odeReview

Review ofOrdinary Dieren tialEquations Denition 1 (a)Adieren tialequation isanequation foranunkno wnfunction thatcontainsthederivativesofthat unkno wnfunction. Forexample y00(t)+y(t)=0isadieren tialequation fortheunkno wnfunction y(t). (b)Adieren tialequation iscalled anordinary dieren tialequation (often shortened to\ODE") if onlyordinary derivativesappear.That is,iftheunkno wnfunction hasonlyasingle independen tvariable. Adieren tialequation iscalled apartial dieren tialequation (often shortened to\PDE") ifpartial derivativesappear.That is,iftheunkno wnfunction hasmore thanoneindependen tvariable. Forexample y00(t)+y(t)=0isanODE while@2u @t2(x;t)=c2@2u @x2(x;t)isaPDE. (c)Theorder ofadieren tialequation istheorder ofthehighest derivativethatappears. Forexample y00(t)+y(t)=0isasecond order ODE. (d)Anordinary dieren tialequation thatisoftheform a0(t)y(n)(t)+a1(t)y(n1)(t)+


+an1(t)y0(t)+an(t)y(t)=F(t) (1) withgivencoecien tfunctions a0(t),


,an(t)andF(t)issaidtobelinear .Otherwise, theODE issaid tobenonlinear .Forexample, y0(t)2+y(t)=0,y0(t)y00(t)+y(t)=0andy0(t)=ey(t)areallnonlinear. (e)TheODE (1)issaidtohaveconstan tcoecien tsifthecoecien tsa0(t),a1(t),


,an(t)areall contants.Otherwise, itissaidtohavevariable coecien ts.Forexample, theODE y00(t)+7y(t)=sint isconstan tcoecien t,while y00(t)+ty(t)=sintisvariable coecien t. (f)TheODE (1)issaidtobehomogeneous ifF(t)isidentically zero. Otherwise, itissaidtobein- homogeneous ornonhomogeneous .Forexample, theODE y00(t)+7y(t)=0ishomogeneous, while y00(t)+7y(t)=sintisinhomogeneous. Ahomogeneous ODE alwayshasthetrivial solution y(t)=0. (g)Aninitial valueproblem isaproblem inwhichoneistondanunkno wnfunction y(t)thatsatises bothagivenODE andgiveninitial conditions, likey(0)=1,y0(0)=0. (h)Aboundary valueproblem isaproblem inwhichoneistondanunkno wnfunction y(t)thatsatises bothagivenODE andgivenboundary conditions, likey(0)=0,y(1)=0. Theorem 2Assume thatthecoecients a0(t),a1(t),


,an1(t),an(t)andF(t)arereasonably smooth, boundedfunctions andthata0(t)isnotzero. (a)Thegeneralsolution totheODE(1)isoftheform y(t)=yp(t)+C1y1(t)+C2y2(t)+


+Cnyn(t) where nistheorderoftheODE(1) theparticular solution, yp(t),isanysolution to(1) C1,C2,


,Cnarearbitraryconstants y1,y2,


,ynarenindependent solutions tothehomogenous equation a0(t)y(n)(t)+a1(t)y(n1)(t)+


+an1(t)y0(t)+an(t)y(t)=0 \Indep endent" justmeansthatnoyicanbewritten asalinearcombination oftheother yj's.For example, y1(t)cannot beexpressedintheformd2y2(t)+


+dnyn(t). January3,2007 Review ofOrdinary Dieren tialEquations 1 (b)Given anyconstants b0,


