If $\frac{d x}{d y}=\left(e^y-x\right)$, where $y(0)=0$, then $y$ is expressed explicitly as

$\ln \left(x+\sqrt{1+x^2}\right)$

We have,

$\frac{d x}{d y}=e^y-x \Rightarrow \frac{d x}{d y}+x=e^y$           ....(i)

This is a linear differential equation with I.F. $=e^y$.

Multiplying both sides of (i) by integrating factor = $e^y$ and integrating with respect to $y$, we obtain

$x e^y=\frac{1}{2} e^{2 y}+C$          ........(ii)

It is given that $y=0$ when $x=0$. Putting these values in (ii), we obtain $C=-\frac{1}{2}$.

Substituting $C=-\frac{1}{2}$ in (ii), we obtain

$x e^y=\frac{1}{2} e^{2 y}-\frac{1}{2}$

$\Rightarrow e^{2 y}-2 x e^y-1=0$

$\Rightarrow e^y=x+\sqrt{x^2+1} \Rightarrow y=\ln \left(x+\sqrt{x^2+1}\right)$