An inverse function and a differential equation What is the inverse function of $f(x)=x+\exp(x)$?
I doubt it's the solution of the differential equation: $$y'+y'\exp(y)-1=0.$$

The derivative of $f(x)$ is $1+\exp(x)$. So by the inverse function theorem
$$(f^{-1})'(x+\exp(x))=\frac{1}{1+\exp(x)}$$
or
$$(f^{-1})'(y)=\frac{1}{1+\exp(f^{-1}(y))}.$$
Thus $g=f^{-1}$ solves the differential equation
$$g'(x)=\frac{1}{1+\exp(g(x))}$$
with the initial condition $g(1)=0$. Straightforward algebra shows that this is equivalent to your differential equation.
