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Im clueless on how to solve the following question...

$xe^y\frac {dy}{dx} = e^y +1$

What i've done is...

$\frac {dy}{dx} = \frac 1x + \frac {1}{xe^e}; \frac {dy}{dx} - \frac {1}{xe^e} = \frac 1x $

Find the integrating factor..

$v(x) = e^{P(x)}; where P(x) = \int p(x)dx \Rightarrow P(x) = \int \frac 1x dx = ln|x| \\v(x) = e^{P(x)} = e^{ln|x|} = x; \\ y = \frac {1}{v(x)} \int v(x)q(x) dx = \frac 1x \int {x}{\frac 1x} dx = 1+c$

I know I made a mistake somewhere. Would someone advice me on this??

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solve rather $x\,dz/dx=z+1$ and then set $z=e^y$ –  user8268 Nov 13 '12 at 9:50
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1 Answer 1

up vote 2 down vote accepted

You have $xe^y\frac{dy}{dx}=e^y+1$, so, by dividing both sides by $x$ and by $e^y+1$, we get $$\begin{align*}\frac{e^y}{e^y+1}\frac{dy}{dx}=\frac1x\hspace{5pt}&\Rightarrow \hspace{5pt}\frac{e^y}{e^y+1}dy=\frac1xdx\hspace{5pt}\Rightarrow \hspace{5pt}\int\frac{e^y}{e^y+1}dy=\int\frac1xdx \\ &\Rightarrow\hspace{5pt}\ln(e^y+1)=\ln (C|x|)\hspace{5pt}\Rightarrow \hspace{5pt}e^y+1=C|x|\end{align*}$$

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