If \(y(\dot{x})\) is the solution of the differential equation \((x+2) \frac{\mathrm{d} y}{\mathrm{~d} x}=x^2+4 x-9, x \neq-2\) and \(y(0)=0\), then \(y(-4)\) is equal to

\(\begin{array}{ll}
& (x+2) \frac{\mathrm{d} y}{\mathrm{~d} x}=x^2+4 x-9 \\
\therefore \quad & \mathrm{~d} y=\frac{\left(x^2+4 x+4\right)-4-9}{x+2} \mathrm{~d} x \\
\therefore \quad & \mathrm{~d} y=\frac{(x+2)^2-13}{(x+2)} \mathrm{d} x
\end{array}\)
Integrating both sides, we get
\(\begin{aligned}
& \int \mathrm{d} y=\int(x+2) \mathrm{d} x-13 \int \frac{1}{x+2} \mathrm{~d} x \\
& y=\frac{(x+2)^2}{2}-13 \log |x+2|+c...(i)
\end{aligned}\)
Given that \(y(0)=0\)
\(\therefore \quad\) from equation (i), we get
\(\begin{aligned}
& \\
\therefore \quad & 0 \\
\therefore \quad & =2-13 \log |0+2|+c ...(ii)\\
& =13 \log (2)-2
\end{aligned}\)
\(\therefore \quad\) from (i) and (ii), we get
\(y(-4)=2-13 \log (2)+13 \log (2)-2=0\)