If the solution y=y(x) of the differential equation [(x^4+2 x^3+3 x^2+2 x+2) d y-(2 x^2+2 …

Question

If the solution $y=y(x)$ of the differential equation $$(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0$$ satisfies $$y(-1)=-\frac{\pi}{4}$$, then $y(0)$ is equal to :

Accepted Answer

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

Solutions · 1