The general solution of the differential equation
$xdy +e^{-y}dx=xe^{x-y}dx$ is (given C is constant of integration)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $e^y-e^x+log x=C$
$xdy +e^{-y}dx=xe^{x-y}dx$
$⇒xdy+\frac{dx}{e^{y}}=\frac{xe^xdx}{e^{y}}$
$⇒xdy=\frac{dx(xe^x-1)}{e^y}$
$⇒\int e^ydy=\int dx(e^x-\frac{1}{x})$
$⇒e^y=e^x-\log x+C$
$⇒e^y-e^x+\log x=C$