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I've had success solving $y' = e^{x - y}$ and $y' = x e^{x - y}$, but this one has me completely stumped. Substituting $z$ for $x - y$, and after separation, I'm faced with integrating $dz / (1 - z e^z$). Algebraic manipulation gives the integral of $z e^z dz / (1 - z e^z)$, which is similarly out of my league.

Here's my work so far

Emanuel Landeholm
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    Where did you encounter this problem? Do you have any reason to believe that there is a nice explicit formula for the solution? – Hans Lundmark Apr 11 '22 at 17:17
  • A Youtuber named Sybermath gave the original problem https://www.youtube.com/watch?v=AltN6CXMwyc y' = e^(x - y) and I just kept moving the goal posts.

    The sore truth is I have no reason whatsoever to believe there may be some "nice explicit" formula for my integral. I'll definitely settle for way less than that! But I can't help thinking Lambert's W whenever I see z e^z...

     – Emanuel Landeholm Apr 11 '22 at 17:21
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    Well, I suspect that you've come about as far one can get. You could perhaps let $w=e^{-z}$ to obtain $x+C=- \int \frac{dw}{w+\ln w}$, but that's very likely a non-elementary integral. You could of course give it a name, say $F(w)$, and then try to say something about the properties of the function $F$, and hence of the solution $w(x)$ (and in turn $z(x)$ and $y(x)$). But it's probably easier to look directly at the ODE for $z(x)$ in order to say something about the solution. – Hans Lundmark Apr 11 '22 at 19:38
  • Just some thoughts. $$e^yy'=xe^x-ye^x$$ $$e^y=(x-1)e^x-\int ye^xdx$$ $$e^y=(x-1)e^x-\int ((ye^x)'-y'e^x)dx$$ $$e^y=(x-1)e^x-ye^x-\int y'e^xdx$$ – clathratus Apr 11 '22 at 19:48
  • Yeah, the integral of dx / ln x is the so called logarithmic integral. – Emanuel Landeholm Apr 14 '22 at 18:02
  • Mathematica can't find a solution so it's probably unlikely you will find a closed form solution – JackT May 13 '22 at 01:24

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