Let $y=y(x)$ be the solution of the differential equation $(x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0$. If $y(0)=0$, then $y(2)$ is equal to

<p>$$\frac{d y}{d x}+\frac{y\left(2 x^3+8 x\right)}{\left(x^2+4\right)^2}=\frac{2}{\left(x^2+4\right)^2}$$</p> <p>$\mathrm{IF}=e^{\int \frac{2 x^3+8 x}{\left(x^2+4\right)^2} d x}$</p> <p>$$\text { Let }\left(x^2+4\right)^2=t \quad \Rightarrow 2\left(x^2+4\right)(2 x) d x=d t$$</p> <p>$$\begin{gathered} =e^{\int \frac{d t}{2 t}}=e^{\log \sqrt{t}}=\sqrt{t}=\left(x^2+4\right) \\ \therefore \quad y\left(x^2+4\right)=\int \frac{2}{x^2+4}+c \\ \Rightarrow y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)+c \\ y(0)=0 \end{gathered}$$</p> <p>$$\begin{aligned} & \Rightarrow 0=0+c \quad \Rightarrow c=0 \\ & \text { put } x=2 \\ & y(8)=\frac{\pi}{4} \quad \Rightarrow \quad y=\frac{\pi}{32} \\ \end{aligned}$$</p>