Question

Find the solution of the differential equation that satisfies the given initial condition. dy/dx = xe^y, y(0) = 0

Answer 1

$\dfrac{\mathrm dy}{\mathrm dx}=xe^y$
$e^{-y}\,\mathrm dy=x\,\mathrm dx$
$-e^{-y}=\dfrac{x^2}2+C$

$y(0)=0$
$\implies-1=C$
$\implies -e^{-y}=\dfrac{x^2}2-1$
$e^{-y}=1-\dfrac{x^2}2$
$-y=\ln\left(1-\dfrac{x^2}2\right)$
$y=-\ln\left(1-\dfrac{x^2}2\right)$