,bn1thereisexactly onefunction y(t)thatobeystheODE(1)andtheinitial conditions y(0)=b0y0(0)=b1


y(n1)(0)=bn1 Example 3(TheRLCcircuit) Asanexample ofthemostcommonly usedtechniques forsolving ODE's, weconsider theRLCcircuit + x(t)R L Ci(t)+ y(t) We'regoing tothink ofthevoltage x(t)asainput signal andthevoltage y(t)asanoutput signal. Thegoal istodetermine theoutput voltage foragiveninput voltage. Inthenotes \The RLCCircuit", wederived theODE LCy00(t)+RCy0(t)+y(t)=x(t) (2) Asaconcrete example, we'lltakeanacvoltage source andchoosetheorigin oftimesothatx(0)=0, x(t)=E0sin(!t).Then thedieren tialequation becomes LCy00(t)+RCy0(t)+y(t)=E0sin(!t) (3) Thisisasecond order, linear, constan tcoecien tODE. Soweknow,fromTheorem 2,thatthegeneral solution isoftheformyp(t)+C1y1(t)+C2y2(t),where yp(t),theparticular solution, isanyonesolution to(3), C1;C2arearbitrary constan tsand y1(t),y2(t)areanytwoindependen tsolutions ofthecorresp onding homogeneous equation LCy00(t)+RCy0(t)+y(t)=0 (3h) Sotondthegeneral solution to(3),weneedtondthree functions: y1(t),y2(t)andyp(t). Finding y1(t)andy2(t):Thebestwaytondy1andy2istoguess them. Anysolution, yh(t),of(3h)has tohavethepropertythatyh(t),RCy0 h(t)andLCy00 h(t)havetocancel eachother outforallt.Wechooseour guess sothatyh(t),y0 h(t)andy00 h(t)areallproportional toasingle function oft.Then itwillbeeasytosee ifyh(t),RCy0 h(t)andLCy00 h(t)allcancel. Hence wetryyh(t)=ert,withtheconstan trtothedetermined. Thisguess isasolution of(3h)ifandonlyif LCr2ert+RCrert+ert=0()LCr2+RCr+1=0()r=RCp R2C24LC 2LCr1;2 (4) Finding y1(t)andy2(t),when R2C24LC6=0:IntheeventthatR2C24LC6=0,thatisR6=2q L C, r1andr2aredieren tandwemaytakey1(t)=er1tandy2(t)=er2t. Finding y1(t)andy2(t),when R2C24LC=0:IntheeventthatR=2q L C,r1=r2.Then wemay takey1=er1t,buter2t=er1tiscertainly notasecond independen tsolution. Sowestillneedtond y2.Hereisatrick(called reduction oforder) fornding theother solutions: lookforsolutions ofthe formv(t)er1t.Hereer1tisthesolution wehavealready found andv(t)istobedetermined. Tosave January3,2007 Review ofOrdinary Dieren tialEquations 2 writing, set=R 2Lsothatr1=r2=.Tosavewriting alsodivide (3h)byLCandsubstitute that R L=2and1 LC=R2 4L2=2(recall thatweareassuming thatR2=4L C).So(3h)isequivalentto y00 h(t)+2y0 h(t)+2yh(t)=0 Subin yh(t)= v(t)et y0 h(t)=v(t)et+v0(t)et y00 h(t)=2v(t)et2v0(t)et+v00(t)et Thuswhen yh(t)=v(t)et, y00 h(t)+2y0 h(t)+2yh(t)= 222+2 v(t)et+ 2+2 v0(t)et+v00(t)et =v00(t)et Thusv(t)etisasolution of(3h)whenev erthefunction v00(t)=0forallt.But,foranyvalues ofthe constan tsC1andC2,v(t)=C1+C2thasvanishing second derivativeso C1+C2t
et= C1+C2t
er1t solves(3h).ThisisoftheformC1y1(t)+C2y2(t)withy1(t)=er1t,thesolution thatwefound rst, andy2(t)=ter1t,asecond independen tsolution. Sowemaytakey2(t)=ter1t. Finding yp(t):Thebestwaytondypistoguess it.Weguess thatthecircuit respondstoanoscillating applied voltage withacurren tthatoscillates atthesame frequency .Sowetryyp(t)=Asin(!t')with theamplitude Aandphase 'tobedetermined. Foryp(t)tobeasolution, weneed LCy00 p(t)+RCy0 p(t)+yp(t)=E0sin(!t) (3p) LC!2Asin(!t')+RC!Acos(!t')+Asin(!t')=E0sin(!t) =E0sin(!t'+') andhence, applying sin(A+B)=sinAcosB+cosAsinBwithA=!t'andB=', 1LC!2
