =—1aresingular points. Comment 59.23. Iftheregion Sintheabove example excludes the lines y==l=1, then each point ofS,asremarked above isanordinary point. Hence byDeﬁnition 59.11, there isoneand only oneparticular solution through each point ofS.Since foreach point ofSthefamily (b) yields aparticular solution through it,wenow know, byComment 58.62, that the1-parameter family (b)is,inS,ageneral solution of(a). If,how- ever, Sincludes thelines y=:l:l,then intheabsence offurther informa- tion, wecannot know whether (b)isageneral solution of(a),i.e.,whether itcontains every particular solution. Forsince (x,=l=1) may besingular points, there may ormay not,byDeﬁnition 59.12, besolutions through these points. Actually, aswesawabove, y==!=1areparticular solutions of(a)notobtainable from thefamily (b). Hence forthisenlarged region S, (b)isnotthegeneral solution of(a). Example 59.24. Determine whether thedifferential equation (a) y’=(v+1)’/v. y#0. hassingular points. Solution. Here f(z,y) ofTheorem 58.5is(y—I-1)2/y and8f(z,y)/0y = (y2-—1)/y2. Letustake forStheregion deﬁned by—B §y<0, IxI<A,where A,andB>1,arepositive numbers, Fig.59.241. Let (x0,y0) beapoint ofS.Since itisaninterior point ofS,itcanbesur- rounded byarectangle contained inS.Inthisrectangle f(a:,y) iscontinu- ousandsatisﬁes (58.56). Itspartial derivative with respect toyisalso continuous and bounded. Therefore, byComment 58.24, f(a:,y) satis- ﬁes(58.51). Hence, each point inthisrectangle, byComment 59.13, isan ordinary point. Since each point ofScanbesimilarly surrounded bya rectangle, allpoints ofSareordinary points. Therefore, byDeﬁnition 59.11, each point inSliesononeandonly oneintegral curve of(a). The n-parameter family ofsolutions of(a)is (b) yf1,+10sIv+1|=w+¢.v¢—1~ Lesson 60 Esvamras 747 Wenow note that (b)excludes particular solutions of(a)that lieonpoints whose ycoordinate is—l. Hence wenow also know that (b)cannot be thegeneral solution of(a)andthat there must beanadditional solution (orsolutions) of(a)that lieonthepoints (:e,— 1).This particular solution (0.0)y=0/' X / ‘(xo»}‘0)/ x=—A\ H / -- / \x=A / sy=—1 y=—B Y Figure 59.241 is,infact, theliney+1=0.This solution together with (b)contain every solution of(a). There arenoothers. Note theroleplayed bythe existence theorem inwarning usthat (b)cannot bethegeneral solution of(a). EXERCISE 59 Determine thesingular points, ifthere areany, ofeach ofthefollowing equations. l.y’=1—y. 2-12.y'=-%»1#0. 3-1/1/’=\/1-2/2. Ir/|§1.v#0~ 4-1/'=\/§,1/;0- 5.y'=;:—i—y» 1—y#0. ANSWERS 59 1.None. 2.None ineach halfplane 1>0,1<0. 3.Lines y=;l;l. 4.Line y=0. 5.None ineach ofthetworegions divided bytheline1—y=0. LESSON 60. Envelopes. Inthis text, wehave described various techniques forﬁnding a1- parameter family ofsolutions f(:v,y,c) =O,forspecial types ofﬁrst order differential equations oftheform F(:v,y, ==0.Ifthis 1-parameter1 748 EXISTENCE Tnaonam: y’=f(1,y). PICARD. CLAIRAUT. Chapter 11 family ofsolutions isnotageneral one, then theproblem ofﬁnding par- ticular solutions, ifthere areany, that arenotobtainable from thefamily isusually acomplex anddiﬁicult one. However, there isonecase where astandard method exists forﬁnding such particular solutions. Themethod weshall describe isclosely associated with thenotion ofan“envolope” of afamily ofcurves. Hence weshall ﬁrstdiscuss themeaning ofanenvelope ofafamily ofcurves. LESSON 60A. Envelopes ofaFamily ofCurves. Consider the family ofcircles (1—c)2+y2=1,with centers at(c,0) andradius 1, Fig. 60.01. Each circle istangent tothelines y=;l=1. Since these two 0) Figure 60.01/\><'><><><“>6?><><\/ lines may belooked atasenclosing thefamily ofcircles, and since the word envelope means awrapper forenclosing something, each hasbeen called anenvelope ofthegiven family ofcircles. More generally, wede- ﬁnetheenvelope ofafamily ofcurves asfollows. Deﬁnition 60.1. Letf(1,y,c) =0beagiven family ofcurves, Fig. 60.2. Acurve Cwillbecalled anenvelope ofthefamily ifthefollowing twoproperties hold. 1.Ateach point oftheenvelope there isaunique member ofthefamily tangent toit. Envelope curve C‘ Members offamily f(x,y,c)=0 Figure 60.2 Lesson 60A ENVELOPES orAFAMILY orCunvas 749 2.Every member ofthefamily istangent totheenvelope atadistinct point oftheenvelope. Deﬁnition 60.21. Afunction f(1,y,c) issaidtobetwice differentiable ifallitsﬁrst andsecond partial derivatives exist. Wenowstate without proof *atheorem which gives asuﬁicient condition fortheexistence ofanenvelope ofafamily f(ac,y,c) =0. Theorem 60.22. Iff(:e,y,c) isatwice differentiable function deﬁned for asetofvalues of:c,y,c, andif,forthissetofvalues, <60-23> rate)=0. =0.and (60.24) g :—‘yf 82]. ¢0, 5-53¢0, 621‘if0:c8c 8y6c then thefamily ofcurves f(:z:,y,c) =0hasanenvelope whose parametric equations aregiven by(60.28). Assume that f(1,y,c) isafunction which satisﬁes thehypothesis of Theorem 60.22 sothat wecanbesure thefamily ofcurves (60.25) f(:c,y,c) =0 hasanenvelope. Inplace ofthedifficult proof oftheabove theorem which wehave omitted, weshall now describe aformal method ofﬁnding thisenvelope. Atthesame time, weshall discover how theequations in (60.23) originated. Noattempt willbemade tojustify eachsteprigorously. Foreach value ofc,weobtain amember ofthefamily (60.25). Hence if c=coandc=co+Ac,where Aca-60issmall, then (60-26) f(w.v.¢o) =0andf(r.1/.60 +Ac)=0 aretwoneighboring members ofthefamily (60.25). ByDeﬁnition 60.1, each ofthese twocurves istangent totheenvelope ofthefamily (60.25). LetuscallP1,P2their respective points oftangency, Fig. 60.29. The twocurves themselves willusually intersect inapoint, called Pinthe ﬁgure, which isnottoofardistant from P1andP2.Since thecoordinates ofthepoint Psatisfy each oftheequations of(60.26), itwillalsosatisfy theequation (60-27) f(1.1/.60 +Ac)"-f(Iv.y.¢o) =0- "‘Aproof canbefound inWilliam F.Osgood, Advanced Calculus. 750 EXISTENCE Tnaonsmz y’=f(z,y). Proxnn. CLAIRAUT. Chapter ll (The point Pliesineach curve; itscoordinates willtherefore make each term zero.) Since Ac¢0,wemaydivide (60.27) byittoobtain (6028, =0_ P2 f(x,y,c0+Ac)=0 Envelope P1 P f(1.y.¢o)=0 Figure 60.29 Bythedeﬁnition ofaderivative, which weassumed exists, f(xry:c0 ‘I’ASL _.f(x1l/160) = , evaluated atc=co. Therefore asAc->0: 6 (a)Theexpression ontheleftof(60.28) approaches 2 - (b)Thepoints P2andPapproach P1. (c)The curve f(1,y,c0 +Ac)=0approaches thecurve f(:z:,y,e0) =0 sothat both aretangent totheenvelope atP1. Since coisanarbitrary value ofc,wehave thus proved that ifafamily ofcurves hasanenvelope, then each point oftheenvelope, by(c)and(a) above and(60.28), must satisfy thetwoequations (60.31) f(1,y,c) =0, =0. These twoequations may beregarded astheparametric equations ofan envelope ofthefamily (60.25). Note thatthey arethesame as(60.23). Comment 60.32. Itmay notalways bepossible toeliminate the parameter cbetween thetwoequations in(60.31), butifitcanbeelimi- nated, then theresulting Cartesian equation y=g(a:)iscalled theelimi- nant of(60.31). Aneliminant, however, may introduce additional loci Lesson 60A ENVELOPES or11FAMILY orCURVES 751 which arenotpart oftheenvelope. Forexample, (a) 1=\/2—I—c, y=\/2—c aretheparametric equations ofthat part ofacircle ofradius twothat lies intheﬁrstquadrant (since :1:andyarepositive). Their eliminant is12+ yz=4,which istheentire circle. Comment 60.33. Not every family hasanenvelope. Forexample, thefamily ofconcentric circles 12+y2=e2hasnoenvelope. Comment 60.84. Theorem 60.22 gives only asufﬁcient condition for theexistence ofanenvelope, notanecessary one. This means that ifa family satisﬁes thehypotheses ofTheorem 60.22, itmust have anen- velope; ifitdoes not,itmay ormay nothave anenvelope. Example 60.35. Find theenvelopes, ifthere areany, ofthefamily of curves (a) y=cos(1+e). Solution. Here f(:c,y,c) =y—cos(re-I-c)and£3‘-fQ‘;1cy—'cl =sin(1+c). The determinant (60.24) istherefore sin(1+c)1 cos(1+e)0 Also 62f/6c2 =cos(1—|—c).If1+c961r/2, then cos(1—I—c);-60. Hence, byTheorem 60.22, thesetsofvalues, excluding 1—I-c=1r/2, that satisfy theparametric equations (c) y——cos(1+c)=0, sin(1—|—c)=0, areenvelopes ofthefamily (a). Writing theﬁrst equation of(c)as cos(1-I-c)=y,then squaring both equations of(c)andadding, weobtain theeliminant,I=-—cos (ze+e). (d) yz=1, y==l=1. The twocurves y=1andy=-1aretheenvelopes ofthefamily (a). Comment 60.36. Ifwewrite (a)as (e) Arccosy=x+c, then f(1,y,c) =Arccosy -—:1:—cand 6f/6c =—1. Equations (60.23) therefore become (f) Arccosy—1—-c=0, ——1=0. 752 EXISTENCE Tnsonnmz y’=f(1,y). PICARD. CLAIR-AUT. Chapter ll Since -1960,theparametric equations arenotsatisﬁed. Yetweshowed above that (a)hastwo envelopes, y==l;1. The trouble isthat now 82f/802 =0forall1,y,sothat oneofthehypotheses ofTheorem 60.22 isnotfulﬁlled. The fact that thefamily (e)hasanenvelope shows that thetheorem gives only asufficient condition, notanecessary one. Example 60.37. Find theenvelopes, ifthere areany, ofthefamily of circles (a) (r—6)’+v’=1. where cisaparameter. Solution. Here f(1,y,c) =(1-—c)’+y2-—1,élf/6c =—2(x -—c). The determinant (60.24) istherefore 29*‘) 2”=4y¢0 i1y¢0sn<1§g=2-60. -20 Hence byTheorem 60.22, thesetsofvalues, excluding y=0,which satisfy theparametric equations (0) (r—c)2+v2—1=0. -—2(1——c)=0, 1-—c=0, areenvelopes ofthefamily (a). Substituting thesecond equation of(c)in theﬁrst, weobtain theeliminant (d) 2/”=1.y==b1- Therefore byTheorem 60.21, y=1and y=—lareenvelopes of(a). SeeFig.60.01. Comment 60.38. Atrouble similar tothat mentioned incomment 60.36, occurs ifwewrite (a)as (e) c=x;l=\/l—y2. Equations (60.23) therefore become (f) c—1=l=\/l—y2=0, 1=0. Since 1960,(f)cannot besatisﬁed. But here 82f/8c2 =0forall1,y,c, sothat oneofthehypotheses ofTheorem 60.22 isnotfulﬁlled. The fact that thefamily (e)hasanenvelope shows again that thetheorem gives only asufficient condition, notanecessary one. Example 60.39. Find theenvelopes ifthere areanyofthefamily (a) yz=2c:c—-e2. Lesson 60A Envsnorss orAFAMILY orCURVES 753 Solution. You canverify that (60.24) issatisﬁed ify960.The parametric equations (60.23) are: (b) y2~—2cx+c2=0 —2:c+2c=0,1=c. Substituting thesecond equation of(b)intheﬁrst, weobtain theelimi- nants (c) y2—2:c2—I—:t2=0, y2—:c2=0, :z:—y=0, 1—I—y=0. Hence x=yandat=—y, y#0,areenvelopes of(a). SeeFig. 60.4. E»=1’/' \ Figure 60.4 Example 60.41. Find theenvelopes, ifthere areany, ofthefamily of curves (=1) (w—c)’=31/’—vi- Solution. Here f(1,y,c) =(1-—c)2—3y2—|—ya,6f/6c =—-2(x —c). The determinant (60.24) istherefore ___ __ 2 2 ”“I6y+3” #0, ify¢0and2; gc-{=2-so. ByTheorem 60.22, setsofvalues, excluding y=0and2,that satisfy the parametric equations (<1) (rv—6)”—3v’+vs=0. 2(x—-e)=0, (1-c)=O 754 EXISTENCE Tnnonmrz y’=f(z,y). P1cA1u>. CLAIRAUT. Chapter ll areenvelopes of(a). Substituting thesecond equation intheﬁrst, we obtain (<1) —3v’+1/“=0. y=0. v=3- The eliminant y=0isexcluded by(b). However, since Theorem 60.22 gives only asufficient condition, wecannot assert without further study that y=0isnotanenvelope. (Actually itisnot, seeExample 60.53.) However, y=3isanenvelope. Example 60.42. Find theenvelopes, ifthere areany, ofthefamily of curves (a) ce"=:c-y-1. Solution. Here f(:t,y,c) =cc”—1+y—I—1,6f/élc =e”.The deter- minant (60.24) istherefore _ 1/ 1'(ll) ‘1C6‘l’1=___e1I;£0, but;9‘3_~;.=0_ ,, c0e Hence oneofthehypotheses ofTheorem 60.22, namely 82f/002 9'50,isnot satisﬁed. The family (a),therefore, may ormay nothave envelopes. (Itactually hasnone, seeExample 60.54.) LESSON 60B. Envelopes ofa1-Parameter Family ofSolutions. Letf(1,y,c) bea1-parameter family ofsolutions ofthedifferential equation y’==f(:c,y). Thefamily may ormay nothave anenvelope. Ifithas,then byDeﬁnition 60.1, ateach point oftheenvelope, there isamember of thefamily ofsolutions that istangent toit.Hence theenvelope, ateach ofitspoints, hasthesame slope y’asanintegral curve. Ittherefore follows that each envelope ofthefamily ofsolutions, excluding those points ofthe envelope where thetangent isvertical, alsosatisﬁes thegiven differential equation y’=f(:e,y). Hence anenvelope ofafamily ofsolutions ofy’= f(x,y) isalsoasolution ofthisdiﬁerential equation. Unless anenvelope iscoincident with anintegral curve ofafamily, itis, byDeﬁnition 59.12, alocus ofsingular points. Foreach point oftheen- velope liesontwointegral curves: theenvelope itself andanintegral curve ofthefamily. Hence anenvelope isaparticular solutioh ofadifferential equation y’=f(:e,y) that cannot beobtained from a1-parameter family ofsolutions byassigning avalue tothearbitrary constant. The converse, however, need notbetrue. Aparticular solution notobtainable from a 1-parameter family ofsolutions isnotnecessarily anenvelope. If,there- fore; wecanﬁnd anenvelope ofa1-parameter family ofsolutions ofa differential equation y’=f(1,y), weshall atthesame time have succeeded inﬁnding aparticular solution oftheequation, notobtainable from the family. Lesson 60B ENVELOPES orAI-PARAMETER FAMILY orSoLU'rxoNs 755 Comment 60.43. ByTheorem 60.22, weknow that iff(1,y,c) =0 isatwice differentiable function andsatisﬁes (60.24) over asetofvalues ofx,y,c, then theparametric equations ofitsenvelope aregiven by (60.44) f(£li,y,C) =0,‘i(%°) =0. Iftheparameter ccanbeeliminated between these two equations, the resulting equation y=g(1), which wecalled theeliminant, gives anen- velope. However, asnoted inComment 60.32, theeliminant mayintroduce extraneous loci. Itisessential, therefore, toverify byactual substitution inagiven differential equation whether each factor oftheeliminant satisﬁes theequation. Ifitdoes, itisaparticular solution notobtainable from the family. SeeExample 60.53 below foraneliminant which isnotasolution. Example 60.5. Find, bymeans ofenvelopes, particular solutions, if there areany, ofthedifferential equation (a) v’=\/1—v2. notobtainable from thefamily ofsolutions (b) y=cos(1—I—c) Solution. InExample 60.35, wefound that y=1andy=—1are envelopes ofthefamily (b). You canverify bydirection substitution that each isasolution ofthegiven differential equation (a). Hence each en- velope isaparticular solution of(a)notobtainable from (b). Example 60.51. Find bymeans ofenvelopes, particular solutions, if there areany, ofthedifferential equation <a> 1'=—-”——‘,7”2» notobtainable from thefamily ofsolutions (b) (1—c)2+y2=1. Solution. InExample 60.37, wefound that y=1andy=--1are envelopes of(b). Since each function satisﬁes (a),each isaparticular solution of(a),notobtainable from thefamily (b). Example 60.52. Find, bymeans ofenvelopes, particular solutions, if there areany, ofthedifferential equation <3») I/I=R-x+ $2 _y2: notobtainable from thefamily ofsolutions (bl yz=201—c2. 