Answer
$$y=e^{-\frac{x^2}{2}-x}$$
Work Step by Step
$y(x+1)+y'=0$
$y'=-y(x+1)$
$\frac{y'}{y}=-(x+1)$
$\int\frac{1}{y}dy=-\int (x+1)dx$
$\ln{y}=-\frac{x^2}{2}-x+C$
$e^{\ln{y}}=y=e^{-\frac{x^2}{2}-x+C}=e^Ce^{-\frac{x^2}{2}-x}$
Let $k=e^C$ so we get
$y=ke^{-\frac{x^2}{2}-x}$
Now applying our initial condition we have
$1=ke^{-\frac{(-2)^2}{2}+2}=ke^0$
Therefore $k=1$
Our Particular solution then becomes
$y=e^{-\frac{x^2}{2}-x}$
