Thank you for reporting, we will resolve it shortly
Q.
If $\frac{d y}{d x}=\left(e^y-x\right)^{-1}$ where $y(0)=0$, then $y$ is expressed explicitly as
Differential Equations
Solution:
We have $\frac{d y}{d x}=\left(e^y-x\right)^{-1} \Rightarrow \frac{d x}{d y}=e^y-x \Rightarrow \frac{d x}{d y}+x=e^y ; $
So I.F. $=e^{\int d y}=e^y$
$\therefore$ General solution is given by $xe ^{ y }=\frac{1}{2} e ^{2 y }+ C \Rightarrow x =\frac{ e ^{ y }}{2}+ Ce ^{- y }$
As $y (0)=0$, so $C =\frac{-1}{2}$
$\therefore x =\frac{ e ^{ y }}{2}-\frac{1}{2} e ^{- y } \Rightarrow e ^{ y }- e ^{- y }=2 x $
$\Rightarrow e ^{2 y }-2 xe ^{ y }-1=0 \Rightarrow 2 e ^{ y }=2 x \pm \sqrt{4 x ^2+4}$
But $e ^{ y }= x -\sqrt{ x ^2+1}$
(Rejected)
Hence $y=\ln \left(x+\sqrt{x^2+1}\right)$