# What is a particular solution to the differential equation dy/dx=(y+5)(x+2) with y(0)=-1?

Jul 9, 2016

$y = 4 {e}^{x \left(\frac{x}{2} + 2\right)} - 5$

#### Explanation:

separate it first

you get

$\frac{1}{y + 5} \frac{\mathrm{dy}}{\mathrm{dx}} = x + 2$

So integrating
$\int \setminus \frac{1}{y + 5} \frac{\mathrm{dy}}{\mathrm{dx}} \setminus \mathrm{dx} = \int x + 2 \setminus \mathrm{dx}$

$= \int \setminus \frac{1}{y + 5} \setminus \mathrm{dy} = \int x + 2 \setminus \mathrm{dx}$

So

$= \ln \left(y + 5\right) = {x}^{2} / 2 + 2 x + C$

feeding in the IV $y \left(0\right) = - 1$

$= \ln \left(- 1 + 5\right) = C = \ln 4$

$\implies \ln \left(y + 5\right) = {x}^{2} / 2 + 2 x + \ln 4$

$\implies \ln \left(\frac{y + 5}{4}\right) = {x}^{2} / 2 + 2 x$

$y = 4 {e}^{x \left(\frac{x}{2} + 2\right)} - 5$