# What is the general equation of dy/dx =e^(x+y)?

Apr 5, 2018

$y = \ln \left(\frac{1}{{C}_{0} - {e}^{x}}\right)$

#### Explanation:

This is a separable differential equation so

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

$- {e}^{-} y = {e}^{x} - {C}_{0}$ or

$y = \ln \left(\frac{1}{{C}_{0} - {e}^{x}}\right)$

Apr 5, 2018

$y = - \ln | C - {e}^{x} |$

#### Explanation:

We have:

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

Which we can write as

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

We can collect terms for similar variables:

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

Which is a separable First Order Ordinary non-linear Differential Equation, so we can "separate the variables" to get:

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

Both integrals are those of standard functions, so we can use that knowledge to directly integrate:

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

And we can readily rearrange for $y$:

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

$\therefore - y = \ln | C - {e}^{x} |$

Leading to the General Solution:

$y = - \ln | C - {e}^{x} |$