# Factoring $y$ out of $y^2 + e^y$ possible?

I'm working on solving a separable first order differential equation where I get down to:

$$\frac12 y^2 + e^y = \frac12x^2 + e^{-x} + c$$

Is it possible to solve for $y$ here? I can't think of a way to factor the left hand side.

Here's the original differential equation:

$$\frac{dy}{dx} = \frac{x-e^{-x}}{y + e^y}$$

Thanks!

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Nope, not really. It's clear $y=-x$ is a solution for $C=0$. Perhaps you'd get more useful responses if you posted the original problem you're working on.. – anon Apr 22 '12 at 23:31
No, you cannot factor $y$ from the left hand side. What is the differential equation you were trying to solve? – Arturo Magidin Apr 22 '12 at 23:31
Ok I posted the original problem. Don't think I made a mistake elsewhere but possible. – dacc Apr 22 '12 at 23:34
Are there no initial conditions? Otherwise it looks like you'll have to be content with this implicit solution. – anon Apr 22 '12 at 23:39
ok thanks, guess that's it. @sdcvvc ah thanks for catching that sign – dacc Apr 22 '12 at 23:41

Nope, not really. It's clear $y=−x$ is a solution for $C=0$. Perhaps you'd get more useful responses if you posted the original problem you're working on.. – anon Apr 22 at 23:31

No, you cannot factor $y$ from the left hand side. What is the differential equation you were trying to solve? – Arturo Magidin Apr 22 at 23:31

Ok I posted the original problem. Don't think I made a mistake elsewhere but possible. – dacc Apr 22 at 23:34

Are there no initial conditions? Otherwise it looks like you'll have to be content with this implicit solution. – anon Apr 22 at 23:39

ok thanks, guess that's it. @sdcvvc ah thanks for catching that sign – dacc Apr 22 at 23:41

Hm, I don't think what I said was relevant. It seems anon is right and either you have a condition giving $C=0$, or the implicit solution is all what you can get... – sdcvvc Apr 22 at 23:45

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