The general solution of the differential equation

$xdy +e^{-y}dx=xe^{x-y}dx$ is (given C is constant of integration)

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

The correct answer is Option (4) → $e^y-e^x+log x=C$

$xdy +e^{-y}dx=xe^{x-y}dx$

$⇒xdy+\frac{dx}{e^{y}}=\frac{xe^xdx}{e^{y}}$

$⇒xdy=\frac{dx(xe^x-1)}{e^y}$

$⇒\int e^ydy=\int dx(e^x-\frac{1}{x})$

$⇒e^y=e^x-\log x+C$

$⇒e^y-e^x+\log x=C$