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

Question

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

Aicharya · Accepted Answer

$\begin{array}{ll} & (x+2) \frac{\mathrm{d} y}{\mathrm{~d} x}=x^2+4 x-9 \\ \therefore \quad & \mathrm{~d} y=\frac{\left(x^2+4 x+4\right)-4-9}{x+2} \mathrm{~d} x \\ \therefore \quad & \mathrm{~d} y=\frac{(x+2)^2-13}{(x+2)} \mathrm{d} x \end{array}$
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 $y(-4)=2-13 \log (2)+13 \log (2)-2=0$

If $y(\dot{x})$ is the solution of the differential equation $(x+2) \frac{\mathrm{d} y}{\mathrm{~d} x}=x^2+4…

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