# What is a solution to the differential equation y'=-2xe^y?

Jul 11, 2016

$y = \ln \left(\frac{1}{{x}^{2} + C}\right)$

#### Explanation:

$y ' = - 2 x {e}^{y}$

this is separable

${e}^{-} y \setminus y ' = - 2 x$

$\int {e}^{-} y \setminus y ' \setminus \mathrm{dx} = - 2 \int x \setminus \mathrm{dx}$

$\int \setminus {e}^{-} y \setminus \mathrm{dy} = - 2 \int x \setminus \mathrm{dx}$

$- {e}^{-} y \setminus = - 2 \left({x}^{2} / 2 + C\right)$

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

${e}^{y} \setminus = \frac{1}{{x}^{2} + C}$

$\ln \left({e}^{y}\right) \setminus = \ln \left(\frac{1}{{x}^{2} + C}\right)$

$y = \ln \left(\frac{1}{{x}^{2} + C}\right)$