# How do you find the general solution to dy/dx=xe^y?

Jul 24, 2016

$y = \ln \left(\frac{2}{C - {x}^{2}}\right)$

#### Explanation:

$y ' = x {e}^{y}$

${e}^{- y} y ' = x$

$\int \setminus {e}^{- y} y ' \setminus \mathrm{dx} = \int \setminus x \setminus \mathrm{dx}$

$\int \setminus {e}^{- y} \setminus \mathrm{dy} = \int \setminus x \setminus \mathrm{dx}$

$- {e}^{- y} = {x}^{2} / 2 + C$

${e}^{- y} = C - {x}^{2} / 2$

${e}^{y} = \frac{1}{C - {x}^{2} / 2}$
$= \frac{2}{C - {x}^{2}}$

$y = \ln \left(\frac{2}{C - {x}^{2}}\right)$