756 Exrsrancs Tnaonam: y’=f(1,y). P1cA1m. CLAmAu'r. Chapter ll Solution. InExample 60.39, wefound that y==|=xareenvelopes of(b). Since these functions satisfy (a),they areparticular solutions of (a),notobtainable from thefamily (b). Example 60.53. Find, bymeans ofenvelopes, particular solutions, if there areany, ofthedifferential equation (a) 90/’)’(2 —v)’=4(3—1/). notobtainable from thefamily ofsolutions (b) (w—C)’=3v’—2/“- Solution. InExample 60.41, wefound twoeliminants, namely y=3 andy=0.The ﬁrst satisﬁes (a),andistherefore aparticular solution of(a),notobtainable from (b). The second y=0,however, does not satisfy (a).Hence, wenowalsoknow that y=0isnotanenvelope of(b). (Inexample 60.41, weremarked that intheabsence offurther information, wecould notknow whether ornoty=0isanenvelope.) Example 60.54. Find, bymeans ofenvelopes, particular solutions, if there areany, ofthedifferential equation 1(3.) y,-'= ﬁr notobtainable from thefamily ofsolutions (b) ce”=1—y——1. Solution. InExample 60.42, wefound that (60.24) ofTheorem 60.22 wasnotsatisﬁed. Hence wedidnotknow whether ornotthefamily (b) hadanenvelope. However, ifweapply theexistence Theorem 58.5 tothe function f(z,y) =1/(:1: —y)of(a),weﬁndthat theonly possible singular points lieontheline 1—y=0.Since thefunction y=1does not satisfy (a),itcannot beanenvelope ofthefamily (b). Inthisexample, therefore, aparticular solution notobtainable from (b),ifthere isone, cannot befound bymeans ofenvelopes. EXERCISE 60 Find theenvelopes, ifthere areany, ofeach ofthefollowing family of curves. 1.y=x—I-c. 5.y=e1—I—3\/1—I—c2,y>0 2.y=1—I—}(x—I-c)2. 6.(y-I—1—c)2=41y. 3.y=(1—c)3. 7.y=sin[(1—c)2]. 4.y1’3 =:t—c. Lesson 61 Tun CLAIRAUT EQUATION 757 Find, bymeans ofenvelopes, particular solutions, ifthere areany, of each ofthefollowing differential equations, notobtainable from thegiven family ofsolutions. 8.1/'=\/1-1+1.i =1+1(z+t->2»; is1+1. 9-1/’=31/2’3.v =(w—¢)3- 10.y=1y'—I-3\/1—I—(y’)2,y =e1—I-3\/l+c2, y>0,x2 <9. ll.y’=2\/(1— y?)\/Arc siny,1=\/Arc siny —I-e. ANSWERS 60 one. =0. = ==l=1. = =1. =0.Seeproblem 3andread =0. mments 60.36,60.38. =\/9W— 12. =\/9W1§. =0,y==1=1. E"!*§"!°l"QQQQQZOP? I-ll-I!"‘.°2°9°Z*‘.°‘=e<==e=:=:1-1== LESSON 61. The Clairaut Equation. Thesimplest type offamily ofcurves isafamily ofstraight lines. From analytic geometry weknow that (61.1) y=m1—I-b, represents such afamily. The slope ofeach lineofthefamily ism,and itsyintercept isb.Thefamily (61.1) isthus a2-parameter family. Wecan form a1-parameter family from itbyrequiring that bbeafunction ofthe slope m,i.e.,that b=f(m). Hence (61.1) becomes (61.12) y=ma:+f(m). Toﬁnd adifferential equation whose 1-parameter family ofsolutions is thefamily oflines (61.12), weproceed aswedidinLesson 4B. Differenti- ation of(61.12) gives y’=m.Substituting thisvalue ofmin(61.12), we obtain thedifferential equation (61-13) v=1/'1+f(y')- Equation (61.13) iscalled Clairaut’s equation.* Comment 61.14. Ifwestart with thedifferential equation (61.13), itssolution (61.12) iseasily found. Replace y’bytheparameter m(or,if you prefer, byc). Forexample, thesolution oftheClairaut equation y=y’:z:+(y’)2 isthe family ofnonvertical lines y=mzt+m2or y=ex—I-c2. ‘Named after theFrench mathematician, Alex Claude Clairaut (1713-1765). 758 EXISTENCE THEOREM: y’=f(z,y). PICARD. CLAIRAUT. Chapter 11 Aswepointed outinLesson 60B, ifthefamily (61.12) hasanenvel- ope,then thisenvelope isaparticular solution of(61.13) notobtainable from thefamily ofsolutions (61.12). ByTheorem 60.22, asufficient condition, among others, fortheexistence ofanenvelope ofthefamily (61.12) isthatpoints oftheenvelope satisfy theparametric equations (61.15) y—ca:—f(c) =0, as+f’(c)=0. Theﬁrstequation in(61.15) hasthesame form as(61.12). Ittherefore satisﬁes Clairaut’s equation (61.13). Theeliminant of(61.15) alsosatisﬁes this ﬁrst equation. (Asolution oftwo simultaneous equations satisﬁes each equation.) Hence, if(61.15) yields aneliminant, then thiseliminant byComment 60.43, may beaparticular solution of(61.13). Itisifit satisﬁes (61.13). Example 61.2. Find a1-parameter family ofsolutions oftheClairaut equation (=1) y=y’w+(y')’- Also investigate forenvelopes ofthefamily ofsolutions. Solution. First wenote by(61.13), that (a)isaClairaut equation withf(y’) =(y')2. Hence byComment 61.14, its1-parameter family of solutions is (b) y=cx+c2. Youcanverify that (b)satisﬁes (60.24). Therefore, byTheorem 60.22, a setofvalues thatsatisﬁes theparametric equations (60.23), (c) y—cx—-c’=0, :z:+2c=0, c=——a:/2, isanenvelope of(b).Substituting thesecond equation of(c)intheﬁrst, weobtain theeliminant 2 4 2 (<1) y=—%+%1-=-1}» which isanenvelope ofthefamily ofnonvertical lines (b). Since this function (d)satisﬁes (a),itisaparticular solution of(a)notobtainable from thefamily (b). Example 61.21. Find a1-parameter family ofsolutions oftheClairaut equation (=1) y=2/w+logu’- Alsoinvestigate forenvelopes ofthefamily ofsolutions. Lesson 61 T1-n-: Cmrmur Eotmrron 759 Solution. First wenote by(61.13) that (a)isaClairaut equation withf(y’) =logy’.Hence byComment 61.14, (b) y=cw+log6, isafamily ofsolutions of(a). Youcanverify that (b)satisﬁes (60.24). Therefore, byTheorem 60.22, thesetofvalues that satisﬁes thepara- metric equations (60.23), (c) y—cx—logc=0 1 1:c+E=O, x=—-6 isanenvelope of(b). Substituting thesecond equation of(c)inthe ﬁrst, weobtain (d) y+1—-log(—x_1)=0, y+1+log(——:c)=0, :z:<0, which isanenvelope ofthefamily ofnonvertical lines (b). Since this function (d)satisﬁes (a),itisaparticular solution of(a)notobtainable from thefamily (b). Example 61.3. Asource oflight orsound which strikes acurve inthe same plane with it,isreﬂected inaﬁxed direction. Find theequation of thiscurve. m=.v'(-r) y=y(x) P(x,y) Reﬂected direction Q 2 1 -_L ml—y'(x) 5' Vm2= _ X 180° —r x O(0,0) r Figure 61.31 Solution. (See Fig. 61.31.) Weplace theorigin atthesource ofthe light orsound, andletthe2:axisbeparallel totheﬁxed reflected direction. Lety=y(z) betheequation ofthecurve weseek. Weassume y(z) is deﬁned anddifferentiable onaninterval I. 760 Exrsrnncs THEOREM: y’=f(z,y). Promo. CLAIRAUT. Chapter ll Call: (a) itheangle theincident rayOPmakes with thenormal totherequired curve, rtheangle thereﬂected rayPQmakes with thenormal tothere- quired curve, y’theslope ofthetangent totherequired curve y=y(z) atP(z,y), mltheslope ofthenormal totherequired curve atP(z,y), (=—1/y’), m2theslope ofOP(=3//2:). Byalawofphysics (b) tani=tanr, andbyaformula ofanalytic geometry 1 ‘§\p-A**‘=2~i<e-__mi m2__ y’ _w1/1/’_ (0) tanl_1+"l1"l2_ 1___— $1/-9 From Fig. 61.31, weseethat tan(180° —1')=-1/y’. Hence, tanr = 1/y’. Therefore by(b),tani=1/y’. Substituting thisvalue oftani in (c),weobtain 1 __ __ I (d) .5/7=$5?!/Z’ 9=My’+y(y')”- Theequation isalmost likeaClairaut equation, butnotquite. If,however, wemultiply itbyy,weobtain (e) 11’=2wyy'+y’(y')2- Now let 2 w u=f. w=2w. wV=%%- Substituting (f)in(e),wehave (g) u=1/w+i(1/)2, which isaClairaut equation inu.Hence its1-parameter family ofsolu- tions is 2 w u=a+%- Substituting in(h)thevalue ofuasgiven in(f),weobtain thefamily of parabolas 2 w w=a+g- Lesson 6l—Exereise 761 Geometric Problems Giving Rise toaClairaut Equation. InLes- sons 13and36D, wesolved geometric problems which gave rise, respec- tively, toaﬁrst order equation and toaspecial type ofsecond order equation. Acurve whose tangents have properties which areindependent ofthepoint atwhich thetangent isdrawn willlead toaClairaut equa- tion. Since thesolution ofaClairaut equation isafamily ofstraight lines, andsince these straight lines willhave theproperties ofthetangent lines, theenvelope ofthefamily willgive thecurve weseek. Example 61.4. Every tangent toacurve hastheproperty that the sum ofitsintercepts hasaconstant value lc.Find thecurve. Solution. ByExercise 13,l(a) and 1(b), thexandyintercepts ofa tangent linearerespectively x—y/y’andy-my’. Hence thecurve must satisfy thecondition (a) w—-%+y—wy'=k, wy’—u+1/1/’—w(y’)2=ky', kl y(y’—1)=wy’(y'—1)+ky’, y=wy’+;T%{' The lastequation in(a)is,by(61.13), aClairaut equation. Hence its solution, byComment 61.14, is (b) y=cx+;%_c—11 :z:c2-—(x+y—-Ic)c+y=0, which isafamily ofstraight lines. You canverify that (b)satisﬁes (60.24). Therefore, byTheorem 60.22, thec-eliminant of (<1) w¢’—(w+y—k)¢+1/=0 2xc—(:v—|—y-lc)=0, c= » may beanenvelope of(b). Substituting thesecond equation of(c)inthe ﬁrst, weobtain theeliminant (w+y—k)’_(w+y—k)’ _<d> 4, 2, +y-0. (w+y—k)2—4w=0, x+y—k==:=2z"2@/"2, at:1:22:1/zyl/2 +y=lc, 1:‘/2 :byl/2 ==|=k‘/2. Since itsatisﬁes (a),itistherequired solution. EXERCISE 61 Find a1-parameter family ofsolutions ofeach ofthefollowing Clairaut equations 1-6. Also investigate forenvelopes. 762 Exrsrnncn Tnnonnmz y’=f(z,y). Prcxan. Cmumtrr. Chapter 11 4-1/=11/’-—(y')3- 1Zl'+\/'1+(1/)1 e”.1-y=my’—(u')2- ay=1r+u+oVi ay=1r~oV@ 7.Solve theequation y=31y’+6y2(y’)2. Hint. Multiply byyz.Then make thesubstitution u=ya,du/dz: =3;/2y’ toobtain theClairaut equation du 2du2 ""a+§hQ'5.y= 6.y=xy' Inproblems 8-13, every tangent toacurve hastheproperty indicated. Find thecurve. 8.Thesumofitsintercepts hasaconstant value 9. 9.Theproduct ofitsintercepts hasaconstant value It. 10.Thelength ofthesegment intercepted bythecoordinate axeshasaconstant value k2.Hint. Setthesquare rootofthesumofthesquares ofitsintercepts equal tok2. Itsdistance from theorigin hasaconstant value k.Hint. First show that theequation ofatangent lineatapoint (z,y) ontherequired curve isgiven byy’X~—Y+y——my’=0,where (X,Y)isapoint onthetangent line. Then usethefactthatthedistance ofapoint (A,B) tothelineaX+bY—l- c=Ois:l:(aA +bB—|- c)/E15. Here A=0,B=0. 12.Thesumofitsdistances from thepoints (a,0) and(—a,0) hasaconstant value k.Inthedistance formula from apoint toaline,asgiven in11,take thepositive square root. Thesegment intercepted bythecoordinate axesforms with these axesaright triangle whose areahasaconstant value k2.ll. 13. Ineach ofproblems 14-16, ﬁndtheClairaut equation whose family of solutions ofstraight lines hastheenvelope indicated. Hint. First ﬁnd :1: andyinterms ofy’andthen solve (61.13) forf(y’). 14.3/2=2:0. 15.2y=2:2. 16.xy=—k. ANSWERS 61 =ex—c2,y=2:2/4. cx+ 1+c2,12+ 4y cx—c2/3,271211 +4 —03,27y2 =42:3. cx+\/1—l—c2,y =\/1—a:2, ca:—e”,y=xloga: —x. =cx+§c2. 12.4(x2+g2)=I02. 1/2=1;yl/2==|=3. 13.21;‘;=ah’. 19°7‘.°‘5"!*9"P!"z~z<=<==:¢=<=<=<=4. 0. =61 = 1/>0. 55 9.4xy =k. 10_$2/3+g2/3 =k4/s_ 2 2 211.x +y =k.14-. y=1/Z-l"ﬁ' ,2 l5.y =y'x—£%- l6.y=y'¢¢2\/W. Chapter 12 Existence and Uniqueness Theorems foraSystem ofFirst Order Differential Equations andforLinear andNonlinear Differential Equations ofOrder Greater Than One. Wronskians. LESSON 62. AnExistence and Uniqueness Theorem fora System ofnFirst Order Differential Equations and foraNonlinear Differential Equation of Order Greater Than One. LESSON 62A. The Existence and Uniqueness Theorem foraSys- tem ofnFirst Order Differential Equations. InTheorem 39.12 we gave asufficient condition fortheexistence ofapower series solution ofa system ofnﬁrstorder equations d _ay?l' 2f1(t.Z/1.2/2. '''11/1|); d ﬂ =f2(t;f/1;?/21 '''1l/fl)!(ll d 7%’: =fn(l,?l1,?l2, '''11/11); satisfying theinitial conditions, (62-11) 3/1(io) =111. 3/2(lo) =112.''',1/n(lo) =G..- Compare itwith thetheorem wenow state, which gives asufficient condi- tion fortheexistence and uniqueness ofasolution of(62.1) satisfying (62.11). 763 764- Orr-nan Exrsrnncn Tnnonnms. Wnonsxrxrx. Chapter 12 Theorem 62.12. Letthefunctions f1,f2,---,f,.oftheﬁrst order system (62.1) becontinuous inaregion Sdeﬁned by -lol §I50, Iyl "‘a1] §kl, lll2‘"a2l§7921"’: ll/11_‘anl §kn- InS,leteach function satisfy aLipschitz condition lfi'(l,?/1,92, '''if/1!) T‘fi(tiy1ry2| '''; § lr2:"'n: where (t,y1,y2, ---,yn)and(t,y1,y2, ---,17,.)areanytwopoints inS. Then aninterval I0:|t—t0|<h,h>O,exists inwhich there isoneand only onesetofcontinuous functions y1(t), y2(t), ---,y,,(t) with continuous derivatives inI0satisfying thegiven system (62.1) andtheinitial conditions (62.1 1). NOTE. The theorem does notalways give amaximum interval ofcon- vergence; read Comment 58.65. Remark. When n=1,thesystem (62.1) reduces toaﬁrst order equation y1'=f1(t,y1). Theorem 58.5, which weproved previously for thisﬁrst order equation, isthus aspecial case oftheabove more general theorem. The proof ofTheorem 62.12, which wehave omitted, although more complicated than theproof ofTheorem 58.5, follows itspattern exactly. Weﬁndtheinterval I0ofTheorem 62.12, forexample, inpractically the same manner wefound theinterval I0ofTheorem 58.5. Since theregion S consists ofclosed intervals andthefunctions fl,f2,---,f,.areeach con- tinuous inS,each isbounded inS.Hence there isapositive number M such that ll‘i(t;f/1:1/2r' ''1?