$$\begin{aligned} & \left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0 \\ & \int d y=\int\left(\frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} d x\right. \\ & \int d y=\int \frac{1}{x^2+1} d x+\int \frac{}{x^2+2 x+2} d x \\ & y=\tan ^{-1}(x)+\tan ^{-1}(1+x)+C \\ & y(-1)=\tan ^{-1}(-1)+\tan ^{-1}(1-1)+C \\ & y(-1)=-\frac{\pi}{4}+C=\left(\frac{-\pi}{4}\right)-\{\text { given }\} \\ & \Rightarrow C=0 \\ & \text { So, } y(x)=\tan ^{-1}(x)+\tan ^{-1}(1+x) \\ & y(0)=\tan ^{-1}(0)+\tan ^{-1}(1+0) \\ & y(0)=\frac{\pi}{4} \end{aligned}$$