Asin(!t')+RC!Acos(!t')=E0cos(')sin(!t')+E0sin(')cos(!t') Matchingcoecien tsofsin(!t')andcos(!t')ontheleftandrighthand sidesgives 1LC!2
A=E0cos(') (5) RC!A=E0sin(') (6) ItisnoweasytosolveforAand' (6) (5)=)tan(')=RC! 1LC!2=)'=tan1RC! 1LC!2 q (5)2+(6)2=)q 1LC!2
2+R2C2!2A=E0=)A=E0p (1LC!2)2+R2C2!2(7) Example 4(Boundary ValueProblems) Bypart(b)ofTheorem 2,aninitial valueproblem consisting ofannthorder linear ODE withreasonable coecien tsandninitial conditions alwayshasexactly one solution. Weshallnowseethataboundary valueproblem mayhavenosolutions atall.Oritmayhave January3,2007 Review ofOrdinary Dieren tialEquations 3 exactly onesolution. Oritmayhaveinnitely manysolutions. Weshallstartbynding allsolutions tothe ODE y00+y=0 (8) Weshallthenimposeboundary conditions. Thefunction y(t)=ertisasolution to(8)ifandonlyif r2ert+ert=0()r2+1=0()r=i where i(whichelectrical engineers often denote j)isasquare rootof1.Thusthegeneral solution tothe second order linear ODE (8)isy(t)=C0 1eit+C0 2eit,withC0 1andC0 2arbitrary constan ts.Wemayrewrite thisgeneral solution interms ofsintandcostbysubstituting in eit=cost+isinteit=costisint Thisgives y(t)=C0 1 cost+isint)+C0 2(costisint)=C1cost+C2sintwhere C1=C0 1+C0 2;C2=i(C0 1C0 2) Notethatthere isnothing stopping C0 1andC0 2frombeingcomplex numbers.Sothere isnothing stopping C1andC2beingrealnumbers. Example 4.aNowconsider theboundary valueproblem y00+y=0 y(0)=0 y(2)=1 (9) Thefunction y(t)satises theODE ifandonlyifitisoftheformy(t)=C1cost+C2sintforsomeconstan ts C1andC2.Afunction ofthisformsatises theboundary condition y(0)=0ifandonlyif 0=y(0)=C1cos0+C2sin0=C1 Afunction ofthisformsatises theboundary condition y(2)=1ifandonlyif 1=y(2)=C1cos2+C2sin2=C1 Thetworequiremen tsC1=0andC1=1areincompatible. Sotheboundary valueproblem (9)hasno solution atall. Example 4.bNextconsider theboundary valueproblem y00+y=0y(0)=0y 2
=0 (10) Thefunction y(t)satises theODE ifandonlyifitisoftheformy(t)=C1cost+C2sintforsomeconstan ts C1andC2.Afunction ofthisformsatises theboundary condition y(0)=0ifandonlyif 0=y(0)=C1cos0+C2sin0=C1 Afunction ofthisformsatises theboundary condition y 2
=0ifandonlyif 0=y 2
=C1cos 2
+C2sin 2
=C2 Sowehaveasolution ifandonlyifC1=C2=0andtheboundary valueproblem (10)hasexactly one solution, namely y(t)=0. January3,2007 Review ofOrdinary Dieren tialEquations 4 Example 4.cFinally consider theboundary valueproblem y00+y=0 y(0)=0 y(2)=0 (11) Thefunction y(t)satises theODE ifandonlyifitisoftheformy(t)=C1cost+C2sintforsomeconstan ts C1andC2.Afunction ofthisformsatises theboundary condition y(0)=0ifandonlyif 0=y(0)=C1cos0+C2sin0=C1 Afunction ofthisformsatises theboundary condition y(2)=0ifandonlyif 0=y(2)=C1cos(2)+C2sin(2)=C1 Sowehaveasolution ifandonlyifC1=0andtheboundary valueproblem (11)hasinnitely many solutions, namely y(t)=C2sintwithC2beinganarbitrary constan t. January3,2007 Review ofOrdinary Dieren tialEquations 5