/n)l <MY = 112: '''/"'1 forevery point (t,y1,y2, ---,yn)inS.By(62.13), (to,a1, ---,an)isan interior point ofS.Itistherefore possible topick anhsuch that the n+1dimensional rectangle R, _tol éh§kg, If/1 _‘a1| < §kl, ''', |y,.—a,.|<Mh §kn, which hasthispoint (t0,a,,a2, ---,an)atitscenter, liesentirely inS.The Misgiven by(62.15); thek’saregiven in(62.13). Wethen ﬁndsetsofsequences ofPicard approximations yllv yl2r '''1ylm y21,3/22, ''",1/21., f/"1; 1/"2; '''1y7l7U Lesson 62B Exrsrrzuca Tnnonnn: Nontrnmn Equxrron orORDER n765 inamanner similar tothat given in(57.33) forasystem oftwoﬁrst order equations. The graph ofeach function ineach approximating set,itis then proved, liesinthen—|—1dimensional rectangle R,andeach setof sequences ofapproximating functions converges uniformly onI:|t—tll| §htotherespective setoffunctions yl(t), y2(t), ---,y,.(t), which onIll isasolution of(62.1) satisfying (62.11). Finally weshow that thissetof functions yl,yg,---,y,,isunique. Example 62.17. Find aninterval, for‘which theapproximating func- tions obtained inExample 57.35 converge totheactual solution. Solution. InthisExample fl(t,:e,y) =ty,f2(t,a:,y) =my.Comparing theinitial conditions ofthis example with (62.11) weseethat to=0, al=1,a2=1.Since flandf2arecontinuous forallt,a;,y, wemay take forSanybounded region, sayonedeﬁned by|t|§5=kl),|x——1|§ 10=lcl,|y—1|§15=kg.Hence §11, §16and (=1) lfil§80. |f2|§176=M. Wemust now choose haccording to(62.16) sothat (b) |t|éh§5, |x—1|<176h §10, |y—1|<176h §15. Ifh=0.056, allthree inequalities in(b)willbesatisﬁed. ThePicard ap- proximations obtained inExample 57.35, willtherefore converge tothe actual solutions x(t), y(t)foratleast |t|<0.056. LESSON 62B. Existence and Uniqueness Theorem foraNon- linear Differential Equation ofOrder n.InTheorem 39.32, wegave a sufﬁcient condition fortheexistence ofapower series solution ofthenth order nonlinear differential equation $5 =f(xvy;1-/Ivy”: '''7:1/(11-1)) satisfying theinitial conditions (62-21) I/($0) =av, I/($0) =al! I/"($10) =<12,‘'‘l l/(n-1)(x0) =an—1- Compare itwith thetheorem wenow state which gives asufficient condi- tion fortheexistence and uniqueness ofasolution of(62.2) satisfying (62.21). Theorem 62.22. Letthefunction finthenthorder diﬁerential equation (62.2) beacontinuous function ofitsarguments x,y,y’, ---,y‘"“‘) ina region Sdeﬁned by (52-23) ll"‘ivol§kc, ly—“cl§kl, |y'—all§/C2, |y// __a2| éks, ___, |y(n-—l) _a"_l| ékm 766 OTHER Exrsrnnca Tnnonrms. Wnorzsxrxu. Chapter 12 andletfsatisfy aLipschitz condition, |f(x:y1:l/1’; ''',1/1(n_n) —f(z)?/21?/2’: '''sll/2(n_1))l éN(|vl —1/2|+I2/1'—1/2'|+|y1” —v2”|+---__|_|y1(n—1) _y2(n—1)|)’ where (xii/11:!/1/1 '''sf/101-1)) and ($412,?/2', '''1f/2(n_D) are any two inS. Then aninterval Io:|x—:00]<h,h>0,exists inwhich there isone andonly onecontinuous function y(z) with acontinuous derivative oforder n satisfying (62.2) andtheinitial conditions (62.21). Proof. Assume that y(z) isasolution of(62.2). Wenow deﬁne new functions yl(x), y2(x), ---,y,.(x) bythefollowing relations (62-25) y(z)=v1(w), '1/(Z)=y1’(w) =um), y"(r) =v1”(w) =uz’(w) =1/a(w). v‘"'"(w) =v1‘""’(r) =y2""”(w) = =v.._1’(w) =um)- Differentiating thefunction inthelastlineof(65.25), weobtain (62-251) y‘"’(@=) =2/1‘")(w) =2/2‘""')($) ='''=1/»_1"($) =y»’(w)- By(65.251) and(62.25), wecantherefore write (62.2) as yr/(Z) =f(xryl;y2ry3; '''I1!»)- Hence, because of(62.25) and (62.26), thesetoffunctions yl,y2, ---,y,, satisﬁes thesystem ofﬁrstorder diﬁerential equations (62-27) y1'(1) =I/201). v-.>’(w) =va(w), I/s'(¢) =y4($). 3/n—1'(27) =3/11(2); 1/»’(w) =f(w,v1.vz.va. ---.u.)- By(6225), l!1($o) =!l($o), ?l2(%o) =1/($0), ''‘l1/n(5'30) =3/(”_D($o)- Wecantherefore replace theinitial conditions (62.21) bytheconditions (6223) 1/1(=vo) =(lo, !l2(1o) =<11. Z/s(f'1o) =<12,''', I/n(9>o) =4111-1- Lesson 62B Exrsrnncn Tnnoanm: Nonnmnxn Eovxrror: orORDER n767 Conversely, ifwestart with thesystem (62.27) andtheinitial conditions (62.28), anddeﬁne (62-281) u(w)=2/1(2). then therelations in(62.25), (62.251), and(62.26) hold; thelastequation of(62.27) becomes identical with (62.2), andtheinitial conditions (62.28) become (62.21). Hence ifthere exists asetoffunctions yl,y2, ---,y,. which isasolution ofthesystem (62.27) satisfying (62.28), then, by (62.281), thefunction yl(x) ofthissetisthesolution y(z) of(62.2) satis- fying (62.21). You canverify that thesystem (62.27) with initial conditions (62.28) satisﬁes thehypotheses ofTheorem 62.12. Hence bythetheorem, an interval I0about xoexists onwhich there isoneandonly onesetofpar- ticular solutions yl,Z/2,''',3/1.S9-tisfying (62.27) and(62.28). Since yl(x) exists andisunique, itfollows, by(62.281), that itsequal y(z) exists and istheunique solution of(62.2) satisfying (62.21). Comment 62.282. Wewish toemphasize that itisalways possible, bymeans of(62.25) and(62.26), toreplace asingle differential equation of order n>1byanequivalent system ofnﬁrst order equations. Example 62.29. Setupasystem ofﬁrstorder equations equivalent to thethird order nonlinear equation (a) v”’(w) =3w’—2/2v" forwhich (b) 1/(0)=1, I/(0) =—1, I/”(0) =2- Solution. Lety(z) beasolution of(a). Assuggested in(62.25), we deﬁne (0) y(t)=v1(w), 1/(1)=1/1'(w) =v2(w), y"(w) =v1"(w) =v2'(w) =213(2)- Differentiating thelastequation in(c),weobtain (<1) y”'(w) =2/1"'(w) =1/a”(w) =ya’(w)- By(d)and(c),wecanwrite (a)as (e) ua’(w) =3Iv2-1112113- Hence, by(c)and(e),wecanreplace thesingle equation (a)bythesystem 768 OTHER Exrsrnncn THEOREMS. WRONSKIAN. Chapter 12 ofﬁrst order equations (f) i/1'(r) =2/2(w), v2'(w) =um). v:-{(1) =3m/2—yr”:/3, andtheinitial conditions (b)by (s)1/1(0)==1/(0)=1,1/2(0)=v’(0)=—1, 1/3(0)=v”(0) =2- Thefunction yl(a:) ofthesolution ofthesystem (f)satisfying (g)willbe thesolution y(z) of(a)satisfying (b). LESSON 62C. Existence and Uniqueness Theorem foraSystem ofnLinear First Order Equations. InTheorem 62.12, norestrictions were placed onthedegree ofthedependent variables inthefunctions fl,f2,---,f,.of(62.1). InTheorem 62.3 below, each ofthedependent variables yl,yz,''',1/1.appears linearly, i.e.,each ofthese variables has exponent one. Theorem 62.3. Given asystem ofnlinear ﬁrst order equations (62.31) =.f11(t)f/1+navy.+--~+r...<vy..+Q10). 9,,’-(2=f21(t)l/1+r..<»>y.+---+r...<vy..+Q20), %=r...<vv.+r...-(vi.+---+r....<vy..+on). where allfunctions f,-,-andQ,-,i=1,2,---,n,j=1,2,---n,arecon- tinuous onacommon interval I.Then there is,onI0,i.e.,Iwithout itsend points, oneandonly onesetofcontinuous functions yl(t), y2(t), ---,y,,(t) with continuous derivatives, satisfying thesystem (62.31) and theinitial conditions (62-32) vl(to) =¢11,"'.!/»(1o) =11... where toisapoint inI0. Remark. InTheorem 39.22, wegave asufficient condition forthe existence ofapower series solution of(62.31) satisfying (62.32). Compare itwith theabove theorem which gives asuﬂicient condition fortheexist- ence anduniqueness ofasolution of(62.31) satisfying (62.32). Proof ofTheorem. The proof ofthetheorem willconsist inshowing that thecontinuity requirement ofthefunctions f,-,~onIisequivalent to therequirement that each function satisfy aLipschitz condition (62.14) Lesson 62C EXISTENCE Tmaonsmz LINEAR SYSTEM 769 inI,provided Iisﬁnite andclosed. IfIisnotﬁnite andclosed, then itis suﬂicient, asweshall show, that each function satisfy thisLipschitz con- dition onaslightly smaller closed subinterval I’contained inI.And since allfunctions ontheright of(62.31) are,byhypothesis, continuous onthis common, closed interval IorI’,thehypotheses ofTheorem 62.12 willbe satisﬁed. Hence itsconclusion willfollow. Let (9') f1(t,I4/1»?/2» ‘‘'1?/n) =fn(¢)1/1 +f12(i)?/2 +'''+fln(t)yn +Q10)» f2(tz/1,112, -~-,y»)=f21(l)y1 +f22(i)!/2 +---+f2»(l)y,. +Q20), f1l(t1y1sy2a '''2yn) =fnl(t)y1 +f1l2(t)y2 +'''+fun“)?/n +Q11“)- Weassume theinterval IofTheorem 62.3 isﬁnite andincludes itsend points. Ifitisnot,wetake aﬁnite closed subinterval I’ofI.Hence inI’, whether itisasubinterval ofIorthewhole interval I,each function f,-,-(t) andQ,-(t) of(62.31) is,byhypothesis, continuous. Allaretherefore bounded inI’.This means that theabsolute value ofeach function foralltinI’ islessthan some positive number. LetNbethelargest ofthese numbers. Then foreach function f,-,-(t) of(62.31) andforevery tinI’, lfif(t)l§N! i=lr2s"'1n; j=112s"'1n‘ Let,1,/1,yz,...,ynandy,,{#72,-~-,1],,betwosetsofvalues ofthede- pendent variables. Therefore by(a) (C) f1'(t1y1sy21 '''1?/1:) =fn(¢)?/1 +f€2(t)y2 'l"''''l"f€n(t)yn + f€(t1y1>y21 '''11]") =fi1(t)yl1 'l'fs2(l)l72 +'''+fin(t)yn 'l"Qi(t)r foreachi =1,2,---,n.Subtracting thesecond equation in(c)from the ﬁrst, weobtain lf1'(t1y1sf/21 '''syn) '~fi(tvy1sy2; '''s =|fn(l)(1/1 —P1)+fi2(¢)(y2 —F2)+'''-l-fm(5)(?l» —POL i=1,2,---,1». Since Ia—bl§Ia]+|b|,andsince, by(b),If,-,-|§N,weobtain from (d) (6) lfi(3,?/1, ''‘if/1|) _.f‘l'(t;yl1 '''1 é lyl _yll +lf='2(i)| |1/2—112i+'--+lfin(t)i ly»—‘Jul §N(l!l1 "l71|+l1/2"172'-F '''+ll!»—?Inl)- Ifyouwillcompare (e)with theLipschitz condition (62.14), youwillﬁnd that both arealike. Hence wehave proved that forthespecial system of linear equations (62.31), thecontinuity requirement ofthefunctions f,-5inI’ isequivalent totheLipschitz condition (62.14) ofTheorem 62.12. 770 Ornnn Ex1s'n-mos Tnnonsns. Wnonsxmu. Chapter 12 Iftheinterval Iofthetheorem isﬁnite andclosed, then I=I’.Ifitis not, then since itisalways possible tosurround each tinIbyaﬁnite, closed subinterval I’contained inI,inwhich thetheorem isvalid, it follows that thetheorem isvalid onI.Inthislastcase I=I0. EXERCISE 62 Thefollowing problems refer tothesolutions ofthesystems inExercise 57.Foreach ﬁnd aninterval forwhich theapproximating solution con- verges. 1.Problem 6. 2.Problem 7. 3.Problem 8. 4.Problem 9. Setupasystem ofﬁrst order equations with appropriate initial condi- tions equivalent toeach ofthefollowing equations. 5-u”(w)=31%’——:1/2,1/(0) =0.1/(0)=1-6-1/”’(r) =2r(;1/)2 -—31/1/”+ wy.2/(0)=1.1/(0) =-1.1/”(0) =2- ANSWERS 62 1.InTheorem 62.12,f1 =1+y,fg =t—-2:2,to =0,a1=2,a2=1.Scan beanybounded region: |t|§I00,Ia:—2|§k1,|y-1|§I62,where ko, k1,k2 arearbitrary. Misthelarger ofM1andM2in|f1|§|t|+|y|<M1, lfgl§t|-|-|:c2|<M2. Choosehso that |t|§h§I60,Ia:—2|<Mh§k1, y—1<Mh§k2. 2.Seeanswer toproblem 1formethod. 3.Thissystem isaﬁrst order linear one.SeeTheorem 62.3: f11(t) =t,fz1(t) =1, Qg(t) =——e‘. Each function iscontinuous forallt.Therefore theapproximat- ingfunctions converge forevery t. 4.InTheorem 62.12, f1=ya,fg=1+z,f3=z—y;to=0,a1=1, dz=0,a3=1.Scanbeanybounded region: |t|§I00,|:c—1|§k1, |y|§I02,I2—-1|§k3,where I60,k1,k3arearbitrary. Misthelarger ofM1, M2,MsiI1|f1l <y2<M1,|f2|éwI+ Izl<M2.lfsléIzl+lyl<Ma- Choose hsothat |t|§h§kg,a:—1|<Mh§k1.|y|<Mh§I62, |z—1|< Mh§ I03. 5-1/1'(w) =1/2.1/z'(w) =3x21/2-—1/1’;y1(0) =0.1/2(0) =1- 6-1/1'(1) =1/2.I/2'01) =ya.1/s’(w) =211/2’—31/11/3+ 11/1;y1(0)=1.1/2(0) ="-1193(0) =2- LESSON 63. Determinants. Wronskians. LESSON 63A. ABrief Introduction tothe Theory ofDeter- minants. Inthis lesson, weshall outline only those essentials ofthe theory ofdeterminants that weshall need forourpurposes. Wehave al- ready given thedeﬁnition ofa2X2and a3X3determinant in Comment 31.35 and (31.72). Forconvenience werecopy them here. (63.1) 111 bl 112b2=@152 "r412511 Lesson 63A DETERMINANTS 771 (63.11) G1 bl C1 I12b262=1115263 —|—1125361 —|—1135162 -4135261 ""(125162; —015362- (13 b3 C3 Weobserve that each term inadeterminant, aswritten ontheright, contains oneandonly oneelement from each rowandeach column. Hence weconclude: Theorem 63.12. Ifeachelement inaroworinacolumn ofadeter- minant iszero,thenthedeterminant iszero. Consider asystem oftwoequations intwounknowns. (63-2) 111$+bl?!=C1. rm+b2?!=62- Multiplying theﬁrst equation byb2,thesecond by—-bl, adding thetwo andthen solving forx,weobtain __6152 —6251_(63.21) av_ialbz_aabl By(63.1), wecanwrite (63.21) as (63.22) 12="EIT ' Note that theelements ofthedenominator determinant arethecoeﬁicients ofatandyin(63.2); theelements ofthenumerator determinant areob- tained byreplacing thecoeﬁicients ofatbytheconstants clandc2and recopying thecoeﬁicients ofy.Similarly youwillﬁnd, ifyousolve (63.2) fory,that (63.23) y=--_, where thenumerator determinant isobtained byreplacing thecoefﬁcients ofybyclandCg,andthedenominator isthesame asthat in(63.22). 772 Oman Exrsrrznca Tnnomzms. Wnonsxmn. Chapter 12 Comment 63.24. From (63.22) and(63.23), weinfer thefollowing: :1 950,thesystem ofequations (63.2) hasoneandonly one2 solution. 2.Ifthedeterminant in1equals zero andthenumerators of(63.22) and (63.23) are5'50,thesystem hasnosolutions. 