# What is a solution to the differential equation dy/dx=e^(x-y)?

Aug 16, 2016

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

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x - y}$
$= {e}^{x} \cdot {e}^{-} y$ we can separate it!

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

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

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

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

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

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