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

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Steve M Share
Nov 16, 2016

$y = - \ln \left(- {e}^{x} + C\right)$, or $\ln \left(\frac{1}{C - {e}^{x}}\right)$

Explanation:

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

So we can identify this as a First Order Separable Differential Equation. We can therefore "separate the variables" to give:

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

Integrating gives us:

$- {e}^{-} y = {e}^{x} + C '$
$\therefore {e}^{-} y = - {e}^{x} + C$
$\therefore - y = \ln \left(- {e}^{x} + C\right)$
$\therefore y = - \ln \left(C - {e}^{x}\right)$, or $\ln \left(\frac{1}{C - {e}^{x}}\right)$

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