3.Ifthedeterminant in1equals zero andthenumerators of(63.22) and (63.23) alsoequal zero, thesystem hasaninﬁnite number ofsolutions. Each ofthese three possibilities isillustrated intheexamples below. Example 63.25. Solve thesystem @ %+%=L :1:——y=4. Solution. By(63.22) and(63.23), ll3| l2ll _4-1_-1-12_13 _1 4__7__(b)”_2 3* 23“s’ ”_2 3* 5 ll—1i ll—li Hence (a)hasanunique solution. Note that thedenominator determinant #0. Example 63.26. Solve thesystem (a) 21¢+3y=1, 40:—|—6y==3. Solution. By(63.22) and(63.23), 13 21 _l36l_—3 __i43i_2 “W-T’ ”*"T§"_6' L..l L.lHence there arenosolutions for:1:andy.Note that thedenominator de- terminant iszero butthenumerator determinants arenot. [The lines in (a)areactually parallel lines.] Example 63.27. Solve thesystem (3,) 2x+3y=1, 4a:+6y=2. Lesson 63A DETERMINANTS 773 Solution. By(63.22) and(63.23) W‘3! I2‘I 26 0 42 0 <'°> ”-W-6’ y-W-6' 46 46 Note that both numerator and denominator determinants arezero. As youcaneasily verify, thesystem (a)hasaninﬁnite number ofsolutions. Give :0anyvalue youwish intheﬁrstequation of(a). Solve itfory.You willﬁndthat these values of2:andywillsatisfy thesecond equation of(a). [The lines in(a)areinfactcoincident] Comment 63.24 isalsoapplicable toasystem ofnequations innunknowns. Forn=3,thesystem becomes (63-3) air+biy+C12=d1, azx—I-bgy+022=dz, a3x+bay+c3z=d3. Itssolution is (63.31) bl61 <11d161 111bl C2 a2d2c2 a2b2 d3 C3 G3 b3 d3:1:=———i- it z=L -bl bl 61 (Z1 bl C11 C1 , y= (11 bg Cg G2 bg C2 (lg b2 b3 b3 b3 Thesystem willhave anunique solution forac,y,and2ifthedenominator 950;nosolution ifthedenominator equals zeroandatleast onenumerator isnotequal tozero; aninﬁnite number ofsolutions ifdenominator andall numerators areequal tozero. Our primary purpose inintroducing determinants istoenable you to prove thefollowing twoimportant theorems. Although forconvenience, we have stated thetheorems forthree unknowns, they areapplicable toasystem ofnequations innunknowns. Theorem 63.4. Thesystem ofequations (63.41) ala:+bly+clz=d1, G223+bzy+C22=dz, (Z3113 +173]] +C32 =d3, 774 Orr-ma Exrsrrzncs Tnnoanms. Wnonsxnm. Chapter 12 hasoneandonly onesolution forx,y,zifthedeterminant formed bytheir coeﬂicients isnotequal tozero. Ithasnone oraninﬁnite number ofsolutions ifthisdeterminant equals zero. Theorem 63.42. Thesystem ofequations (63.43) alas+bly+clz=0, 112$—|—52?!+62?=0, ‘139?-I"ball+63?=0» always hasthetrivial solution :1:=0,y=0,2=0.Ithasaninﬁnite num- berofnontrivial solutions, i.e.,ithasaninﬁnite number ofsetsofvalues of x,y,and2satisfying (62.43), where atleast oneof2:,y,zisnotzero, ifand only ifthedeterminant formed bytheir coeﬂicients iszero. LESSON 63B. Wronskians. InLesson 19A, when wediscussed the linear dependence andindependence ofasetoffunctions, westated that wewould intime produce theorems bywhich their linear dependence or independence could bedetermined. Since these theorems areclosely asso- ciated with theconcept ofaWronskian, weshall ﬁrst discuss thissubject below. Deﬁnition 63.5. The Wronskian* ofasetoffunctions f1(:z:), f2(:z:), ---,f,,(:c), each ofwhich possesses derivatives oforder n—1,isdeﬁned tobethefollowing determinant. (63-51) f1(w) f2(w) ''-f..(w) fi’(1>) f2’(w) ---f»’(w) f1”(w) f2”(1) ''-ft/'(w).-------¢¢.---.---- f1‘""“(w)f2‘""1’(w) '''f..‘"_”(r) Note thattheelements ofthedeterminant arethegiven setoffunctions andtheir derivatives through order n—1.Wedenote theWronskian (63.51) by W(.fl:.f21 ''‘fur. Ifn=2,theWronskian offlandf2is,by(63.52) and (63.51), (63.53) W<r..r.; n=f‘f’=r.r.'—f.r.'.1.’fr ‘Pronounced Vronskian andnamed forHoéné Wronski, aPolish mathematician. Lesson 63B Wnonsxrxns 775 Deﬁnition 63.54. Afunction f(:v) issaid tovanish identically on aninterval I:a§to§b,ortobeidentically zero onI,ifforevery :1:inI,f(z) =0.Weindicate thisfactbywriting f(z) E0. Theorem 63.55. Ifasetoffunctions fl,f2,---,f,,,each ofwhich possesses aderivative oforder n—1,islinearly dependent onaninterval I:a§x§b,then itsWronskian vanishes identically onI. Proof. Since thesetofgiven functions f1,f2,---,f,,islinearly de- pendent onI,there exist, byDeﬁnition 19.1, constants c1,C2,---,onnot allzero such that <9») ¢1f1 ‘l’Czfz +'''‘l’cnfn =0, forevery asinI.Bydifferentiating (a)(n—1)times, weobtain f1c1+f2c2+"'+fncn=0; f1'¢1-l'f2'62 +'''-l'f1/6n =0, f1(n-—1)c1 +f2(n—1)c2 +____|_fn(n——1)cn =O_ Thesystem (b)isasetofnequations, eachequal tozeroforevery xinI. Thec’sineach equation arenotallzero. There istherefore asetofc’s notallzero that satisﬁes each equation in(b). Hence thissetofc’sisa solution ofthesystem ofequations (b). Ittherefore follows byTheorem 63.42, that thedeterminant (0) fr f2 fn fl’ f2’ f»' > ----.--.¢--.----- f1(n_-1) f2("_n '''f»(n_n which istheWronskian ofthegiven setoffunctions, iszero. Comment 63.56. Theorem 63.55 gives anecessary condition forthe dependence ofasetoffunctions. Itsaysthatifasetoffunctions isde- pendent onI,then itsWronskian isidentically zero onI.Unfortunately itisnotasufficient condition. IftheWronskian ofthesetoffunctions is identically zeroonI,onecannot conclude that thesetislinearly dependent onI.Itneed notbe.Here isanexample. Letf1(a:) =$2,f2(:z:) =:v|:e|, Fig. 63.57. Lettheinterval Iinclude :1:=0.By(63.53), theWronskian ofthefunctions is (a) $2Zlxl 20:a:|a:|’ +=x3|:v|' —|—xzlxl —2x2|:c|. 776 Oman Exrsrsncn Tmeonsns. Wnonsxmn. Chapter 12 When 2:>0,|a:|=av.Therefore |x|’=2:’=1.The right side of(a) thus becomes :03+13-—2x3=0.When :1:=0, =:1:=O,.and again theright side of(a)iszero. When :0<0, =—a:and therefore I:e|’=—x’ =——1. The right side of(a)now becomes :c3(—1) —:03—I- 2:c3=0.Wehave thus shown that theWronskian ofthetwogiven func- 2=x2 xxlxl (Q0) xm Figure 63.57 tions isidentically zero onaninterval Iwhich includes :1:=0.However, weshall now show that thetwofunctions arelinearly independent onI. Form with them thelinear combination clxz +c2a:|:c| =0.If2:=1, then c1-1-C2=0,andif:1:=—1,then cl-—-c2=0.The only solution ofthispair ofequations isc1=0,c2=0.Hence byDeﬁnition 19.1, the functions :02and:e|:e|arelinearly independent. Ontheother hand, iftheWronskian ofasetoffunctions isnotequal to zero foronly one:1:inaninterval I,then theset,byTheorem 63.55, must belinearly independent onI.Inshort, noconclusion canbedrawn as regards thelinear dependence orindependence ofasetoffunctions onI, from thefact that itsWronskian isidentically zero onI.But ifthe Wronskian ofthesetisnotequal tozero foronly onevalue of:cinI,then thesetmust belinearly independent onI. InExercise 63,5, youwillﬁndanecessary andsuﬁicient condition for thelinear independence ofasetoffunctions onaninterval. EXERCISE 63 1.Determine, foreach ofthefollowing systems, whether ithasone, none, or inﬁnitely many solutions. (a)1+4y=1. (d)1»+2y=0.2:c—y=3. 2:2:+4y=0. 2:c+2y=10. a:—y=0. (0) = W2: =01 -2:c—y=3. a:—3y=0. Lesson 63-—Exei-cise 777 SetuptheWronskian ofeach ofthefollowing setoffunctions. (a)1,e‘. (b)1,2:. (c)2:,e‘. (d)2:,xe’. (e)1:,re’,32:. (f)sin2:,sin22:. (g)sin2:,cos2:. (h)ax,bx+c,c940. (i)e",e",a;éb. (i)1—cos22:,2:2:—2cos22:. (k)1,rt,2:2. (1)2:,e’,me’,2e’. (m)1,e‘,e2’. (n)2:,e‘,xe‘. (o)1,sin22:, cos21, (p)sin31:,sin2:—Q;sin32:. Determine, byComment 63.56, which ofthesetsin2arelinearly independent over asuitable interval. Forthose setswhose Wronskian iszero, useDeﬁni- tion19.1todetermine which aredependent andwhich areindependent. Prove that anytwononconstant functions that differ byaconstant are linearly independent. Theorem 63.55 gives only anecessary condition forthelinear independence ofasetoffunctions. Thefollowing theorem gives anecessary andsuﬂicient condition forthelinear independence ofasetoffunctions. Asetoffunctions f1(x), f2(a:), ---,f,,(a:), each continuous onaninterval I:a§x§b,is linearly dependent onI,ifandonlyifthedeterminant b b b /11% /amid.» /finds b b b o=land.» fyfas [;.;..d¢ =0 b b b I/,./14¢ /Ifnfzdx fffai onI.Thedeterminant Giscalled theGrammian ofthesetoffunctions. Usethistheorem toprove that (a)2:,22:,32:isalinearly dependent setonI:—-1§:2:§1, (b)e’,e2‘isalinearly independent setonI:—1§2:§1, (c)ac, isalinearly independent setonI:—1§2:§l. Prove Theorem 63.4. Prove Theorem 63.42. ANSWERS 63 (a)one. (b)Inﬁnitely many. (c)None. (d)Inﬁnitely many. (e)Inﬁnitely many. (f)1:=0,1;=0. (11)‘Ie”=e._ (d)1re‘ =Z28. 0e’ 1:cez—|— ez (b),12: (e)2:we” 3:2:=1. 01 1a:e'+ ez 3=0. Ore’+2e’ 0 (c)2:e=(x_De,‘ (f)sin:2:sin2:2: =_2Sin3$_ 1ez cos:0:2cos2:: 778 OTHER EXISTENCE THEOREMS. Wnonsxum. Chapter 12 (g)sin:4:cosa: =__1 (1)W(z,e',:ce”,2e’; 1:)=0. cosa:—sin 2: (h) ax b:c—|—c =_ac. (m)2e3'. (i) =(b_a),,<=»+b>z' (11)622(1)! —2)- (j) :c—cos2a: 2a:—2cos2z =0. (0)0. 1+2sin2a: 2(1+2sin2x) (k)1IZ2 (p)0. 0121 =2. 002 3.Linearly independent setsare(a),(b),(c),(d),(f),(g),(h),(i),(k),(m),(n). Linearly dependent setsare(e),(j),(1),(0),(p)—see Example 21.32. LESSON 64-. Theorems About Wronskians and theLinear Independence ofaSetofSolutions ofa Homogeneous Linear Differential Equation. Intheprevious lesson, weproved that ifasetoffunctions islinearly dependent onI,then their Wronskian isidentically zero onI.Wealso showed byanexample, that theconverse need notbetrue. If,however, thesetoffunctions arensolutions ofahomogeneous linear diﬂerential equation oforder n,namely of (64-1) f»(w)y‘") +fn-1(1)y("_” +'"+f1(Iv)y’ +fo(=v)1/ =0. where f0(re),f1(x),---,f,,(:c) areeach continuous functions onaninterval I,andf,,(:c) 960when :0isinI,then, asweshall show (see Comment 64.15 below) thevanishing oftheir Wronskian isanecessary andsuﬁicient condition forthelinear dependence ofthesetonI. Theorem 64.11. Ifeachfunction yl,7/2,---,y,,isasolution of(64.1) onaninterval I:a§x§b,then theWronskian ofthissetoffunctions is either identically zeroonI,oritisnotzeroforanyatinI. Proof. Weshall prove thetheorem only forn=2.Call y1,y¢ the twosolutions of(64.1) when n=2.Since each function satisﬁes alinear equation ofthesecond order, their ﬁrst and second derivatives exist. By(63.53), yi t/2 (a) Wu/1.1/2; w)= =viz/2'—1/2111’-?l1' 2/2' Lesson 64- LINEAR INDEPENDENCE orASETorSoi.U'rioNs 779 Differentiating both sides of(a)with respect toav,weobtain (b)W’(yi,z/2; w)=I/ll/2”+vi'v2’ —vi’?/1' —vzvi” =viz/2” —uni”- Byhypothesis, each ofthefunctions y,and1/2isasolution of(64.1) with n=2.Hence, (9) f2(x)l/1” +f1($)1/1' +fo($)y1 ==0» f2($)1/2" -l"fi($)1/2' -l-fo($)y2 =0- Since weassumed f2960,wemay divide each equation in(c)byitto obtain (d) ylll=._ , 1,211=__ . f2 f2 Substituting these values in(b)gives, with thehelp of(63.53), (e) W'(v1.v2; w)=—v1 +11-.» =—%one—um’)=—j.'-§W(1/it/2; itBy(e),therefore (r) W'+5W= 0,f2 anequation which islinear inW.Itssolution is (64-12) W(vi.v2; w)=cem’. where F(:c) =—f(f1 dx/fg). Since, by(18.86), theexponential function em‘) isnever zero, weconclude from (64.12) that, ifc=0,then W=0 forevery rcinI;ifc#0,then W¢0forevery soinI. Remark. Formula (64.12) isknown asAbel’s formula. Theorem 64.13. Ifeachfunction yl,yz,---,y,,isasolution of(64.1) inaninterval I:a§so§b,andif W(y1i1/2| '''1I/ni x)=01 atapoint at=1:0inI,thenthesetoffunctions islinearly dependent onI. Proof. Weshall prove thetheorem only forn=2.Byhypothesis, there isapoint moinI,forwhich (a) W(vi.v2;r<>) =0. where ylandygaretwosolutions of(64.1) with n=2.Hence, by(63.53) 780 OTHER EXISTENCE THEOREMS. WRONSKIAN. Chapter 12 and(a), 3/1($o) 1/2($o) =0 2/i'($0) 1/2'(1?o) Using theelements ofthedeterminant (b)ascoefﬁcients, weform thetwo equations (<1) lli($o)¢1 —|—3/2($o)¢2 =0. 1/1'(1?o)¢1 +1/2'($o)¢2 =0- Since thedeterminant whose elements arethecoefficients ofclandcgof(c) is,by(b),zero, itfollows, byTheorem 63.42, that there isaninﬁnite num- berofpairs ofnontrivial solutions of(c)forclandcg.With anyonesuch pair, form thelinear combination (d) 3/a($) =¢11li(1'I) —|—62?/2($), where ylandygarethetwogiven solutions. Therefore, byTheorem 19.3, y3(:v) isalso asolution of(64.1) with n=2.By(d)and (c),y3(:t°) = ¢1yi(Io) +621/2($o) =0and1/a'($0) =¢1y1'(930) 'l'¢2y2'($o) =0- Wehave thus proved that y3(x) of(d)isasolution of(64.1), with n=2,i.e.,itisasolution of (8) f2(w)y" +f1(v)v’ +fo(fv)1/ =0. satisfying theconditions (f) 3/s($o) =0» ?/a'($o) =0- Butyoucaneasily verify that y(z) -E0isalsoasolution of(e)andthat thissolution satisﬁes y(:z:0) ==0,y’(a:0) =0.Bytheuniqueness Theorem 19.2 (which westillhave notproved butshall soon, seeTheorem 65.2) thetwosolutions y3(x) of(d)andy(z) =0areidentical. Hence by(d) (8) ¢1y1($) +621/2($) =0- Weremarked after (c),that clandcgarenontrivial solutions of(c)and hence notboth arezero. Therefore byDeﬁnition 19.1, thefunctions y1 andygarelinearly dependent. Comment 64.141. IftheWronskian ofasetofsolutions of(64.1) is zeroatapoint moinI,thenitiseasytoshow, byTheorem 64.13, thatthe Wronskian isidentically zero onI.Weprove thestatement, which is equivalent toTheorem 64.11, forn=2.Letyl,ygbetwosolutions of (64.1). Since W=0atapoint 2:0inI,thesolutions, byTheorem 64.13 Lesson 64 LINEAR Innsrnnnnncs or11SmorSOLUTIONS 781 arelinearly dependent. Therefore (91) 61?/i +62112 E0 onI,where clandcgarenotboth zero. Differentiation of(a)gives (bl Cilli' ‘l’62112’ E0- Since clandcgarenotboth zero, itfollows byTheorem 63.42, that (C) vv‘1 2E0. 1/1’ 112I Butthedeterminant in(c)is,by(63.53), theWronskian ofylandyg. NOTE. Although theabove proof ismuch shorter than theone in Theorem 64.11, thelatter proof was given inorder toobtain (64.12), anequation which willbeneeded later. Comment 64.15. Theorems 63.55 and64.13 together give anecessary andsuﬁicient condition forthelinear dependence ofasetoffunctions that aresolutions ofthelinear differential equation (64.1). Theﬁrstsays that if thesetisdependent onI,then itsWronskian isE0onI.The second says thatifitsWronskian iszeroforonly one2:inI(and therefore by Theorem 64.11 itisE0onI)then thesetislinearly dependent onI. Hence younow have atesttodetermine whether asetofnsolutions ofa homogeneous linear differential equation (64.1) islinearly dependent or independent. IfitsWronskian iszero forone:2:inI,thesetislinearly dependent; ifitsWronskian isnotequal tozero foronezcinI,thesetis linearly independent. Example 64.2. Show that ifml,mg,mgaredistinct, then thesetof functions (a) vi=8"". 