If y(x) is the solution of the differential equation $(\mathrm{x}+2)\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2+4\mathrm{x}-9, \mathrm{x}\ne -2$ and $\mathrm{y}\left(0\right)=0,$ then $\mathrm{y}(-4)$ is equal to :

$\begin{array}{l}(x+2)\frac{dy}{dx}=x^2+4x+4-13\\ \Rightarrow (x+2)\frac{dy}{dx}={(x+2)}^2-13\\ \Rightarrow \frac{dy}{dx}=(x+2)-\frac{13}{x+2}\\ \Rightarrow \int dy=\int (x+2)dx-13\int \frac{1}{x+2}dx\\ \Rightarrow y=\frac{x^2}{2}+2x-13\log \left|(x+2)\right|+c\end{array}$

$\begin{array}{l}Putx=0\\ \Rightarrow y\left(0\right)=0-13\log 2+c\\ \Rightarrow c=13\log 2\\ y=\frac{x^2}{2}+2x-13\log (x+2)+13\log 2\\ Putx=-4\\ \Rightarrow y(-4)=8-8-13\log \left|(-2)\right|+13\log 2=0\end{array}$