The general solution of the differential equation $xdy +e^{-y}dx=xe^{x-y}dx$ is (given C is constant of integration)

$e^x+log\, x -2xy =C$ $x^2+e^{-y}+xe^x=C$ $xe^y-ye^x+log \left(\frac{x}{y}\right)=C$ $e^y-e^x+log x=C$

$e^y-e^x+log x=C$