Let y = y(x) be the solution of the differential equation (x^2 + 1)y' - 2xy = (x^4 + 2x^2 …

Question

Let y = y(x) be the solution of the differential equation $(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x$,
$y(0) = 1$. Then $ \int\limits_{-3}^{3} y(x) \, dx $ is :

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \left(x^2+1\right) \frac{d y}{d x}-2 x y=\left(x^4+2 x^2+1\right) \cos x \
& \frac{d y}{d x}-\left(\frac{2 x}{x^2+1}\right) y=\frac{\left(x^2+1\right)^2 \cos x}{c x^2+1}=\left(x^2+1\right) \cos x
\end{aligned}\
\&\text { (Linear D.E) }\
\&\mathrm{P}=\frac{-2 \mathrm{x}}{\mathrm{x}^2+1}, \mathrm{Q}=\left(\mathrm{x}^2+1\right) \cos \mathrm{x}
\end{aligned}$$

$$\begin{aligned}
& \text { I.F }=\mathrm{e}^{\int P d x}=\mathrm{e}^{\int \frac{-2 x}{x^2+1} d x}=\frac{1}{x^2+1} \
& y \cdot \frac{1}{x^2+1}=\int\left(x^2+1\right) \cos x \cdot \frac{1}{x^2+1} d x \
& \frac{y}{x^2+1}=\sin x+c \Rightarrow y \cos =1 \Rightarrow c=1
\end{aligned}$$

$$\begin{aligned}
& y=\left(x^2+1\right)(\sin x+1) \
& \int_{-3}^3 y d x=\int_{-3}^3\left(x^2+1\right)(\sin x+1) \
& d x=\int_{-3}^3 x^2 \sin x+x^2 \sin x+1 d x \
& \Rightarrow \int_{-3}^3 x^2 \sin x d x+\int_{-3}^3 x^2 d x+\int_{-3}^3 \sin x d x+\int_{-3}^3 1 d x \
& =0+18+0+6=24
\end{aligned}$$

Solutions · 1