First order differential equation confusion I have a differential equation $$y' + e^{y'}-x=0$$ that I have simplified like so $$e^{y'}=x-y'$$ $$\ln e^{y'}=\ln (x-y')$$ $$y'= \ln (x-y')$$ but I do not know how to solve this further to obtain the general solution. I have done first order linear differential equation strategies so far. How should I get about doing this question with the strategies I have? 
 A: Your (revised) simplification doesn't really simplify because it separates the $y'$s from each other. The productive thing to do would be to rewrite it to
$$ x = y'+e^{y'} $$
and work from there.
First, let $g(u) = u+e^u$; then the equation becomes $x = g(y')$ or $y'=g^{-1}(x)$. In this form we get the solution
$$ y = \int_1^x g^{-1}(t)\, dt + C$$
This integral is not nice, but (noting that $g^{-1}(1)=0$, which is why I chose $1$ as the lower limit) we can flip it around to
$$ y = xg^{-1}(x) - \int_0^{g^{-1}(x)} g(u) \,du + C $$
and $g$ is easy enough to integrate so we get
$$ y = xg^{-1}(x) - \frac{g^{-1}(x)^2}2 - e^{g^{-1}(x)} + C_1 $$
(with $C_1=C+1$). With a bit of help from Wolfram Alpha we find $g^{-1}(x) = x - W(e^x)$ where $W$ is the Lambert W function. We can plug this in to get
$$ y = \frac{x^2}2 - \frac{W(e^x)^2}2 + e^{x-W(e^x)} + C_2$$
By the definition of $W$ we have $e^{-W(e^x)} = W(e^x)/e^x$, so this further simplifies to
$$ y = \frac{{x^2} - (W(e^x)-1)^2}{2} + C_3 $$
If still not exactly pretty, this is at least a "closed form" modulo the $W$ function.
A: I also suppose a typo somewhere. Using a CAS, I did not get any solution beside $$y=c+\int_1^x \left(t-W\left(e^{t}\right)\right) \, dt$$ which is not very pleasant.
Even $y' + e^{y}-x=0$ is not very pleasant, the solution being $$y=\frac{1}{2} \left(x^2-2 \log \left(c+\sqrt{\frac{\pi }{2}}
   \text{erfi}\left(\frac{x}{\sqrt{2}}\right)\right)\right)$$
