Transcription of community project mathcentrecommunityproject
1 Mathcentre community projectencouraging academics to share maths support resourcesAll mccp resources are released under an Attribution Non-commerical Share Alike licencecommunity projectSolving Differential Equations with Integrating Factorsmccp-dobson-0111 IntroductionSuppose we have the first order differential equationdydx+P y=QwherePandQare functions involvingxonly. For exampledydx+3yx=exx3ordydx 3yx+ 1= (x+ 1) can solve these differential equations using the technique of anintegrating FactorWe multiply both sides of the differential equation by the integrating factorIwhich is defined asI=e P SolutionMultiplying our original differential equation byIwe get thatdydx+P y=Q Idydx+IP y=IQ (Idydx+IP y)dx= IQ dx Iy= IQ dxsinceddx(Iy) =Idydx+IP yby the product bothIandQare functions involving onlyxin most of the problems you are likely to meet, IQ dxcan usually be found.
2 So the general solution to the differential equation is found by integratingIQand then re-arranging the formula to makeythe find the general solution of the differential equationdydx+3yx= Katy DobsonUniversity of LeedsAlan SlomsonUniversity of Leedsmathcentre community projectencouraging academics to share maths support resourcesAll mccp resources are released under an Attribution Non-commerical Share Alike licencewe first find the integrating factorI=e P dx=e 3xdxnow 3xdx= 3 lnx= lnx3henceI=elnx3= we multiply the differential equation byIto getx3dydx+ 3x2y=exso integrating both sides we havex3y=ex+cwherecis a constant.
3 Thus the general solution isy=ex+ find the general solution of the differential equationdydx 3yx+ 1= (x+ 1)4we first find the integrating factorI=e P dx=e 3x+1dxnow 3x+ 1dx= 3 ln(x+ 1) = ln(x+ 1) 3henceI=eln(x+1) 3= (x+ 1) 3=1(x+ 1) multiplying the differential equation byIwe get1(x+ 1)3dydx 3y(x+ 1)4= (x+ 1)so integrating both sides we havey(x+ 1)3=12x2+x+cwherecis a the general solution isy= (x+ 1)3(12x2+x+c).ExercisesFind the general solution +2yx= yx= xe + 2xy= 2yx= 3x3 Answers1. y=c cosxx22. y=x(e x+c)3. y=12+ce x24. y=32x4+ Katy DobsonUniversity of LeedsAlan SlomsonUniversity of Leeds
