# What is a solution to the differential equation e^x(y'+1)=1?

Jul 10, 2016

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

#### Explanation:

you can tell just by looking that this is separable, so we separate it out

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

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

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

integrate both sides

$\int \setminus y ' \setminus \mathrm{dx} = \int \setminus {e}^{-} x - 1 \setminus \mathrm{dx}$

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