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Question Number 6415 by sanusihammed last updated on 26/Jun/16

$$y^{{}^{\prime }}\left(t\right)=C+\frac{v\left(t\right)y\left(t\right)}{1+v\left(t\right)t}$$$$$$$$Here,tisthevariable$$$$y\left(t\right)isthefunction$$$$y^{{}^{\prime }}\left(t\right)isitsfirstderivative$$$$v\left(t\right)isananotherfunctioninthesamevariablet$$$$Cisaconstant.$$$$$$$$Findanarbitrarysolution$$$$\left(1\right)youcaneitherfindanypairoftandy\left(t\right)asasolutionto$$$$this$$$$\left(2\right)youcantakeanyvalueforthefunctionv\left(t\right)$$$$\left(2a\right)v\left(t\right)canbeaconstantfunction$$$$\left(2b\right)v\left(t\right)canbeanegativeincreasinglyvaryingfunction.$$

Answered by Chaeris27 last updated on 26/Jun/16

$$Youshouldfirstputy\left(t\right)and{y}^{\prime }\left(t\right)onthesameside$$$$y^{\prime }\left(t\right)=C+\frac{v\left(t\right)y\left(t\right)}{1+v\left(t\right)t}$$$$y\left(t\right)(D-I\frac{v\left(t\right)}{1+v\left(t\right)t})=C$$$$Letu\left(t\right)=\frac{v\left(t\right)}{1+v\left(t\right)t},thenwehave$$$$y^{\prime }\left(t\right)-u\left(t\right)y\left(t\right)=C$$$$WecanfindaparticularsolutionforC=0$$$$y^{\prime }\left(t\right)=u\left(t\right)y\left(t\right)$$$$\frac{y^{\prime }\left(t\right)}{y\left(t\right)}=u\left(t\right)$$$$\frac{d}{dt}ln\left(y\left(t\right)\right)=u\left(t\right)$$$$ln\left(y\left(t\right)\right)=\int u\left(t\right)dt$$$$y\left(t\right)=e^{\int u\left(t\right)dt}$$$$substitutingu\left(t\right)backwehave$$$$y\left(t\right)=e^{\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}$$$$ToendjustaddCbacktogetgeneralsoljtion$$

Commented by sanusihammed last updated on 26/Jun/16

$$Thankssomuch$$

Commented by Chaeris27 last updated on 26/Jun/16

$$Ihavecheckedandmysolutionis$$$$correctbutaddingCbackishard:$$$$$$$$Firstly,thegeneralsolutionforC=0$$$$is:$$$$y\left(t\right)=c_1\cdot e^{\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}$$$$Itappearsthatifyoutrytocancelout$$$$thevariablepartoftheequationtoget$$$$tosimplerproblemyouhave$$$$e^{-\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}y\left(t\right)=c_1$$$$e^{-\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}{y}^{\prime }\left(t\right)=\frac{v\left(t\right)}{1+v\left(t\right)t}{c}_1$$$$soifyoumultiplytheoriginalequation$$$$by{e}^{-\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}youget$$$$e^{-\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}{y}^{\prime }\left(t\right)-e^{-\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}\frac{v\left(t\right)}{1+v\left(t\right)t}y\left(t\right)={Ce}^{-\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}$$$$Orwithu\left(t\right)=\frac{v\left(t\right)}{1+v\left(t\right)t}$$$$e^{-\int u\left(t\right)dt}{y}^{\prime }\left(t\right)+(-u\left(t\right)e^{-\int u\left(t\right)dt}y\left(t\right))=e^{-\int u\left(t\right)dt}C$$$$LetU\left(t\right)=e^{-\int u\left(t\right)dt}then{U}^{\prime }\left(t\right)=-u\left(t\right)e^{-\int u\left(t\right)dt}$$$$whichwecansubstituteinourequation$$$$U\left(t\right)y^{\prime }\left(t\right)+U^{\prime }\left(t\right)y\left(t\right)=\frac{d}{dt}\left(U\left(t\right)y\left(t\right)\right)$$$$\frac{d}{dt}\left(U\left(t\right)y\left(t\right)\right)=e^{-\int u\left(t\right)dt}C$$$$U\left(t\right)y\left(t\right)=\int e^{-\int u\left(t\right)dt}Cdt+c_1$$$$=C\cdot \int e^{-\int u\left(t\right)dt}dt+c_1$$$$(c_{1}comesfromthefactthatwehave$$$$justintegrated,itcanbeanynumber)$$$$FinallydividebyU\left(t\right)toget$$$$y\left(t\right)=\frac{1}{U\left(t\right)}(C\cdot \int e^{-\int u\left(t\right)dt}dt+c_1)$$$$Substituteu\left(t\right)andU\left(t\right)(notethat$$$$\frac{1}{U\left(t\right)}=\frac{1}{e^{-\int u\left(t\right)dt}}=e^{\int u\left(t\right)dt})$$$$$$$$y\left(t\right)=e^{\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}\cdot (C\cdot \int e^{-\int \frac{v\left(t\right)}{1+v\left(t\right)t}dt}dt+c_1)$$$$Soyouhaverightherethemost$$$$generalsolutionavailable$$

Commented by sanusihammed last updated on 26/Jun/16

$$Great.ireallyappreciate.$$

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