Question

Solve the ODE xe^x2+ydx=ydy

Answer 1

$xe^{x^2+y}dx=dy\\ xe^{x^2}e^ydx=dy\\ xe^{x^2}dx=e^{-y}dy\\ \text{Integrating both side, we get}\\ put x^2=t\\ \implies 2xdx=dt\\ \frac{1}{2}\int e^tdt=\int e^{-y}dy\\ \frac{1}{2}e^t+C=-e^{-y}\\ \text{This is the required answer.}$