community project mathcentre community project

1 Mathcentrecommunityprojectencouragingaca demicstosharemathssupportresourcesAllmcc presourcesarereleasedunderaCreativeCommo nslicencecommunity projectSecond order Ordinary differential Equationsmccp-richard-2 IntroductionPrerequisites:Inordertomaket hemostofthisresource,youneedtoknowaboutt rigonometry,di erentiation, (x),its rstderivativeandsecondderivative:ad2ydx2 + bdydx+ cy = f(x)(1)Wewillonlylookatequationswherethe coe cientsa,bandcareconstant;wewillnottreati nthishandoutthecaseofcoe EquationsIff(x) = ,welookatthecharacteristicequation,obtai nedbyreplacingd2ydx2,dydxandybyr2,rand1i ntheODE:ar2+ br + c = 0 Wedistinguishbetween3cases:thecasewhenth erootsofthecharacteristicequationaredist inctandreal, :realanddistinctro otsr1andr2 Thenthesolutionsofthehomogeneousequation areoftheform:y(x) = Aer1x+ , (x0) = andy (x0) = ,thenAandBwillbeuniquelyde nedby:Aer1x0+ Ber2x0= Ar1er1x0+ Br2er2x0= Exampled2ydx2+ 5dydx+ 6y = 0 Thecharacteristicequationis:r2+ 5r + 6 = 0andtherootsare 5 25 4 62= 3or :y(x) = Ae 3x+ Be MorgianeRichardUniversityofAberdeenShazi aAhmedUniversityofGlasgowmathcentrecommu nityprojectencouragingacademicstosharema thssupportresourcesAllmccpresourcesarere leasedunderaCreativeCommonslicenceCase2.

2 Complexro otsIftherootsarecomplexthentheycanbewrit tenasr + jsandr js(withjtheimaginarynumber,j2= 1)andthesolutionsofthehomogeneousequatio nsareoftheform:y(x) = erx(Aejsx+ Be jsx)whichcanalsobewrittenas= erx(Ccos(sx) + Dsin(sx))Asbefore,theconstantsAandB(orCa ndD)willbede + 4dydx+ 9y = 0 Thecharacteristicequationis:r2+ 4r + 9 = 0andtherootsare 4 16 4 92= 2 + 5jor 2 :y(x) = e 2x(Ae 5xj+ Be 5xj)or= e 2x(C cos ( 5x) + D sin ( 5x))Case3:equalro otsr1=r2=rIfthecharacteristicequationhas onerootonlythenthesolutionsofthehomogene ousequationareoftheform:y(x) = Aerx+ BxerxExampled2ydx2+ 4dydx+ 4y = 0 Thecharacteristicequationis:r2+ 4r + 4 = (r + 2)2= :y(x) = Ae 2x+ Bxe 2xSecond order ODEs with Right-Hand SideIftheright-handsideinEquation(1)isno t0,thenthesolutionscanbefoundasfollows: First, ndtheformofthesolutionofthecorresponding homogeneousequationkeepingtheconstantsAa ndBassuch:thisiscalledthecomplementaryso lutionyc(x); second , ndaparticularintegraloftheODEyp(x).

3 ThenthesolutionsoftheODEareoftheform:y(x ) = yc(x) + yp(x).Atthisp ointonly, ndaparticularintegraloftheODE:themethodo fundeterminedcoe MorgianeRichardUniversityofAberdeenShazi aAhmedUniversityofGlasgowmathcentrecommu nityprojectencouragingacademicstosharema thssupportresourcesAllmccpresourcesarere leasedunderaCreativeCommonslicenceUndete rminedco e :f(x)particularintegralkCkxCx + Dkx2Cx2+ Dx + Ek sin xork cos xC cos x + D sin xk sinh xork cosh xC cosh x + D sinh xekxCekxerx,whererisarootofthecharacteri sticequationCxerxorCx2erxTheconstantsCan dDarefoundby`plugging'theparticularinteg ralintheODE,whichwillleadtoconditionstha tde 5dydx+ 6y = 2 sin 4xWe rst 5r + 6 = 0andtherootsare5 25 4 62= :yc(x) = Ae3x+ Be2xThen,we 4x,welookforaparticularintegralintheform yp(x) = C cos 4x + D sin :d2ypdx2 5dypdx+ 6yp= 2 sin 4xWehave:dypdx= 4C sin 4x + 4D cos 4xd2ypdx2= 16C cos 4x 16D sin 4xPuttingbackintheODE.

