If the solution $y=y(x)$ of the differential equation $\left(x^4+2 x^3+3 x^2+2 x+2\right) \mathrm{d}…
If the solution $y=y(x)$ of the differential equation $\left(x^4+2 x^3+3 x^2+2 x+2\right) \mathrm{d} y-\left(2 x^2+2 x+3\right) \mathrm{d} x=0$ satisfies $y(-1)=-\frac{\pi}{4}$, then $y(0)$ is equal to :
$\frac{\pi}{2}$
$-\frac{\pi}{2}$
0
$\frac{\pi}{4}$
Solution
$\begin{aligned} & \int d y=\int \frac{\left(2 x^2+2 x+3\right)}{x^4+2 x^3+3 x^2+2 x+2} d x \\ & y=\int \frac{\left(2 x^2+2 x+3\right)}{\left(x^2+1\right)\left(x^2+2 x+2\right)} d x \\ & y=\int \frac{d x}{x^2+2 x+2}+\int \frac{d x}{x^2+1} \\ & y=\tan ^{-1}(x+1)+\tan ^{-1} x+C \\ & y(-1)=\frac{-\pi}{4} \\ & \frac{-\pi}{4}=0-\frac{\pi}{4}+C \Rightarrow C=0 \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x \\ & y(0)=\tan ^{-1} 1=\frac{\pi}{4}\end{aligned}$