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$$xy'-y=x(1+e^{\frac{y}{x}})$$ Please give me a hint on how to solve this. If I'm not mistaken, this is a Bernoulli equation, but I can't seem to solve it using the substitution $z=y^{\frac{1}{1-a}}$. Using variation of constants also didn't help.

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  • $\begingroup$ @TIWARI You might mean $y=tx$. $\endgroup$ – Did Jun 25 '15 at 12:31
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Hint:

Try $u=\frac{y}{x}$. Then $y'=u'x+u$.

By the way, this is not Bernoulli.

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$y\backprime -\frac { y }{ x } =1+{ e }^{ \frac { y }{ x } }$ $\frac { y }{ x } =t$ , $\Rightarrow y=xt$ $\Rightarrow y\prime =t\prime x+t$ $\Rightarrow y\prime =t\prime x+t$ $ \Rightarrow \int { \frac { dt }{ 1+{ e }^{ t } } =\int { \frac { dx }{ x } } } $

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  • $\begingroup$ May I suggest to forget completely the odd \backprime and \prime. $\endgroup$ – Did Jun 25 '15 at 12:31

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