4 ( 16C cos 4x 16D sin 4x) 5 ( 4C sin 4x + 4D cos 4x) + 6 (C cos 4x + D sin 4x) = 2 sin 4xRe-arrangingcosandsin:( 16C 20D + 6C) cos 4x + ( 16D + 20C + 6D) sin 4x = 2 sin 4x( 10C 20D) cos 4x + ( 10D + 20C) sin 4x = 2 sin 4xThelastequationmustbetrueforanyvalueof x,sowemusthave:{ 10C 20D = 020C 10D = 2So:{C =225D = MorgianeRichardUniversityofAberdeenShazi aAhmedUniversityofGlasgowmathcentrecommu nityprojectencouragingacademicstosharema thssupportresourcesAllmccpresourcesarere leasedunderaCreativeCommonslicenceSoapar ticularintegraloftheODEisyp(x) =225cos 4x 125sin 4xandthegeneralsolutionsoftheODEareofthe form:y(x) =225cos 4x 125sin 4x + Ae3x+ Be2xVariationofparametersThismethodismor egeneralandwillworkforanyfunctionf(x)int heright-handsideofEquation(1),althoughit maylookintimidatingat rstsight!Firstlet'srewritethecomplementa rysolutionoftheODEintheform:yc(x) = Ay1(x) + By2(x)withy1(x) = er1x,y2(x) = er2xorxer1xifr1= r2withr1,r2rootsofthecharacteristicequat ionThenaparticularintegralofEquation(1)i s:yp(x) = y1(x) y2(x)f(x)W(y1,y2)dx + y2(x) y1(x)f(x)W(y1,y2)dxwithWtheWronskian:W(y 1,y2) = y1(x)y 2(x) y 1(x)y2(x)Exampled2ydx2 2dydx+ y =exx2+ 1 First,let's 2r+1 = 0, (r 1)2= 0, :yc(x) = Aex+ BxexTo ndaparticularintegraloftheODE,wecalculat etheWronskian:with:y1(x) = exand:y2(x) = xexW(y1,y2) = y1(x)y 2(x) y 1(x)y2(x)= ex(1 + x)ex exxex= e2xThenaparticularintegraloftheODEis:yp( x) = ex xexe2xexx2+ 1dx + xex exe2xexx2+ 1dx= ex xx2+ 1dx + xex 11 + x2dx x1 + x2dx =12ln (1 + x2)and 11 + x2dx = arctan x= ex 12ln (1 + x2) + xex arctan xThegeneralsolutionoftheODEis.}}

5 Y(x) = Aex+ Bxex 12exln (1 + x2) + xexarctan MorgianeRichardUniversityofAberdeenShazi aAhmedUniversityofGlasgowmathcentrecommu nityprojectencouragingacademicstosharema thssupportresourcesAllmccpresourcesarere leasedunderaCreativeCommonslicenceExerci ses(a)d2ydx2+ 7y = 0(d)d2ydx2+ 4dydx+ 5y = 2e 2xwithy(0) = 1,dydx(0) = 2(b)d2ydx2+ 2dydx+ y = e 2x(e)d2ydx2+ 4dydx+ 4y = 2 cos2x(c) 3d2ydx2 2dydx y = 2x 3 (f)d2ydx2+ 2dydx+ y = 4 sinh xAnswers(a) y = Aei 7x+ Be i 7x(d) y = e 2x(2 cos x)ory = C cos ( 7x) + D sin ( 7x)(b) y = Ae x+ Bxe x+ e 2x(e) y = (A + Bx)e 2x+14+18sin (2x)(c) y = Aex+ Be 1/3x 2x + 7(f) y = (A + Bx x2)e x+ MorgianeRichardUniversityofAberdeenShazi aAhmedUniversityofGlasgow