How to solve the equation ? the solution in the book is y=ex. y'=e^{xy'/y}.

idontknow Dec 2015 1,084 169 Earth Jan 28, 2020 #1 How to solve the equation ? the solution in the book is \(\displaystyle y=ex\). \(\displaystyle y'=e^{xy'/y}\).

How to solve the equation ? the solution in the book is \(\displaystyle y=ex\). \(\displaystyle y'=e^{xy'/y}\).

skipjack Forum Staff Dec 2006 21,479 2,470 Jan 28, 2020 #2 $y = ce^{x/c}$, where $c$ is a non-zero constant. The graphs of the above solutions have a common tangent line with equation $y = ex$, corresponding to a singular solution of the differential equation. Reactions: idontknow and topsquark

$y = ce^{x/c}$, where $c$ is a non-zero constant. The graphs of the above solutions have a common tangent line with equation $y = ex$, corresponding to a singular solution of the differential equation.

idontknow Dec 2015 1,084 169 Earth Jan 29, 2020 #3 My approach after hours : set \(\displaystyle x=e^{t} \) ; \(\displaystyle \; y'=\frac{dy}{dt}\cdot \frac{dt}{dx}=e^{e^t dy/ydx}\). \(\displaystyle e^{-t} dy/dt =e^{dy/ydt }\) ; \(\displaystyle \; dy/dt=e^{dy/ydt +t } \; \) ; \(\displaystyle \; \ln dy =\ln dt + \ln e^t +\ln e^{dy/ydt}=\ln e^t dt +\ln e^{dy/ydt}\). \(\displaystyle \ln \frac{dy}{dy/ydt} =\ln e^t dt \; \) ; \(\displaystyle \; yd(C_1 t)=e^t d(C_2 t)\). \(\displaystyle y=\frac{C_2 }{C_1 } e^t =ce^t=cx \). \(\displaystyle (cx)'=e^{x(cx)'/cx}=e\). \(\displaystyle c=e\) and the general solution: \(\displaystyle y=ex\).

My approach after hours : set \(\displaystyle x=e^{t} \) ; \(\displaystyle \; y'=\frac{dy}{dt}\cdot \frac{dt}{dx}=e^{e^t dy/ydx}\). \(\displaystyle e^{-t} dy/dt =e^{dy/ydt }\) ; \(\displaystyle \; dy/dt=e^{dy/ydt +t } \; \) ; \(\displaystyle \; \ln dy =\ln dt + \ln e^t +\ln e^{dy/ydt}=\ln e^t dt +\ln e^{dy/ydt}\). \(\displaystyle \ln \frac{dy}{dy/ydt} =\ln e^t dt \; \) ; \(\displaystyle \; yd(C_1 t)=e^t d(C_2 t)\). \(\displaystyle y=\frac{C_2 }{C_1 } e^t =ce^t=cx \). \(\displaystyle (cx)'=e^{x(cx)'/cx}=e\). \(\displaystyle c=e\) and the general solution: \(\displaystyle y=ex\).

skipjack Forum Staff Dec 2006 21,479 2,470 Jan 29, 2020 #4 idontknow said: \(\displaystyle \ln \frac{dy}{dy/ydt} =\ln e^t dt\) Click to expand... That should be \(\displaystyle \ln \frac{dy}{e^{dy/ydt}} =\ln e^t dt \, \). You can't obtain the book's solution as the general solution, because it's a singular solution. Reactions: idontknow

idontknow said: \(\displaystyle \ln \frac{dy}{dy/ydt} =\ln e^t dt\) Click to expand... That should be \(\displaystyle \ln \frac{dy}{e^{dy/ydt}} =\ln e^t dt \, \). You can't obtain the book's solution as the general solution, because it's a singular solution.

idontknow Dec 2015 1,084 169 Earth Jan 30, 2020 #5 skipjack said: That should be \(\displaystyle \ln \frac{dy}{e^{dy/ydt}} =\ln e^t dt \, \). Click to expand... Yes , my mistake , I will post the correct version later.

skipjack said: That should be \(\displaystyle \ln \frac{dy}{e^{dy/ydt}} =\ln e^t dt \, \). Click to expand... Yes , my mistake , I will post the correct version later.

