Solution of the equation $\frac{dy}{dx}=e^{x-y}\left(e^x-e^y\right)$ is

We have,

$\frac{dy}{dx}=e^{x-y}\left(e^x-e^y\right)$

$\Rightarrow e^y\frac{dy}{dx}+e^y{e}^x=e^{2x}$

Put $e^y=v$

$\therefore e^y\frac{dy}{dx}=\frac{dv}{dx}$

$\therefore \frac{dv}{dx}+v{e}^x=e^{2x}$

This is linear equation 

$\mathrm{I}.\mathrm{F}==e^{\int e^{x}dx}=e^{e^x}$

$\therefore$ Required solution is 

$v·e^{e^x}=\int e^{e^x}·e^{2x}dx+C$

$v·e^{e^x}=e^{e^x}\left(e^x-1\right)+C$

$\Rightarrow e^y·e^{e^x}=e^{e^x}\left(e^x-1\right)+C$

$\Rightarrow e^y=e^x-1+c{e}^{-e^x}$