1/2=6"“. ya=6"”. which arethree solutions ofathird order homogeneous linear equation, arelinearly independent onI:—oo<:0<oo. Solution. By(63.51), theWronskian ofthegiven functions is emlz emzl emgz m1e'”‘” mge'"" m3e""' ml2e:n1: m22emgz m32em3z By(63.11), (b)isequal to (0) @(m‘+m’+mml(m2ms2 '"msmzz) -'(mdm/32 -m3m12) +(m1m22 -m2m12)l» 782 OTHER EXISTENCE THEOREMS. WRONSKIAN. Chapter 12 which canbewritten as (d) e(m‘+m'+m3)z(m1 '-m2)(m-2 —ms)(ms -mi)- Since wehave assumed ml,mg,mgaredistinct andsince, by(18.86), the exponential function e("‘1+"'ﬂ+’”=)’ isnever zero, itfollows that (d)cannot bezero foranyre.Hence theWronskian (b)isnotzero foranyso.By Comment 64.15, thesetofsolutions in(a)istherefore linearly independent onI:-—oo <:1:<oo. Remark. The above proof can beextended toshow that ifml, mg,---,m,,aredistinct, then thesetofnfunctions yl=e"‘1", yg= e"‘1‘, ---,y,l=e"'»", each ofwhich isasolution of(64.1), islinearly inde- pendent onI:—oo <to<oo. Example 64.21. Show that thesetoffunctions (11) 3/1=9”, 1/2=$6“, which arethesolutions ofy"—2ay’ +azy=0,arelinearly independent onI:--00 <x<oo. Solution. By(63.53), theWronskian ofthegiven functions is (b) e“” we“ M43 axed: + ell=_axe2ax+ e2ax _axe2az =e2o.x. Since e2“"9-E0forevery :0,thegiven functions areindependent onI: —oo <:6<co. Example 64.22. Show that thesecond solution yg(:c) of(23.28) ob- tained bythereduction oforder method described inLesson 23B, isinde- pendent ofthesolution yl(x). Solution. LetF(a:) =fl(x)/fg(:c) and 6-1‘F(z)d:c (=1) 9($)=IT div- Then, by(23.28), yg=ylgand, by(63.53), theWronskian ofthetwo solutions is (b) W(1/1.:/2; w)=y‘W’- vi’via’+2/1'0 =1/120'+2/1v1'a —vi’:/la =1/129’- Lesson 65 EXISTENCE Tnsomm: LINEAR EQUATION, ORDER n783 By(a)andTheorem 58.3 -Irma: (<1) !/(95) =2'7 ' Substituting (c)intheright of(b),weobtain (<1) W<y..y.; o=e-“'"“"'- Since theexponential function isnever zero, itfollows, byComment 64.15, that thegiven functions arelinearly independent solutions onanyinterval Iover which thefunctions ylandygaredeﬁned. EXERCISE 64 1.Show thatthesetoffunctions 1,x,2:2which arethree solutions ofy"’=0 arelinearly independent. 2.Show thateach ofthefunctions yl=sin2:—-§sin3:2:andyg=sin32:,isa solution ofy”+(tanx ~—2cot:c)y’ =0,butthat y=cit/1'l'czygisnot thegeneral solution ofthedifferential equation. Hint. Show that thetwo functions arenotlinearly independent; seeExercise 63.3(p). 3.Prove that y=clsinx+ cgcosx isageneral solution ofy”+y=0. Hint. Show thateach function satisﬁes theequation andthatthetwofunc- tions arelinearly independent. 4.Prove thaty=(01+cg2:)e“ isageneral solution ofy”—2y’—l—y=0. 5.Prpve thaty=cle‘+cgez‘ +c3:ce2" isageneral solution ofy”’—5y”+ 8y——4y=0. LESSON 65. Existence and Uniqueness Theorem forthe Linear Differential Equation ofOrder n. Weareready atlasttoprove theexistence anduniqueness Theorem 19.2 forthelinear differential equation oforder n.Since ontheinterval Iin which weshall beinterested, f,l(:c) 9'!0,wecandivide (19.21) byf,,(:c) to obtain alinear differential equation (65-1) 2/‘")(rv) +f.._i(r)v("'” +---+f1(1¢)y' +fo(¢v)1/ =Q(I)~ Theorem 65.2. Ifthecoeﬂicients f0(x), fl(x), ---,f,,_l(:z:) andQ(x) in thelinear diﬂerential equation (65.1) areeach continuous functions of:1:ona common interval I,thenforeach point 2:0inIandforeach setofconstants ao,al,---,a,,_l, there isoneandonly onefunetion y(z) thatsatisﬁes (65.1 ) andtheinitial conditions (65-21) ll(1o) =110, 1/($0) =<11," ‘il/(n_U($o) =¢ln-1- Remark 65.22. InTheorem 37.51, wegave asufficient condition for theexistence ofapower series solution of(65.1) satisfying (65.21). Compare 784 Oman EXISTENCE THEOREMS. WRONSKIAN. Chapter 12 itwith theabove theorem, which gives asufficient condition fortheexist- enceanduniqueness ofasolution of(65.1) satisfying (65.21). Proof oftheTheorem. Assume y(z) isasolution of(65.1). Wenow deﬁne new functions yl(a:), yg(:v), ---,y,,(:c) bythefollowing relations. (65-23) v(v)=v1(v). v’(w)=vi'(w) =v2(w). v”(w)=vi”(w) =vz’(w) =vat»). y(n—1)(x) =y1(n—1)(x) =y2(n—2)(x) =_y3(n-3)(x) =°''=I/n-'-1'(x) = Differentiating thefunction inthelastlineof(65.23), weobtain (65-231) v""(v) =v1""(w) =v2‘”"’(v) ='-'=v.._i"($) =v..’(w)- By(65.231) and(65.23), wecantherefore write (65.1) as (65-24) vi/(1) +f.._1(r)v..(w) +---+f1(Iv)1/2(5) +fo($)l/1(5) =Q(1)- Solving (65.24) fory,,’(x), weobtain (65-25) vi/(f'=) =_.fn—1I/n — —fit/2-fovi+Q- Hence because of(65.23) and(65.25), thesetoffunctions yl,yg,---,y,,, satisﬁes thesystem oflinear ﬁrstorder equations (65.26) 3/i'(1’) =ll2($)l 1/2'($) =ys($)lQ/n._.1I.(x.) .............................. yﬂ'(x) =—f.._1(w)v-.(v) -'"—f1(r)v2(w) —f0(1)?/1(1=) +Q(Iv)- By(55-23). ll1($o) =!l(f¢o)l l/2(50) =1/($0), ''',I/n(370) =y(”'_1)($o)- Wecantherefore replace theinitial conditions (65.21) bytheconditions (6527) ll1($o) =aol 3/2($o) =alt'''1I/n($o) =(111-1- Conversely ifwestart with thesystem ofequations (65.26) and the initial conditions (65.27), anddeﬁne (65-28) v(=v)=vi(rv). then therelations in(65.23), (65.23l), and(65.24) hold; thelastequation of(65.26) becomes identical with (65.1), andtheinitial conditions (65.27) Lesson 65 EXISTENCE Tneonsuz LINEAR Eqmmon, ORDER n785 become (65.21). Hence ifthere exists asetoffunctions y1(x), y2(a:), ---, y,,(a:) thatisasolution ofthesystem (65.26) satisfying (65.27), then, by (65.28), thefunction y1(x) ofthissetisthesolution y(z)of(65.1) satisfy- ing(65.21). Allthefunctions fo(a:), f1(x), ---,f,,_1(x), Q(z) inthelastequation of (65.26), are,byassumption, continuous onI.Thefunctions g2,ya,---,y,, arealsocontinuous since their derivatives exist. Therefore thehypotheses ofTheorem 62.3aresatisﬁed. Hence, bythetheorem, there is,onI,or I0ifIisclosed, oneandonlyonesetofparticular solutions y1(:v), y2(x), ---,y,,(a:) that satisﬁes (65.26) and (65.27) where 2:0isapoint inIo. And since y1(:c) exists andisunique, itfollows that itsequal y(z)also exists andistheunique solution of(65.1) and(65.21). Example 65.3. Setupasystem oflinear ﬁrst order equations forthe second order linear equation (=1) 1/”—3y’+2y=w andshow thatthesolution y1(x) oftheresulting system isthesame asthe solution y(z) of(a). Solution. Lety(z) beasolution of(a). Assuggested in(65.23), we deﬁne (b) y(w)=y1(r), y’(w)=y1'(w)=212(10)- Diﬁerentiating thelastequation of(b),weobtain (C) 1/”(=v) =1/1"($) =y2'(¢)- By(c)and(b),wecanwrite (a)as (<1) 1/2'(w) —3yz(w) +21/1(w) =w- Hence, by(b)and(d),wecanreplace thesingle equation (a)bythesys- temoflinear ﬁrst order equations (8) y1’(¢) =1/z(fv), y2’(w) =3y:—22/1+x. Inoperator notation wecanwrite (e)as (fl D?/1 -"3/2=0, 21/1+(D—3)y;=w- Multiplying theﬁrstequation in(f)byD—3andadding ittothesecond, weobtain (8) (D2 —3D+2)!/1 =7% Z/1”_31/1' +2?/1=$- 786 O'r1-nan Exrsrsncn Tusonnms. Wnonsxum. Chapter 12 Comparing thesecond equation in(g)with (a),weseethat thesolution y1(x) ofthesystem (f)willbethesame asthesolution y(z) of(a). InTheorem 19.3, westated that ahomogeneous linear differential equa- tion oforder nhasnlinearly independent solutions, butproved thispart ofthetheorem only forthecase when thecoeﬂicients intheequation are constants. Wenow prove thestatement fornonconstant coefficients. Theorem 65.4. Iff0(x), f1(ac), ---,f,,(a:) areeach continuous functions ofaconacommon interval I,thenthehomogeneous linear diﬁerential equation (65-41) 1/‘"’+fn_i(w)y‘"_" +---+f1(=v)y’ +f<>(w)y =0, hasnlinearly independent solutions yl,y2,---,yn. Proof. Weconsider ﬁrst aspecial setofsolutions, g1(:z:), g2(a:), ---, g,,(:c) of(65.41), each satisfying respectively theinitial conditions (=1)91(w<>) =1,91’(1o) =0,9i”(w0) =0,---,g1‘"_”(wo) =0, g2($o) =0,g2'($o) =1,9/($0) =0,''',92("_1)(1o) =0, 9s($0) =0,Q:/($0) =0,93”($0) =1,''',9a(”_1)($o) =0, 91-($0) =0,y»'(wo) =0,yn”(w<>) =0,---,gn‘""’(wo) =1, where acoisapoint inI.ByTheorem 65.2, each function g1,g2,--~,gn exists. Weform alinear combination ofthisspecial setofsolutions, setit equal tozero, andtake itssuccessive derivatives. Wethus obtain (b) ¢191(r) +620261) +---+¢».¢1»(w) =0, ¢1g1'(w) +czgz’(w) +---+c..g1.’(w) =0, c1g1”(w) +czyz”(w) +--~+Cay,/'(rv) =0, ¢i9i(n_l)(-7?) +¢292(n_l)(3’3) 'l''''+¢n9n(n_l)(33) =0- Letrc=wo.Then theﬁrstequation in(b),bytheﬁrstcolumn ofvalues in (a),simpliﬁes toc,=0;thesecond equation in(b),bythesecond column ofvalues in(a),simpliﬁes toc2=O,---,thelastequation in(b),bythe lastcolumn ofvalues in(a),simpliﬁes toc,,=0.Wehave thus shown that each constant cl,C2,---,anin(b)andinparticular each such constant in theﬁrst equation of(b)iszero forxoinI.Therefore byDeﬁnition 19.1, thesetoffunctions g1,g2,---,gnis,onI,alinearly independent set. Since each function is,byassumption, asolution of(65.41), they form collectively asetofnlinearly independent solutions of(65.41). Lety(z) beasolution of(65.41), andletxobeapoint inI.ByTheorem 19.3, thefunction (<1) h(=v)=1/(wo)91(w) +y'(Io)92(@v) +---+y"‘_”(=vo)Qn(1), Lesson 65 EXISTENCE THEOREM! LINEAR EQUATION, ORDER n787 isalsoasolution of(65.41) since itisalinear combination ofnindependent solutions. [Remember thecoeﬂicients in(c)areconstants.] Taking suc- cessive derivatives of(c),weobtain (<1) WI)=1/(@o)9i'(1v) +y'(Io)92’(1) +---+y("_”(Io)9n'(@), h”(1) =y(1o)91"(1) +y'(Io)y2”(I) +---+y("_”(Io)9n"(w), h‘”_"(=v) =y(ro)a1‘"‘“(w) +y’(ro)92‘"“’(r) +~'-+y‘"_"(wo)y»‘"“’(r)- Letx=xo.Then (c),bytheﬁrst column ofvalues in(a),simpliﬁes to h(x0) =y(:c0); theﬁrst equation in(d),bythesecond column ofvalues in(a),simpliﬁes toh’(:c0) =y'(x0), -~-,thelastequation in(d),bythe lastcolumn ofvalues in(a),simpliﬁes toh"_1(x0) =y("-”(:c0). Wehave thus shown that thetwosolutions h(:c) andy(z) of(65.41), andtheir ﬁrst n—1derivatives areequal when a:=xo. Hence, bytheuniqueness Theorem 65.2, thetwosolutions h(z) andy(z)areidentical. Wecanthere- forereplace h(a:) byy(z) in(c)toobtain (8) 1/(w)=y(1o)a1(w) +1/’(1vo)a2(w) +---+y‘"_"(rvo)g»(w), where xoisapoint inI.Since y(z) isanarbitrary solution of(65.41), it follows that every solution of(65.41) canbeexpressed asalinear combi- nation ofthespecial setofnlinearly independent solutions g1,g2,---,gn. Let1/1(1), 1/2(1):), ---,y,,(x) bensolutions of(65.41), and letxobea point inI.Therefore by_(e), (f)m(m)=y1(ro)91(r) +y1’(w<>)92(r) +---+y1‘""’(w0)gn(w), y2($) =y2($o)91($) +!l2'($o)Q2($) +'''‘l’y2<"_1)($0)Qn(1'3)» um)=1/n($0)Q1($) +y»'(r0)a2(w) +---+yn‘"“’(@o)a-(1), areeach valid equations. The setoffunctions g1,Q2,---,g,,arelinearly independent. Hence thesystem (f)canbesolved forg1,g2,--~,gnin terms ofy,,yz,---,y,,.This means, byTheorem 63.4, that thedetermi- nant formed bythecoeﬁicients oftheg’sisnotzero. Therefore, (8) y1(xo) ?l1'($0) 1/1%-l)($0) I/2(xo) I/2'($o) 1/2(n_1)($0) #0- 1/n(x0) I/1/($0) y1.‘"‘“(wo) Atheorem ofdeterminants now permits ustointerchange rows and 788 OTHER Ex1s'rENcE THEOREMS. Waonsxnm. Chapter 12 columns toobtain from (g)theequivalent determinant (11) !l1($o) 1/2($o) ‘''t/»($0) Z/1'($o) 3/2'($0) '''2/1/($0) 9'50- !/1(n_1)($o) 2/2(n_1)($o) 3/»(n_1)($o) ByDeﬁnition 63.5, thedeterminant (h)istheWronskian ofasetofsolu- tions yl,1/2,---,ynof(65.41) evaluated atx=mo,where moisapoint inI.And since thisWronskian isnotzero foronea:inI,thesetofsolu- tions, byComment 64.15, islinearly independent onI.Wehave thus proved theexistence ofasetofnlinearly independent solutions of(65.41). Theorem 65.5. Ifyl,yz,---,y,,arenlinearly independent solutions of(65.41), then (65-51) 1/.=cm+cm+---+6.11.. isageneral solution of(65.41), i.e.,every solution of(65.41) canbeobtained from (65.51) byaproper choice oftheconstants cl,C2,---,c,.. Proof. Weprove thetheorem forn=2.Letyl,ygbetwolinearly independent solutions of (a) y"+f1(w)z/' +fo(=v)y =0- Therefore, byTheorem 19.3, (b) ye=611/1 'l'621/2 isalsoasolution of(a).Assume y3isasolution of(a)notobtainable from (b). Intheproof ofTheorem 64.11, weshowed that foreach twosolutions of(a),equation (64.12) isvalid. Therefore, by(64.12), (C) W(y1!y2ix) =c12eF(z)) W(y1.ya;rv) =men’). W(y2.y3;w) =czsem’. where cm,cu,ande23areconstants. Multiplying theﬁrstequation in (c)byya,thesecond byyz,thethird byyl,andadding theresulting equations, weobtain (<1) 1/3W(y1.z/2; 1)+1/2W(y1,1/3; w)+1/1W(1/2,113; w) =(1/3612 -l"32613 -l"?/1¢2a)@F(”)- By(63.53), theleftsideof(d)is (e) 3/s(1/11/2' —yzyi’) +1/2(yn/3’ —yaw’) +yi(y21/3’ ~—2/ayz’). Lesson 65 EXISTENCE THEOREM; LINEAR EqUA'rroN, ORDER n789 which reduces tozero. Hence theright sideof(d)must alsobezero. And since, by(18.86), eF(1);-60forevery 0:,itfollows that (f) H3612 +?/2613 +111623 =0- The solutions yl,ygare,byassumption, linearly independent. Therefore, byComment 64.15, theﬁrst Wronskian in(c)isnotzero, i.e.,cl;sf0. Hence dividing (f)bycm,weobtain ._=_§1§ _CE(E) 1/3 cw2/2 C121/1, from which weseethat yaisobtainable from (b)byaproper choice ofthe constants clandC2in(b). Hence ourassumption that y3isnotobtain- able from (b)isfalse. Theorem 65.6. Letyl,y2,---,y,,benlinearly independent solutions ofthehomogeneous linear diﬁerential equation (65.41), andlet ya='C1]/1 +C2]/2 +‘''+c,,y,, beitscomplementary function. Lety,,beaparticular solution ofthenon- homogeneous linear diﬁerential equation (66.1). Then (65-62) y=y.+l/P isageneral solution of(66.1), i.e.,(66.62) includes every solution of(66.1). Proof. ByTheorem 65.2, atleast oneparticular solution y,,of(65.1) exists. Assume gisasolution of(65.1) notobtainable from (65.62). Since y,,andgareeach solutions of(65.1), wehave (a) y.‘"‘>+f.._1(r)vp‘""" +---+room.=co). 9"“+f.._1(w)v""" +---+fo(w)s1 =Q(x)- Subtracting theﬁrst equation from thesecond, weobtain <9~y.><">+r._.<@>o —1/.>‘""“+---+n<x><g —1/.)=0. Hence y=g—y,isasolution ofthehomogeneous linear equation (65.41). Therefore, byTheorem 65.5, (g—y,,)canbeobtained from (65.61) byaproper choice oftheconstants cl,C2,---,cn.Therefore, (<=) a—up=an/1+am+ +any... 9=1/.»+(<m/1+ am+---+any»)- Wehave thus shown that thesolution gcanbeobtained from (65.62). Hence ourassumption that gisnotobtainable from (65.62) isfalse. 790 OTHER EXISTENCE THEOREMS. WRONSKIAN. Chapter 12 Allstatements made inTheorems 19.2, 19.3, and Comment 19.41, in connection with alinear differential equation oforder n,have now been proved. EXERCISE 65 1.Setupanequivalent system oflinear ﬁrstorder equations foreach ofthe following. Then show that solution y1(:::) oftheresulting system willbe identical with thesolution y(z) ofthegiven equation. (a)vi’-—y’,+3v=$2+1- (C)1/”+22/’+311=tan1-(b)y'+3y --2y=e"+:c. (d)y”’—5y"+8y'—4y=0. 