Respuesta :

JцstinBieber
[tex]\frac{dy}{dx} = xe^{-y}\ \Rightarrow\ \frac{dy}{e^{-y} } = x dx\ \Rightarrow\ \\ \int \frac{dy}{e^{-y} } = \int x dx \Rightarrow \int e^y dy= \int x dx \Rightarrow \\ e^y = \frac{x^2}{2} + C \implies y = \ln\left(\frac{x^2}{2} + C \right)[/tex]