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

Aug 1, 2016

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

#### Explanation:

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

so this is separable

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

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

$\int \setminus \frac{d}{\mathrm{dx}} \left({e}^{y}\right) \setminus \mathrm{dx} = \int \setminus {e}^{x} \setminus \mathrm{dx}$

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

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