2.Setupanequivalent system oflinear ﬁrstorder equations foreach ofthe following. on111',"4:/+2y=x+1*.1/<0)=1.1/<0)=-1-,<1»)1/')+81/"—1/,41/=,e=.1/<0)=1.rm)=0.,y'<0> =1.(6)y“~—3y"'+ 2/+5v—62/=0.y(0)=0.11(0)=1.1/”(0) =2.y!n(0) =3. ANSWERS 65 1-(11)1/1'(r) =ya (0)y1'(I) =1/2, y2’(w) =yz—3y1+$2+1- 1/2’(w) =-22/2 —3yi+tan@- (b)1/1’(w) =ye. (d)1/{(1) =1/2.1/2'($) =—3z/z+2z/1+e’+ 1- 1/2'(1) =ya, t!a'($) =5y3-'3y2+41/1~ 2-(=1)1/i’(r) =l/2: :'12'(I) =492—2111+1+12.1/1(0) =1,112(0) =—1- (b)y1’(1) =1/2. 1/2'(1) =ya, ya’(w) =—3y3+ 1/2—an+e’.1/1(0)=1,1/2(0)=0.2/3(0)=1-(0)yi’(w) =ye. 1/a’(1) =1/4.y2'($) =ya, 1/4'(I) =3?/4—ya"-5y2+61/1, :1/1(0) =0,1/2(0) =1,713(0) =2,y4(0) =3- Bibliography Intended forstudents whowish toadvance beyond thematerial ofthistext. Bessel Funetions—G. N.Watson. ATreatise ontheTheory ofBessel Functions. New York: Macmillan Co., 1944. Celestial Mechanics—F. R.Moulton. AnIntroduetion toCelestial Mechanics. New York: Macmillan Co.,1914. (Dover, 1970) General Theory——E. L.Ince. Ordinary Diﬂerential Equations. London: Longmans, Green &Co.,1927. (Dover, 1956) Laplace Transforms-—-R. V.Churchill. Modern Operational Mathematics inEngineering. New York: McGraw-Hill, 1944. Legendre Functions—E. W.Hobson. The Theory ofSpherical and Ellipsoidal Harmonics. Cambridge, England: Cambridge University Press, 1931. Mechanics——W. F.Osgood. Mechanics. New York: Macmillan Co., 1946. Numerical Methods—F. B.Hildebrand. Introduction toNumerical Analysis. New York: McGraw-Hill, 1956. Perturbation Theory——F. R.Moulton. AnIntroduetion toCelestial Mechanics. New York: Macmillan Co., 1914. (Dover, 1970) 791 Index Note: Aboldfaced number refers toaproblem inthe exercises. Abel’s formula, 779 Absolute value ofacomplex number, 198 Acceleration, deﬁnition of,138 duetogravity, 140 inpolar coordinates, 461 inrectangular coordinates, 460 Accretion problems, 122-124 Adams’ method, y’=f(z,y), 681 second order equation, 712-3 Airresistance, 141 Airplane problems, seePursuit curves Algebra, ofcomplex numbers, 200-201 fundamental theorem of,197,198 ofoperators, 255-259 Ampliﬁcation ratio, 363 Amplitude, 317 damped, 351 modulated, 345 slowly varying, 345 Analog computation, 375-376 Analytic function, 537,548 onaninterval, 537 Analytic geometry, review of,62 Angular momentum, 472 Apogee, 490 Approximations, toe,639-9, 644-8 tointegrals, seeNumerical methods tolog2,639-10, 644-9 topi(1r),639-8, 644-7 byPicard method, seePicard’s method ofsuccessive approximations bypolynomial interpolation, seePoly- nomial interpolation Arbitrary constants, 129,130 Arclength ofacurve, 111 Arctan:0,expansion of,533,537 Archimedes’ principle, 335 Argument (Arg) ofacomplex number, 199 Arithmetical errors, 636 Associative lawforoperators, ofaddition, 256 ofmultiplication, 257 Atmospheric pressure, 186-188 Axis ofabeam, 383 Backward differences (V,del), 668-670 Beams, axis of,383 bending of,383-389 bending moment of,384 cantilever, 384 condition ofcontinuity ofcurve of,387 795Beams, (continued) condition ofcontinuity ofslope of,387, 388 elastic curve of,384 modulus ofelasticity of,385 neutral surface of,384 simple, 384 simply supported, 384 Young’s modulus of,385 Beats, 346 Bending ofbeams, seeBeams Bending moment, 384 Bernoulli equation, 95-96 Bessel equation, 579,583,609-622 equations leading toa,615-619 ofindex 1/2,579 ofindex zero, 583 Bessel functions, 611-622 oftheﬁrstkind, J;,(x), 611-615 J06”), J1($)1 614 J._k(z), 613 modiﬁed Bessel functions oftheﬁrst kind, 619 properties ofBessel functions ofthe ﬁrst kind, J,,(:::), 619-622: integral, 620-622; orthogonal, with respect to weight function 2:,622;zeros, 619-620 ofthesecond kind, —N,,(:z:), 612 Beta function, 306 Binomial series, 599 Biological problem, 447-451 Body falling inwater, 144 Bounded region, 14 Bridge, over gorge, 362 troops crossing a,362 Buoyancy, 335 Cable, hanging, 507-512 suspension, 507-514 Calorie, 185 Cantilever beam, 384 Capacitance, 370 Capacitor, 369 inseries, 455-3 Carbon—l4 test, 5 Catenary, 510 Cave, prehistoric, 2 Center ofattraction, 470 Central force, 380, 470 Seealso, Particle inmotion inspace subject toacentral force 796 INDEX Centrifugal force, 380 Centripetal force, 380 Chain around acylinder, 188-189 Chain sliding from atable, 190 Characteristic equation, 212-220, 444 deﬁnition of,212 roots of:realanddistinct, 213-214; real butsome multiple, 214-217; imagi- nary, 217-220 Charcoal problem, 2 Charge, electric, 370 Circuit, closed electric, 369 Seealso, Electric circuits Clairaut equation, 757-760 geometric problems giving riseto,761 parabolic reﬂector, 759-760 Closed electric circuit, 369 Coeﬂicient, offriction, 160 ofinductance, 370 ofresistance, 348 ofsliding friction, 160 ofstatic friction, 160 Combination, linear, 205 Commutative lawforoperators, ofaddi- tion, 256 ofmultiplication, 258 Complementary function, 210 Complete elliptic integral oftheﬁrstkind, 333 Complex electric circuits, 451-455 Complex functions, 201-203 exponential, 201 hyperbolic, 203 trigonometric, 201 Complex numbers, 197-201 absolute value of,198 algebra of,200-201 argument (Arg) of,199 conjugate of,198 deﬁnition of,197 imaginary part of,197 polar form of,199 realpart of,197 rectangular form of,199 Complex variables, 201-203 solution byuseof,230 Compound interest, 128-5, 6 Computation, analog, 375-376 Condition ofcontinuity, ofacurve, 387 ofaslope, 387, 388 Conductivity, thermal, 185 Conic, equation inpolar coordinates, 483 Conjugate ofacomplex number, 198 Conservation, ofangular momentum, 472 ofenergy, 329,475 Conservative ﬁeld, 473 Construction ofatable ofLaplace trans- forms, 302-306, 309-311Continuing formulas, Adams, 681,712-3 improvement ofpolygonal method, 641-643 polygonal, seePolygonal method Runge-Kutta, seeRunge-Kutta method Taylor series, 645-652 third degree interpolating polynomial, 676 Continuing methods, 632 Continuous function, 732 onaninterval, 730 atapoint, 730 inaregion, 731 Convergence, ofimproper integral, 292, 294 interval of,532 ofpower series, 531-532 ratio testfor,532 ofasequence offunctions, 728,732 uniform: ofasequence offunctions, 729,732; ofaseries offunctions, 732-733 Coordinates ofapoint, 6 Corrector formulas, 680 ﬁfth degree polynomial, 676 Simpson, 675 sixth degree polynomial, 677 three~eights rule, 681 trapezoidal, 675 Weddle’s rule, 677 Corrector methods, 632 Cosine, hyperbolic, 203 Cosine 2:,cosine z,expansions of,201 Coupled springs, 440-443 Critically damped, 350 Cumulative errors, 636,690 Current, electric, 369,370 induced, 185 steady state, 372 transient, 372 Curvature, radius of,528 Curve, elastic, 384 Curves, envelopes ofafamily of,748- 754 ofpursuit, 168-175, 523-525 Cycloid, 335 D,differential operator, 251 D-", inverse operator, 270 Damped, amplitude, 351 critically, 350 frequency, 351 harmonic motion, 348-353 motion, 347-353, 359-364; deﬁnition, 347; forced, 359-364; free (damped harmonic), 348-353 period, 351 periodic, 351 Damping factor, 351 Decomposition problems, 131-132 Decrement, logarithmic, 356 Degenerate systems, 413-415 Dependence, seeLinear dependence of functions Determinant, deﬁnition of,399, 770,771 Grammian, 777 ofasystem ofequations, 400 theory of,770-774 Wronskian, 774 Diagram, lineal element, 39 Differences, backward, 668-670 ﬁnite, 659-661 forward, 659-668 tables of,660,662,663,670 Diﬁerential, exact, 72 operator, 251 total, 72 Differential, ofafunction, 47-51 ofoneindependent variable, 48 oftwoindependent variables, 50 Differential equations, Bernoulli, 95-96 Bessel, 579,583, 609-622 Clairaut, 757-760 deﬁnition ofordinary, 20 Euler, 247 exact, 70-78, 248-26: deﬁnition of,73; recognizable, 80-82 existence ofsolutions of,seeExistence anduniqueness theorems explicit solution ofa,22 ﬁnding a,from itsn-parameter family, 31-33 ﬁnding anintegrating factor, 84-90, 94-95, 248: forﬁrst order, 84-90; forlinear ﬁrstorder, 94-95; forlinear second order, 248 ofﬁrstorder, seeFirst order equations Gauss’s, 587 general solution of,28-37: deﬁnition, 35 Hermite, 607 with homogeneous coefficients, 57-60 homogeneous inx,505 hypergeometric, 589 implicit solution of,24 initial conditions, 36 integral curve of,deﬁnition, 39 integrating factor of,seeIntegrating factors Laguerre, 624-630 Legendre, 586,591-605, 606-13 with linear coefficients, 63-68 linear: ofﬁrst order, 91-95; ofhigher order, seeLinear diﬁerential equations miscellaneous ﬁrstorder, 101-103 multiplicity ofsolutions of,28 n-parameter family ofsolutions of,30 nonlinear, seeNonlinear equations numerical solutions of,seeNumerical methods order of,21INDEx 797 Diﬁerential equations, (continued) ordinary, 20 partial, 20 particular solution of,35 recognizable exact, 80-82 Riccati, 97,247-22, 23,24with separable variables, 51-55 singular, point of,43:solution of,34 solution ofa,deﬁnition, 22 systems of,seeSystem ofequations Tschebyscheff, 589 variables separable, 51-55 Dilution problems, 122-124 Dirac 5-function, 344 Direction ﬁeld, 39 construction ofa,38-41 isoclines ofa,40-41 Distributive lawofmultiplication for operators, 258 Divergence ofimproper integrals, 292 Domain ofindependent variable, 9,11 e,anapproximation to,639-9, 644-8 e‘,e‘,expansion of,201 Elastic curve, 384 Elastic helical springs, 324-326 parallel, 330-13 inseries, 330-14 Elasticity, modulus of,385 Electric circuits, complex, 451-455 ﬁrstorder, 184-185, 377-7, 10 simple, 369-375 Electromotive force, 369 induced, 185 Element, line, 39 lineal, 39 ofaset,6 Elementary functions, 17 Eliminant, 750 Ellipse, equation inpolar coordinates, 483 Elliptic integral oftheﬁrstkind, complete, 333 incomplete, 334 Energy, inverse square law, 487-488 kinetic, 329,475 lawofconservation of,329,475 potential, 329,473 Envelopes ofafamily, ofcurves, 747-754 deﬁnition of,748 eliminant of,750 ofsolutions, 754-757 Equation, indicial, 574 Equilibrium position, 317 Equivalent triangular system, 405,406, 416 Error, inﬁfth degree polynomial, 679 function, 670 inanimprovement ofpolygonal method, 644-3 inMilne method, 688-689 798 INDEx Error, (continued) First order equations, (continued) inNewton’s interpolation formulas, 670-671 inpolygonal method, 636-638 inpolynomial interpolation, 670-671, 679 inRunge-Kutta, 657-658 inSimpson’s rule, 679-684 inTaylor series, 537,649-652, 653-4- inthird degree polynomial, 679 intrapezoidal rule, 679,683 inWeddle’s rule, 679 Errors, arithmetical, 636 cumulative, 636,690 formula, 636,690 general comment on,636, 690-691, 703 rounding off,636,690 Escape velocity, 148 Euler equation, 247 Exact differential, 72 Exact differential equation, 70-78, 248-25 definition of,73,248-25 necessary andsufficient condition for, 73,248-25 recognizable, 80-81 solution of,73,76 Existence anduniqueness theorems, 720 ﬁrstorder equation y’=f(z,y),734-743 linear equation oforder n,783-786 nonlinear equation oforder n,765-767 system: ofnﬁrstorder equations, 763- 765; ofnlinear ﬁrst order equations, 768-770 Explicit solution, 21-23 deﬁnition of,22 Exponential complex function, 201 Exponential shift theorem, forinverse operators, 277 forpolynomial operators, 260 F-2region ofatmosphere, 155 Factorial function (nl), 306-308, 596 Factors, integrating, seeIntegrating factors ofpolynomial operators, 258 Falling bodies, seeVertical motion Faltung theorem, 303 Family, ofcurves, envelopes of,748-754 n-parameter, ofsolutions, 30,31-33: envelopes of,754-756 Field, offorce, 473 conservative, 473 direction, 39 slope, 39 Fifth degree interpolating polynomial, 676, 678,679 Finite differences, 659-661 First order equations, Bernoulli, 95-96 Clairaut, 757-760exact, 70-78 existence anduniqueness theorem for, 734-743 with homogeneous coelﬁcients, 57-60 integrating factors of,84-90, 94-95 linear, 91-95 with linear coefficients, 63-68 miscellaneous, 101-103 perturbation theory, 713-715 problems giving riseto,seeProblems, ﬁrstorder equations recognizable exact, 80-82 Riccati, 97,247-22, 23,24 with separable variables, 51-55 solution ofy’=f(z,y): bynumerical methods, seeNumerical methods; by Picard’s method, 720-723; byseries method, 548-553 First order, processes, 137 systems, seeSystem ofﬁrstorder equations Flow through anoriﬁce, 183 Force, central, seeParticle inmotion in space subject toacentral force centrifugal, 380 centripetal, 380 damping, 349,350,351 electromotive, 369 ﬁeld, 473 ﬁeld of,473 frictional, 160 function, 473 ofgravity, 140 impressed (forcing function), 338,360 induced electromotive, 185 intermittent, 344 inversely proportional: tocube ofdis- tance, 521-522; tosquare ofdistance 481-488, 494 moment of,381 proportional todistance, 476-479 Forced motion, with damping, 359-364 undamped, 338-342 ' Forcing function, 338 Formula errors, 636,679 Forward differences, (A,delta), 659- 668 Fourth degree interpolating polynomial, 681-2 Fractions, partial, 283-284 ofinverse operators, 287,289 Free motion, damped, 347-353 undamped, 313-329 Seealso, Simple harmonic motion Frequency, damped, 351 impressed, 339 natural (undamped), 318,319 normal, 443 resonance, 373,443 Frequency, (continued) undamped resonant, 341 Friction, coefficient of,160 sliding, 160 static, 160 Frictional force, 160 Frobenius, method, 572; seealso, Solution about aregular singularity series, 572 f(a), deﬁnition of,11 f(b,y), meaning of,18-10 f(:v,a), deﬁnition of,13 Function ofoneindependent variable, 6-11, 14-17, 48 deﬁnition of,6,9 deﬁnition off(a), 11 differential ofa,48 elementary, 17 implicit, 14-17 range of,9 Function oftwoindependent variables, 11-14, 18,50,57-58 deﬁnition of,11 deﬁnition: off(x,a), 13;off(b,y), 18- 10 differential ofa,50 domain ofdeﬁnition ofa,11 homogeneous, oforder n,57,58 range of,11 Functions, analytic, 537: onaninterval, 537 Bessel, seeBessel functions beta, 306 