Second OrderDifferential Equations_`

19.3IntroductionIn this Section we start to learn how to solve second order dierential equations of a particular type:those that are linear and have constant coecients. Such equations are used widely in the modellingof physical phenomena, for example, in the analysis of vibrating systems and the analysis of electricalcircuits.The solution of these equations is achieved in stages. The rst stage is to nd what is called a com-plementaryfunction. Thesecondstageistondaparticularintegral. Finally,thecomplementaryfunctionandtheparticularintegralarecombinedtoformthegeneralsolution._`

PrerequisitesBeforestartingthisSectionyoushould. . .understandwhatismeantbyadierentialequationunderstandcomplexnumbers( 10)_

Learning OutcomesOncompletionyoushouldbeableto. . .recognisealinear,constantcoecientequationunderstandwhatismeantbythetermsauxiliaryequationandcomplementaryfunctionndthecomplementaryfunctionwhentheauxiliaryequationhasreal,equalorcomplexroots30 HELM(2008):Workbook19: DierentialEquations1. Constant coefcient second order linear ODEsWe now proceed to study those second order linear equations which have constant coecients. Thegeneralformofsuchanequationis:ad2ydx2+ bdydx+ cy= f(x) (3)wherea, b, careconstants. Thehomogeneousformof(3)isthecasewhenf(x) 0:ad2ydx2+ bdydx+ cy= 0 (4)Tondthegeneral solutionof(3), itisrstnecessarytosolve(4). Thegeneral solutionof(4)iscalled the complementary function and will always contain two arbitrary constants. We will denotethissolutionbyycf.ThetechniqueforndingthecomplementaryfunctionisdescribedinthisSection.TaskStatewhichofthefollowingareconstantcoecientequations.Statewhicharehomogeneous.(a)d2ydx2+ 4dydx+ 3y= e2x(b) xd2ydx2+ 2y= 0(c)d2xdt2+ 3dxdt+ 7x = 0 (d)d2ydx2+ 4dydx+ 4y= 0Yoursolution(a)(b)(c)(d)Answer(a)isconstantcoecientandisnothomogeneous.(b)ishomogeneousbutnotconstantcoecientasthecoecientofd2ydx2isx,avariable.(c)isconstantcoecientandhomogeneous. Inthisexamplethedependentvariableisx.(d)isconstantcoecientandhomogeneous.Note: A complementary function is the general solution of a homogeneous, linear dierential equation.HELM(2008):Section19.3: SecondOrderDierentialEquations312. Finding the complementary functionTondthecomplementaryfunctionwemustmakeuseofthefollowingproperty.If y1(x) and y2(x) are any two (linearly independent) solutions of a linear, homogeneous second orderdierentialequationthenthegeneralsolutionycf(x),isycf(x) = Ay1(x) + By2(x)whereA, Bareconstants.Weseethatthesecondorderlinearordinarydierential equationhastwoarbitraryconstantsinitsgeneralsolution. The functions y1(x) and y2(x) are linearlyindependent if one is not a multipleoftheother.Example5Verifythat y1=e4xandy2=e2xbothsatisfytheconstant coecient linearhomogeneousequation:d2ydx2 6dydx+ 8y= 0Writedownthegeneralsolutionofthisequation.SolutionWheny1= e4x,dierentiationyields:dy1dx= 4e4xandd2y1dx2= 16e4xSubstitutionintotheleft-handsideof theODEgives16e4x 6(4e4x) + 8e4x, whichequals0, sothaty1= e4xisindeedasolution.Similarlyify2= e2x,thendy2dx= 2e2xandd2y2dx2= 4e2x.Substitution into the left-hand side of the ODE gives 4e2x6(2e2x) +8e2x, which equals 0, so thaty2= e2xis also a solution of equation the ODE. Now e2xand e4xare linearly independent functions,so,fromthepropertystatedabovewehave:ycf(x) = Ae4x+ Be2xisthegeneralsolutionoftheODE.32 HELM(2008):Workbook19: DierentialEquationsExample6Findvaluesofksothaty= ekxisasolutionof:d2ydx2 dydx 6y= 0Hencestatethegeneralsolution.SolutionAssuggestedwetryasolutionoftheformy= ekx. Dierentiatingwenddydx= kekxandd2ydx2= k2ekx.Substitutionintothegivenequationyields:k2ekx kekx 6ekx= 0 thatis (k2 k 6)ekx= 0Theonlywaythisequationcanbesatisedforallvaluesofxisifk2 k 6 = 0thatis,(k 3)(k + 2) = 0sothatk= 3ork= 2. Thatistosay,ify= ekxistobeasolutionofthedierentialequation,kmustbeeither3or 2. Wethereforehavefoundtwosolutions:y1(x) = e3xand y2(x) = e2xThesearelinearlyindependentandthereforethegeneralsolutionisycf(x) = Ae3x+ Be2xTheequationk2 k 6 = 0fordeterminingkiscalledtheauxiliaryequation.TaskBysubstitutingy= ekx,ndvaluesofksothatyisasolutionofd2ydx2 3dydx+ 2y= 0Hence,writedowntwosolutions,andthegeneralsolutionofthisequation.Firstndtheauxiliaryequation:YoursolutionAnswerk2 3k + 2 = 0HELM(2008):Section19.3: SecondOrderDierentialEquations33Nowsolvetheauxiliaryequationandwritedownthegeneralsolution:YoursolutionAnswerTheauxiliaryequationcanbefactorisedas(k 1)(k 2) = 0andsotherequiredvaluesofkare1and2. Thetwosolutionsarey= exandy= e2x. Thegeneralsolutionisycf(x) = Aex+ Be2xExample7Findtheauxiliaryequationofthedierentialequation:ad2ydx2+ bdydx+ cy= 0SolutionWetryasolutionoftheformy= ekxsothatdydx= kekxandd2ydx2= k2ekx.Substitutionintothegivendierentialequationyields:ak2ekx+ bkekx+ cekx= 0 thatis (ak2+ bk + c)ekx= 0Sincethisequationistobesatisedforallvaluesofx,thenak2+ bk + c = 0istherequiredauxiliaryequation.Key Point 5The auxiliary equation of ad2ydx2+ bdydx+ cy= 0 is ak2+ bk + c = 0 where y= ekx34 HELM(2008):Workbook19: DierentialEquationsTaskWritedown,butdonotsolve,theauxiliaryequationsofthefollowing:(a)d2ydx2+dydx+ y= 0, (b) 2d2ydx2+ 7dydx 3y= 0(c) 4d2ydx2+ 7y= 0, (d)d2ydx2+dydx= 0Yoursolution(a)(b)(c)(d)Answer(a)k2+ k + 1 = 0 (b)2k2+ 7k 3 = 0 (c)4k2+ 7 = 0 (d)k2+ k = 0Solving the auxiliary equation gives the values of k which we need to nd the complementary function.Clearlythenatureoftherootswilldependuponthevaluesofa, bandc.Case1 Ifb2> 4actherootswill bereal anddistinct. Thetwovaluesofkthusobtained,k1andk2, will allowustowritedowntwoindependentsolutions: y1(x)=ek1xandy2(x)=ek2x,andsothegeneralsolutionofthedierentialequationwillbe:y(x) = Aek1x+ Bek2xKey Point 6Iftheauxiliaryequationhasreal,distinctrootsk1andk2,thecomplementaryfunctionwillbe:ycf(x) = Aek1x+ Bek2xCase 2 On the other hand, if b2= 4ac the two roots of the auxiliary equation will be equal and thismethod will therefore only yield one independent solution. In this case, special treatment is required.Case3 If b2