# What is a solution to the differential equation dy/dx=x+y?

Oct 19, 2016

$y = {C}_{1} {e}^{x} - x - 1$

#### Explanation:

Let $u = x + y$

$\implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(x + y\right) = 1 + \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} - 1$

Thus, making the substitutions into our original equation,

$\frac{\mathrm{du}}{\mathrm{dx}} - 1 = u$

$\implies \frac{\mathrm{du}}{u + 1} = \mathrm{dx}$

$\implies \int \frac{\mathrm{du}}{u + 1} = \int \mathrm{dx}$

$\implies \ln \left(u + 1\right) = x + {C}_{0}$

$\implies {e}^{\ln \left(u + 1\right)} = {e}^{x + {C}_{0}}$

$\implies u + 1 = {C}_{1} {e}^{x} \text{ }$ (where ${C}_{1} = {e}^{{C}_{0}}$)

Substituting $x + y = u$ back in,

$\implies x + y + 1 = {C}_{1} {e}^{x}$

$\therefore y = {C}_{1} {e}^{x} - x - 1$