How do you solve the separable differential equation dy/dx = e^(x-y)?

Mar 8, 2018

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

Explanation:

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

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

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

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

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

${e}^{y} = {e}^{x} + C$ ; $C \setminus \equiv$ constant

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

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