# How do you find the general solution to dy/dx+e^(x+y)=0?

Jul 24, 2016

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

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{x + y} = 0$

this is separable

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

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

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

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

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

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

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

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

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