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: $$y'+y'\exp(y)-1=0.$$
 A: Notice:
The inverse function of $y=x+\exp(x)$ is: $x=y-\text{W}\left(\exp(y)\right)$

$$y'(x)+y'(x)\exp(y(x))-1=0\Longleftrightarrow$$
$$y'(x)=\frac{1}{\exp(y(x))+1}\Longleftrightarrow$$
$$y'(x)\left(\exp(y(x))+1\right)=1\Longleftrightarrow$$
$$\int y'(x)\left(\exp(y(x))+1\right)\space\text{d}x=\int1\space\text{d}x\Longleftrightarrow$$
$$\exp(y(x))+y(x)=x+\text{C}\Longleftrightarrow$$
$$y(x)=x-\text{W}\left(\exp(x+\text{C})\right)+\text{C}$$
So you have to set the initial condition on: $y(1)=0$
$$1-\text{W}\left(\exp(1+\text{C})\right)+\text{C}=0\Longleftrightarrow$$
$$\text{C}=0$$
So:
$$y(x)=x-\text{W}\left(\exp(x)\right)$$
With $\text{W}(z)$ is te product log function
A: 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.
A: $$y'(1+e^y)=1$$
Integration, with constant $c$ :
$$y+e^y=x+c$$
$$x+c-y=e^y$$
$$(x+c-y)e^{-y}=1$$
$$(x+c-y)e^{(x+c-y)}=e^{x+c}$$
Let $Y=(x+c-y)$ and $X=e^{x+c}$
$$Ye^Y=X$$
With the Lambert W function :
$$Y=W(X)$$
$$x+c-y=W\left(e^{x+c}\right)$$
$$y=x+c-W\left(e^{x+c}\right)$$
