# How do you find the particular solution to y(x+1)+y'=0 that satisfies y(-2)=1?

Nov 11, 2016

$y = {e}^{- \frac{1}{2} {x}^{2} - x}$

#### Explanation:

We have $y \left(x + 1\right) + y ' = 0$ which is a First Order separable DE,which can be separated as follows;

$y \left(x + 1\right) + y ' = 0$
$y ' = - y \left(x + 1\right)$
$\therefore \int \frac{1}{y} \mathrm{dy} = \int - \left(x + 1\right) \mathrm{dx}$

Integrating gives us;

$\ln y = - \frac{1}{2} {x}^{2} - x + C$
And $y \left(- 2\right) = 1 \implies \ln 1 = \left(- \frac{1}{2}\right) \left(4\right) - \left(- 2\right) + C$
$\therefore 0 = - 2 + 2 + C \implies C = 0$

So the particular solution is

$\ln y = - \frac{1}{2} {x}^{2} - x$
$\therefore y = {e}^{- \frac{1}{2} {x}^{2} - x}$