complementary, 210 complex, seeComplex functions continuous, 730, 731: onaninterval, 730; atapoint, 730; inaregion, 731 Dirac 6,344 elementary, 17 elliptic, 333, 334 error, 670 factorial (nl), 306-308, 596 force, 473 forcing, 338,360 gamma, 306-309 homogeneous, oforder n,57,58 hypergeometric, 587 Legendre, 594: ofsecond kind, Qk(x), 597 linear dependence of,seeLinear dependence offunctions linear independence of,seeLinear independence offunctions orthogonal, 602 polynomial interpolating, 662 remainder, 670 sequence of,728, 729,732 series of,732-733 unit impulse, 344 vector point, 473INDEX 799 Fundamental theorem, ofalgebra, 197,198 ofcalculus, 70 Gamma function, 306-309 deﬁnition of,307 Gauss’s equation, 587 General solution, ofadiﬁerential equation, 28-37 deﬁnition ofa,35 ofaﬁrstorder linear equation, 93 ofahomogeneous linear equation, 210, 788 ofanonhomogeneous linear equation, 210, 789 Geometric problems leading to,Clairaut equation, 761 ﬁrstorder equation, 107-111 special types ofsecond order equations, 528-530 Grammian, 777 Graphical solutions, 38-44, 424-438 Gravitation, Newton’s universal lawof, 491 Gravitational constant, 139,491 Gravity, force of,140 speciﬁc, 153 Growth problems, 131-132 Halley’s comet, 492 Hanging cable, 507-514 Harmonic motion, damped, 348-353 simple, seeSimple harmonic motion undamped, seeSimple harmonic motion Harmonic oscillators, 323-329, 377 elastic helical spring, 324-326 simple pendulum, 327-329 Heat, speciﬁc, 130 steady state ﬂow of,185-186 Helical spring, seeElastic helical springs Hermite, equation, 607 polynomials, 607 Homogeneous function, order ofa,57,58 Homogeneous linear equations, seeLinear differential equations Homogeneous inx,equation, 505 Hooke’s law,324 Horizontal motion, 160-162 Hyperbolic, cosine, 203 sine, 203 tangent, 203 Hypergeometric, equation, 587 function, 587 series, 587 i(current), 369,370 i(imaginary unit), 197 Identically zero, meaning of,775 Imaginary number, part ofacomplex number, 197 pure, 197 Impedance ofacircuit, 372 800 INDEX Implicit function, 14-17 deﬁnition of,16 Implicit solution ofanequation, 24-27 deﬁnition of,24 Impressed, frequency, 339 force (forcing function), 338 Improper integrals, convergence of,292, 294 divergence of,292 Improvement ofpolygonal method, 641-643 Impulse function, unit, 344 Impulsive response ofasystem, 344 Inclined motion, 164-166 Incomplete elliptic integral oftheﬁrst kind, 334 . Independence offunctions, seeLinear independence offunctions Independent solutions oflinear equation, number of,208, 786 Indicial equation, 574 Induced, current, 185 electromotive force, 185 Inductance, coefficient of,370 mutual, 454 Induction, mathematical, 250 Inductor, 369 Inertia, 381 moment of,381 rotational, 381 Inﬁnite series, seeSeries Initial conditions, 33-37 deﬁnition of,36 number of,36 Input ofasystem, 341,360,372 Integrable combinations, 81 Integral, elliptic oftheﬁrstkind, complete, 333 incomplete, 334 Integral, improper, 292,294 convergence of,292 divergence of,292 Integral, Riemann, 70 Integral curve, deﬁnition of,39 Integrals, approximations to,see Numerical methods Integrating factors, 82-90 deﬁnition of,82 ofﬁrstorder equation, 84-90, 94-95 ofsecond order linear equation, 248 Interest problems, 126-127 Intermittent force, 344 Interpolation, seePolynomial inter- polation Interval, 6 Interval ofconvergence, 532 Inverse Laplace transform, deﬁnition of, 296 linear property of,296Inverse operators, 268-282 deﬁnition of,269,270 exponential shift theorem for,277 meaning of,269-271 partial fraction expansion of,287 series expansion of,272-277 solution oflinear equation by,272-282, 288-291 Inverse square law,481-488, 491, 494 Irregular singularity, 572 Isoclines ofadirection ﬁeld, 40-41 Isogonal trajectory, 115-117 deﬁnition of,115 Isothermal surface, 185 Kepler’s laws, 492 Kinetic energy, 329,475 Kirchhoff’s, ﬁrstlaw,451 second law,370 Kutta, seeRunge-Kutta formulas Laguerre equation, 624-630 Laguerre polynomials, Lk(:c), 625-630 properties of,627-630: analogue of Rodrigue’s formula, 627-629; inte- gral property, 629-630; orthogonal with respect toweight function e", 630 Laplace transforms, construction oftable of,302-306, 309-311 deﬁnition of,294 Faltung theorem, 303 inverse, 296 properties of,295-296 solution ofalinear equation by, 296-302 solution ofasystem by,418-420 tables of,306,310 Law ofconservation, ofangular momen- tum, 472 ofenergy, 329,475 Law ofthemean, 731 ofuniversal gravitation, 491 Legendre equation, 586,591-605, 606-13 comment onsolution of,593-594 functions, 594: ofthesecond kind, Qt(w), 597 Legendre polynomials, P;,(x), 597-605 properties of,598-605: coeﬂicients in binomial series expansion, 598-600; orthogonal property, 602-604; other integral properties, 604-605; recur- sion formula, 601; Rodrigue’s for- mula, 602; values ofP,,(0), P,,(1), P,,(—1), 600-601 Libby, Dr.Willard F.,5 Line elements, 39 Lineal element diagram, 39 Lineal elements, 39 Linear coefficients, 62-69 Linear combinations, deﬁnition of,205 Linear dependence, offunctions, 205, 775—Comment 63.56, 777—~5 deﬁnition of,205 ofsolutions, 781——Comment 64.15 Linear differential equations, Bessel, see Bessel equation characteristic equation of,212-220: deﬁ- nition, 212;roots imaginary, 217-220; roots realanddistinct, 213-214; roots realbutsome multiple, 214-217 complementary function of,210 deﬁnition of,92,196 Euler, 247 exact, 248 existence theorem for,seeExistence and uniqueness theorems ﬁrst order, 91-95: deﬁnition of,92; general solution of,93;integrating factor for,94;solution of,92 form ofsolution of,211-220 fundamental theorems for,207,208 Gauss's, 587 general solution of,93,210,788 Hermite, 607 homogeneous, with constant coeffi- cients, 211-220: deﬁnition of,196; with nonconstant coefficients, 241- 246 hypergeometric, 587 integrating factors for,94,248 Laguerre, 624-630 Legendre, seeLegendre equation nlinearly independent solutions of,208, 786 nonhomogeneous, 221-246: with con- stant coefficients, 221-240; deﬁnition of,196;with nonconstant coefficients, 236-237, 241-246ordinary point of,570 reduction toasystem, 784-785 singularity of,570: irregular, 572; regu- lar,571 solution of,by:complex variables, 230- 231; inverse operators, 272-282, 288- 291; Laplace transforms, 296-302; method ofFrobenius, 572-584; par- tial fraction expansion ofinverse operators, 288-291; polynomial op- erators, 262-265; power series, 537- 546; reduction oforder, 242-246; un- determined coeﬁicients, 221-230; va- riation ofparameters, 233-240 systems of,seeSystem entries Tschebyscheff, 589 uniqueness theorem for,seeExistence anduniqueness theoremsINDEX 801 Linear independence, offunctions, 205, 775—Comment 63.56, 777-5 deﬁnition of,205 ofsolutions, 781-Comment 64.15 Linear property of,inverse operators, 296 Laplace transformation, 295 polynomial operators, 253 Linearization ofﬁrst order systems, 424- 438 Lipschitz condition, 731,734,764,766 Liquid, ﬂowing through anoriﬁce, 183 rotating inacylinder, 193-194 Logarithmic decrement, 356 Maclaurin series, 535 Magniﬁcation ratio, 363 Mass, variable, 191-193 Mathematical induction, 250 Mean, lawofthe,731 Method of,Frobenius, 572; seealso, Solu- tionabout aregular singularity reduction oforder, 242-246 , undetermined coefficients, 221-230 variation ofparameters, 233-240 Milne method, comment onerror in,688- 689 forsecond order equation, 710-711 forsystem oftwoﬁrstorder equations, 703 forthird order equation, 712—4- fory’=f(z,y), 684-689 Modulated amplitude, 345 Modulus, ofelasticity, 385 Young's, 385 _ Moment, bending, 384 offorce, 381 ofinertia, 381 Momentum, angular, 472 Motion, ofacomplex system, 189-191 damped: forced, 359-364; free(damped harmonic), 348-353 forced: damped, 359-364; undamped, 338-342 free: damped (damped harmonic), 348- 353; undamped, seeSimple harmonic motion horizontal, 160-162 inclined, 164-166 ofaparticle: onacircle, 316-317; in space, seeParticle inmotion inspace subject toacentral force; ona straight line, 138-166, 314-316 period of,318,333,351,477 planetary, 491-492 ofaprojectile, 463-465 simple harmonic, seeSimple harmonic motion stable, 339 steady state, 361 802 INDEX Motion, (continued) transient, 361 undamped: forced, 338-342; free, see Simple harmonic motion unstable, 340 vertical, seeVertical motion Multiplicity ofsolutions, 28-31 Mutual inductance, 454 Natural (undamped) frequency, 318, 319 Neutral surface, 384 Newton’s, ﬁrstlawofmotion, 138 interpolation formulas, 663-671: back- ward, 669; error in,670, 671; for- ward, 667 lawofuniversal gravitation, 491 proof ofinverse square law,494-495 second law‘pfmotion, 138,459 Nonhomogeneous linear equation, deﬁni- tionof,196 Nonlinear equations, existence and uniqueness theorem for,765-767 numerical solution of,seeNumerical methods reduction toasystem, 766-767 series solution of,562-567 Seealso, Special types ofsecond order equations Normal, coordinates, 444 frequencies, 443 n-parameter family ofsolutions, deﬁnition of,30 ﬁnding adifferential equation from an, 31-33 Number, complex, 197 pure imaginary, 197 real, 197 Numerical methods, 631-718 Adams’, 681,712-3 choosing sizeofh,691-692 continuing, 632 corrector, 632 decreasing h,692-694 difference tables, 660, 662,663,670 errors, seeError; Errors ﬁfth degree polynomial, 676, 678, 691, 703 ﬁnite differences, 659-661 forﬁrstorder equation y’=f(z,y), 632- 658, 681-6, 684-688, 690-701, 713- 715 fourth degree polynomial, 681-2 illustrative example andsummary, 694- 701 improvement ofpolygonal method, 641- 643 increasing andreducing h,692-694 Milne, seeMilne methodNewton’s interpolation formulas, 663- 671: backward, 669; forward, 667 perturbation theory, 713-718: ﬁrstorder equation, 713-715; second order equa- tion, 716-718 Picard’s, seePicard’s method ofsucces- siveapproximations polygonal, 632-638 polynomial interpolation, seePoly- nomial interpolation reducing andincreasing h,692-694 Runge-Kutta, seeRunge-Kutta for- mulas forasecond order equation, 707-711, 712-3, 715-718 series, 645-652 Simpson’s rule, 677,678,684 sixth degree polynomial, 677 starting methods, 632 Seealso, Starting methods summary andanexample, 694-701 forasystem of:three ﬁrst order equa- tions, 726-2; twoﬁrst order equa- tions, 702-706, 723-726 Taylor series, 645-652 third degree polynomial, 676,678 forathird order equation, 712-4- three-eighths rule, 681 trapezoidal rule, 675,678, 682 Weddle’s rule, 677,678, 691,703 Oceanic pressure, 186-188 Operator, differential, 251 Seealso, Inverse operators; Poly- nomial operators Order, ofadifferential equation, 21 ofahomogeneous function, 57,58 Ordinary differential equation, deﬁnition of,20 Ordinary point, ofy’=f(z,y), 43,744-747 ofalinear differential equation, 570 Oriﬁce, flow through an,183 Orthogonal, deﬁnition of,functions, 602 property: ofBessel functions, 622; of Laguerre polynomials, 630; of Legendre polynomials, 602-604 Orthogonal trajectory, inpolar coordi- nates, 118 inrectangular coordinates, 117 Oscillator, harmonic, 323-329, 377 Output ofasystem, 341,360,372 Overdamped, 349 Parabolic reﬂector, 759-760 Parameters, number of,30 variation of,233-240 Parasite problem, 447-451 Partialdifferential equation, meaning of,20 Partial fraction expansion, 283-284 ofinverse operators, 287,289 Particle, moving, inacircle, 316-317 inaplane, seeParticle inmotion in space subject toacentral force inrotating tube, 380 onastraight line, 138-166, 314-316 Particle inmotion inspace subject toa central force, 470-496, 521-522 force inversely proportional tocube of distance, 521-522 force inversely proportional tosquare ofdistance, 481-488, 494: determin- ingconstants ofintegration, 484-486; energy considerations, 487-488; plan- etary motion, 491-492 force proportional todistance, 476-479 Kepler's laws, 492 lawofconservation: ofangular momen- tum, 471-472; ofenergy, 475 moves inplane, proof that particle, 470-471 period of,477, 492 satisﬁes lawofconservation: ofangular momentum, 471-472; ofenergy, 475 special central force problem, 521-522 sweeps outequal areas inequal times, proof that particle, 472 Particular solution ofadifferential equa- tion, 33-37 deﬁnition of,35 Pendulum, simple, 327-329 actual period of,333 amplitude of,328 period independent ofamplitude, 334-34 weight ofwire notnegligible, 331-29 Perigee, 490 Period, 318,477 actual, ofasimple pendulum, 333 damped, 351 Perturbation method, 713-718 ﬁrstorder equation, 713-715 second order equation, 715-718 Phase andphase angle, 318 Picard’s method ofsuccessive approxima- tions, 720-726 forﬁrstorder equations, 720-723 forasystem: ofthree ﬁrst order equa- tions, 726—9; oftwoﬁrstorder equa- tions, 723-726 Plane analytic geometry, areview of, 62-63 Plane motion ofaparticle, seeParticle in motion inspace subject toacentral force Plane motion ofaprojectile, 463-465 Planetary motion, 491-492Innax 803 Point, ordinary, seeOrdinary point singular, 43,744-747 Polar form ofacomplex number, 199 Polygonal method, 632-638 comment onerrors in,636-638 animprovement of,641-643 Polynomial interpolation, 661-684 Adams’, 681,712-3 error in,670-671, 679 ﬁfth degree polynomial, 676,678,679 function, 662 Newton’s: (backward) formulas, 669; (forward) formulas, 667 Simpson’s rule, 675,678,679,684 sixth degree polynomial, 677 third degree polynomial, 676,678,679 three-eighths rule, 681 trapezoidal rule, 675,678,679,682 Weddle’s, 677,678,679 Polynomial operators, 251-265 algebraic properties of,255-259 associative law: ofaddition, 256; ofmultiplication, 257 commutative law: ofaddition, 256; ofmultiplication, 258 deﬁnition of,251 distributive lawofmultiplication, 258 exponential shift theorem for,260: corollaries to,261 factoring of,258 linear property of,253 