# e^x (y'+1)=1 ? using Separation of Variables

Apr 25, 2018

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

#### Explanation:

We have:

${e}^{x} \left(y ' + 1\right) = 1$

Which we can write as:

$y ' + 1 = {e}^{- x}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- x} - 1$

This is a separable DE, so we can "separate the variables":

$\int \setminus \mathrm{dy} = \int \setminus \left({e}^{- x} - 1 \setminus\right) \mathrm{dx}$

Both integrals are standard function, so we can immediately integrate:

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

Which is the General Solution.