Question

Solve $y(x^{2}y+e^x)dx=e^{x}dy$

Answer 1

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

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

$\frac{1}{y^2}\frac{dy}{dx}=\frac{x^2}{e^x}+\frac{1}{y}$

$\frac{-1}{y}=t$

$\frac{1}{y^2}\frac{dx}{dy}=\frac{dt}{dx}$

$\frac{dt}{dx}+t=\frac{x^2}{e^x}$