idontknow Dec 2015 1,084 169 Earth Jan 30, 2020 #6 \(\displaystyle y'\) must be a power of e. we can set \(\displaystyle y=e^{\phi (x)} \). \(\displaystyle (e^{\phi})'=e^{\frac{x(e^{\phi})'}{y}}\; \) ; \(\displaystyle \; e^{\phi} \phi ' =e^{xe^{\phi}\phi ' /e^{\phi} }=e^{x\phi '}\). \(\displaystyle \ln e^{\phi} +\ln \phi ' =x\phi ' \; \) ; \(\displaystyle \; \ln \phi ' =x\phi ' -\phi \; \) ; \(\displaystyle \; \frac{d}{dx} \ln \phi ' = \frac{d}{dx} [x\phi '-\phi] \). \(\displaystyle \frac{\phi ''}{\phi '} =x\phi '' +\phi ' -\phi ' =x\phi '' \; \) ; \(\displaystyle \; \phi_1 '' [x-\frac{1}{\phi_2 '}]=0\). (we need only \(\displaystyle \phi _1 \)) \(\displaystyle \phi _1 '' =0 \: \implies \: \phi_1=\ln y = Cx \: \implies \: y=e^{Cx}=cx\). \(\displaystyle (cx)'=e^{x(cx)'/cx}=e \: \implies \: y=ex . \) The book does not mention about the general solution , just y=ex.

\(\displaystyle y'\) must be a power of e. we can set \(\displaystyle y=e^{\phi (x)} \). \(\displaystyle (e^{\phi})'=e^{\frac{x(e^{\phi})'}{y}}\; \) ; \(\displaystyle \; e^{\phi} \phi ' =e^{xe^{\phi}\phi ' /e^{\phi} }=e^{x\phi '}\). \(\displaystyle \ln e^{\phi} +\ln \phi ' =x\phi ' \; \) ; \(\displaystyle \; \ln \phi ' =x\phi ' -\phi \; \) ; \(\displaystyle \; \frac{d}{dx} \ln \phi ' = \frac{d}{dx} [x\phi '-\phi] \). \(\displaystyle \frac{\phi ''}{\phi '} =x\phi '' +\phi ' -\phi ' =x\phi '' \; \) ; \(\displaystyle \; \phi_1 '' [x-\frac{1}{\phi_2 '}]=0\). (we need only \(\displaystyle \phi _1 \)) \(\displaystyle \phi _1 '' =0 \: \implies \: \phi_1=\ln y = Cx \: \implies \: y=e^{Cx}=cx\). \(\displaystyle (cx)'=e^{x(cx)'/cx}=e \: \implies \: y=ex . \) The book does not mention about the general solution , just y=ex.

skipjack Forum Staff Dec 2006 21,479 2,470 Jan 30, 2020 #7 idontknow said: \(\displaystyle \phi _1 '' =0 \: \implies \: \phi_1=\ln y = Cx \: \implies \: y=e^{Cx}=cx\). Click to expand... That's not quite right. If you correct it, you can get the general solution I gave. Solving for $\phi_2$ leads to the singular solution $y = ex$. Reactions: idontknow

idontknow said: \(\displaystyle \phi _1 '' =0 \: \implies \: \phi_1=\ln y = Cx \: \implies \: y=e^{Cx}=cx\). Click to expand... That's not quite right. If you correct it, you can get the general solution I gave. Solving for $\phi_2$ leads to the singular solution $y = ex$.

idontknow Dec 2015 1,084 169 Earth Feb 3, 2020 #8 \(\displaystyle \ln \phi ' =x\phi ' -\phi \) is known as Clairaut's DE : \(\displaystyle \: s(t)=ts'(t)+f(s'(t))\). Clairaut's Differential Equation -- from Wolfram MathWorld Last edited: Feb 3, 2020

\(\displaystyle \ln \phi ' =x\phi ' -\phi \) is known as Clairaut's DE : \(\displaystyle \: s(t)=ts'(t)+f(s'(t))\). Clairaut's Differential Equation -- from Wolfram MathWorld