oforder n,251 P(D +a),deﬁnition, 259 P(D)y, deﬁnition, 252 product ofh(x)by,256 product oftwo, 257 solution: oflinear equations by,262- 265; ofsystems oflinear equations by,398-417 sum oftwo, 255 Polynomials, Hermite, 607 interpolation by,seePolynomial inter- polation Laguerre, 625-630 Legendrie, 597-605Tschebyscheﬁ, 589 Potential, 473 energy, 329, 473 Power series, seeSeries; Solution byseries methods Pressure, atmospheric andoceanic, 186- 188 Principal value of0,199 Principle ofsuperposition, 211,254 Problem involving acentrifugal force, 380 Problems, ﬁrst order equations, 107-195, 377-7, 10 accretion, 122-124 804 Inonx Problems, ﬁrstorder equations, (continued) atmospheric pressure, 186-188 body falling inwater, 144 chain around acylinder, 188-189 decomposition, 131-132 dilution, 122-124 electric circuit, 184-185, 377-7, 10 ﬁrstorder processes, 137 ﬂow ofheat, steady state, 185-186 flowthrough anoriﬁce, 183-184 geometric, 107-111 growth, 131-132 horizontal motion, 160-162 inclined motion, 164-166 interest, 126-127 isogonal trajectory, 115-117 moon, 151-6, 7,3;156-38, 39 motion ofacomplex system, 189- 191 oceanic pressure, 186,187 orthogonal trajectories, 117-120 pursuit curves, 168-175 raindrop, 143 relative pursuit curves, 177-182 rocket motion, 191-193 rope around acylinder, 188-189 rotation ofaliquid inacylinder, 193-194 second order processes, 134-137 steady state ﬂowofheat, 185-186 straight linemotion, 138-166 temperature, 129-130 variable mass, 191-193 vertical motion, 139-151 Problems, linear second order equations, 313-389 bending ofbeams, 383-389 circle, particle moving oncircumference of,316-317 damped harmonic motion, 347-353: deﬁnition of,347 damped motion: forced, 359-364; free, 347-353 elastic helical spring, 324-326: inparal- lel,330-13; inseries, 330-14 electric circuit, simple, 369-375 forced motion with damping, 359-364 forced undamped motion, 338-342 freedamped motion, 347-353 freeundamped motion, 313-329 Seealso, Simple harmonic motion particle moving: oncircumference ofa circle, 316-317; onastraight line, 314-316 pendulum, simple, seePendulum problem involving acentrifugal force, 380 rolling bodies, 381-383 simple electric circuit, 369-375Problems, linear second order equations, (continued) simple harmonic motion, 313-329 Seealso, Simple harmonic motion straight line, particle moving on,314- 316 twisting bodies, 383 undamped motion, 313-342: examples of,323-329; forced, 338-342; free, 313-321 Problems, special types ofsecond order nonlinear equations, 506-530 central force, 521-522 geometric, 528-530 pursuit, 523-525 special central force, 521-522 suspension cable, 507-514 Problems, system ofequations, 440-492 biological, 447-451 central force, seeParticle inmotion in space subject toacentral force electrical, 451-455 mechanical, 440-443 particle inmotion inspace, seeParticle inmotion inspace subject toacentral force plane motion: ofaprojectile, 463-465; ofaparticle, seeParticle inmotion in space subject toacentral force planetary motion, 491-492 projectile inplane, 463-465 Product oftwooperators, 257 Projectile, motion ofa,463-465 P_roof bymathematical induction, 250 Properties of,Bessel function ofﬁrstkind, 619-622 Laguerre polynomials, 627-630 Laplace transforms, 295-296 Legendre polynomials, 598-605 polynomial operators, seePolynomial operators Pure imaginary number, 197 Pursuit curves, 168-175, 523-525 relative, 177-182 Radioactive material, 2 Radius ofcurvature, 528 Raindrop problem, 143 Range ofafunction, 9,11 Ratio testforconvergence, 532 Real number system, 197 Real part ofacomplex number, 197 Rectangular form ofacomplex number, 199 Recursion formula, 577 forBessel equation, 610 forHermite polynomial, 607-18 forLaguerre equation, 625 forLegendre polynomial, 601 Reduction oforder method, 241-246 Region, bounded, 14 F-2ofatmosphere, 155 simply connected, 71 Regular singularity, 571 atco,586 Relative pursuit curve, 177-182 Remainder function, 670 Resistance, coefficient of,348 electric, 370 Resistors, 369 inseries, 454-2 Reonance, 373,443 undamped, 341 Resonant frequency, undamped, 341 Response ofasystem, impulsive, 344 Riccati equation, 97,247 Riemann-integrable, 70,731 Riemann integral, 70 Rocket motion, 191-193 Rodrigue’s formula, 602 Rolling bodies, 381-383 Roots, inconjugate pairs, 204-9 imaginary, 217-220 ofindicial equation, 574-584 realanddistinct, 213-214 realbutsome multiple, 214-217 Rope around acylinder, 188-189 Rotating tube, particle ina,380 Rotation ofaliquid, 193-194 Rotational inertia, 381 Rounding offerrors, 636,690 Runge-Kutta formulas, comment on error in,657-658 forsecond order equation, 709 forsystem oftwoﬁrstorder equations, 702-703 fory’=f(z,y), 653-658 Satellites, 488-491, 496 Second order equations, 500-504, 707-711, 712-1, 3 numerical solution of,707-711: by Adams’ method, 712-3; error in, 712-1; byMilne’s method, 710-711; byRunge-Kutta method, 709 special types ofnonlinear, 500-504 Second order processes, problems, 134- 137 Separable variables, 52 Separated variables, 52 Sequence offunctions, convergence ofa, 728, 732 uniform convergence ofa,729,732 Series, binomial, 599 convergence of,531,532 forcos.1:andcosz,201 deﬁnes afunction, 533 determination ofcoefficients ofa,534INDEX 805 Series, (continued) differentiation ofa,534 fore‘ande‘,201 equality of,534 expansion ofinverse operators by, 272-277 Frobenius, 572: solution ofhomo- geneous linear equation by,572-584; solution ofnonhomogeneous linear equation, 585-18 offunctions, 732-733: converges uni- formly, 732,733 hypergeometers, 587 interval ofconvergence of,532 Maclaurin, 535 partial sums of,732 power, 531 ratio testforconvergence of,532 forsin:0andsin2,201 solutions by,seeSolution byseries methods fortan1:,742 Taylor, 535: with remainder, 537 Set,elements ofa,6 meaning ofa,5-6 Simple beam, 384 Simple electric circuit, 369-375 Simple harmonic motion, 313-329 amplitude, 317 deﬁnition of,314 description ofthemotion, 315,317 elastic helical spring, 324-326 equilibrium position, 317 examples ofbodies moving in,323-329 frequency, natural (undamped), 318, 319 harmonic oscillators, 323-329 natural (undamped) frequency, 318,319 particle moving: oncircumference ofa circle, 316-317; onastraight line, 314-316 pendulum, 327-329 period, 318 phase andphase angle, 318 simple pendulum, 327-329 spring, helical, 324-326 Simply, connected region, 71 supported beam, 384 Simpson’s rule, 675,678,679,684 Simultaneous equations, seeentries under System Sine, hyperbolic, 203 Sine x,sin2,expansions of,201 Singular point ofy’-f(z,y), 43,744-747 Singular solution, 34 Singularity ofalinear equation, 570 irregular, 572 regular, 571 regular atco,586 806 Inonx Sixth degree interpolating polynomial, 677 Sliding friction, 160 Slope ﬁeld, 39 Slowly varying amplitude, 345 Solution about aregular singularity, 572- 584,585, 586 ofahomogeneous linear equation, where roots ofindicial equation: differ by aninteger, 578-582; donotdiffer by aninteger, 574-578; equal, 583-584 atco,586 ofanonhomogeneous linear equation, 585-18 Solution byseries methods, 537-584 byFrobenius series, seeSolution about aregular singularity byTaylor series: ofalinear equation, 537-546; ofanonlinear equation of order n,562-567; ofasystem ofﬁrst order equations. 555-559; ofasystem oflinear ﬁrst order equations, 559- 562; ofy’==f(z,y), 548-553 Solution ofasystem ofequations, 394-420 ofaﬁrstorder linear system, 396-397 ofaﬁrstorder system, 394-396 ofalinear system with constant coeffi- cients, 398-420: byLaplace trans- forms, 418-420; bypolynomial opera- tors, 398-417 bynumerical methods, 702-706 byPicard’s method, 723-726, 726-9 byseries method, 555-562 Seealso, each heading under System Solutions ofdifferential equations, by choice ofmethod, 99 bycomplex variables, 230-231 deﬁnition of,22 existence of,seeExistence andunique- nesstheorems explicit, 21-23 general, 28-37: deﬁnition of,35 implicit, 24-27 byintegrating factors, 82-90, 94 byinverse operators, 272-282, 288-291 byLaplace transforms, 296-302 bylineal element diagram, 39 linear combination of,208 bymethod ofFrobenius, 572-584 multiplicity of,28-31 number ofindependent, 208,786 bynumerical methods, 631-718 byoperators, 262-265, 272-282, 288-291 n-parameter family of,30,31-33 bypartial fraction expansion ofinverse operators, 288-291 particular, 35 byPicard’s method ofsuccessive approximations, 720-726Solutions ofdiﬁerential equations, (continued) bypolynomial operators, 262-265 bypower series, 537-553, 562-567 byreduction oforder, 242-246 byseparation ofvariables, 51-55 byseries, 537-553, 562-567 singular, 34 steady state, 361,372 bysubstitution andother means, 101 bysuccessive approximations, 720-726 transient, 361,372 byundetermined coeﬁicients, 221-230 uniqueness of,seeExistence andunique- nesstheorems byvariation ofparameters, 233-240 Wronskian of,781—Comment 64.15 Solve adifferential equation, meaning of, 22,81Special central force problem, 521-522 Special types ofsecond order equations, 500-504 2absent, 503-504 xandy’absent, 500-501 yabsent, 502-503 Speciﬁc, gravity, 153 heat,130weight, 512 Speed, ofescape, 148 terminal, 142 Spring, helical, seeElastic helical springs Spring constant, 324 Sputnik, 491-7 Stable motion, 339 Starting formula, 680 Starting methods, 632 polygonal method, seePolygonal method: animprovement of,641-643 Runge-Kutta, seeRunge-Kutta formulas Taylor series, 645-652 Static friction, 160 Steady state, current, 372 ﬂow ofheat, 185-186 motion, 361 solution, 361,372 Stiffness coefficient ofaspring, 324 Stiﬁness constant, torsional, 383 Straight linemotion, 138-166, 314-316 Subnormal, 112-1 Substitution, solving anequation by,101 Subtangent, 112-1 Successive approximations, seePicard’s method ofsuccessive approximations Sum oftwooperators, 256 Superposition principle, 211,254 Surface, isothermal, 185 neutral, 384 Suspension cable, 507-514 System ofequations, meaning ofasolution ofa,393 Seealso, System ofﬁrstorder equa- tions; System oflinear equations with constant coefficients; Sys- temoflinear ﬁrstorder equations System ofﬁrstorder equations, deﬁnition ofa,394 errors innumerical solution of,703 existence anduniqueness theorem for, 763-765 linearization of,424-438 Milne formula fora,703 Picard’s method ofsolution ofa,723- 726,726-9 problems giving risetoa,seeProblems, system ofequations Runge-Kutta formula fora,702-703 series method ofsolution ofa,555-562 solution ofa:bynumerical methods, 702-706; byPicard’s method, 723- 726, 726-9; byseries methods, 555- 562 special types ofsecond order equations giving risetoa,500-504 System oflinear equations with constant coefficients, deﬁnition ofa,398 degenerate, 413-445 determinant of,399,400 equivalent triangular, 405-413 general solution ofa,398,399 problems giving risetoa,seeProblems, system ofequations solution ofa:byLaplace transforms, 418-420; byoperators, 398-417 ofthree equations, 415-420 System oflinear ﬁrstorder equations, 396-397 deﬁnition ofa,396 existence anduniqueness theorem fora, 768-770 problems leading toa,seeProblems, system ofequations solution ofa:bynumerical methods, 702-706; byPicard’s method, 723- 726,726-9; byseries methods, 555- 562 Seealso, System ofﬁrstorder equa- tions; System oflinear equations with constant coefficients Table, chain sliding from, 190 Tables, ofdifferences, 660,662,663, 670 ofLaplace transforms, 306,310 Tangent, hyperbolic, 203 Tan:0,expansion of,742 Taylor series, 535 with remainder, 537 review of,531-537INDEX 807 Taylor series, (continued) solution by,645-652: comment onerror in,649-652, 653-4; creeping up process, 646-647; direct substitution, 646;Seealso, Solution byseries Tchebycheff, seeTschebyscheff's Temperature problems, 129 Terminal velocity, 142 Thermal conductivity, 185 Third degree interpolating polynomial, 676, 678,679 Three-eighths rule, 681 Time constant, 351 Torque, 381 Torsional, resistance constant, 499 stiffness constant, 383,498 Total differential, 72 Tractrix, 176- 6 Trajectories, 115-120 isogonal, 115-117: deﬁnition of,115 orthogonal, 117-120: deﬁnition of,117; inpolar coordinates, 118;inrectangu- larcoordinates, 117 Transform, Laplace, seeLaplace trans- forms Transient, current, 372 motion, 361 solution, 361,372 Trapezoidal rule, 675, 678,679,682 Triangular ytem, equivalent, 405,406, 416 Trigonometric functions ofcomplex numbers, 201 Tschebyscheﬁ"s, equation, 589 polynomials, 589 Twice differentiable function, deﬁnition of,749 Twisting bodies, 383 Undamped frequency, 318-319 Undamped motion, 313-342 examples of,323-329 free, seeSimple harmonic motion forced, 338-342 Undamped, resonance, 341 resonant frequency, 341 Underdamped, 351 Undetermined coefficients, 221-230 Uniform convergence, ofasequence of functions, 729,732 ofaseries offunctions, 732-733 Uniqueness theorems, seeExistence and uniqueness theorems Unit impulse function, 344 Universal gravitation, Newton’s lawof, 491 Unstable motion, 340 Vanish identically, deﬁnition of,775 808 INDEX Variable mass, 191-193 Voltage drop, 369 Variables, separable, 52 separated, 52 Weddle’s rule, 677,678,679, 691,703 Variation ofparameters, 233-240 Weight, speciﬁc, 512 Varying amplitude, slowly, 345 Wronskian, Abel's formula for,779 Vector, point-function, 473 deﬁnition of,774 quantities, 164,459 determinant of,774 Velocity, ofescape, 148 determining linear dependence andinde- formulas inpolar coordinates, 461 pendence by,775-Comment 63.56 terminal, 142 determining linear dependence andinde- Vertical motion, 139-151 pendence ofsolutions by,781-Com- body near earth’s surface: airresistance, ment 64.15 142-145; noresistance, 140-141 theorems about, 778-780 body farfrom earth’s surface, 146-151 Vqlt, 369 Yo\mg's modulus, 385