Cannot solve the DE

Dec 2015
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How to solve the equation ? the solution in the book is \(\displaystyle y=ex\).

\(\displaystyle y'=e^{xy'/y}\).
 

skipjack

Forum Staff
Dec 2006
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$y = ce^{x/c}$, 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.
 
Dec 2015
972
128
Earth
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

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Dec 2006
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\(\displaystyle \ln \frac{dy}{dy/ydt} =\ln e^t dt\)
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.
 
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Dec 2015
972
128
Earth
\(\displaystyle y'\) 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
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\(\displaystyle \phi _1 '' =0 \: \implies \: \phi_1=\ln y = Cx \: \implies \: y=e^{Cx}=cx\).
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 $y = ex$.
 
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