Q31 / 75 · JEE Main 2025 (7 Apr Shift 2) · Mathematics
Let $y=y(x)$ be the solution of the differential equation $\left(x^2+1\right) y^{\prime}-2 x y=\left(x^4+2 x^2+1\right) \cos x$, $y(0)=1$. Then $\int_{-3}^3 y(x) d x$ is :
- Correct answerA24
- B36
- C30
- D18
Solution
$\begin{aligned}<br/>& \left(x^2+1\right) \frac{d y}{d x}-2 x y=\left(x^4+2 x^2+1\right) \cos x \\
& \frac{d y}{d x}-\left(\frac{2 x}{x^2+1}\right) y=\frac{\left(x^2+1\right)^2 \cos x}{\mathrm{cx}^2+1}=\left(\mathrm{x}^2+1\right) \cos x \\
& \text { (Linear D.E) } \\
& P=\frac{-2 x}{x^2+1}, Q=\left(x^2+1\right) \cos x \\
& \text { I.F }=e^{\int P d x}=e^{\int \frac{-2 x}{x^2+1} d x}=\frac{1}{x^2+1} \\
& y \cdot \frac{1}{x^2+1}=\int\left(x^2+1\right) \cos x \cdot \frac{1}{x^2+1} d x \\
& \frac{y}{x^2+1}=\sin x+c \Rightarrow y \cos =1 \Rightarrow c=1 \\
& y=\left(x^2+1\right)(\sin x+1) \\
& \int_{-3}^3 y d x=\int_{-3}^3\left(x^2+1\right)(\sin x+1) \\
& d x=\int_{-3}^3 x^2 \sin x+x^2 \sin x+1 d x \\
& \Rightarrow \int_{-3}^3 x^2 \sin x d x+\int_{-3}^3 x^2 d x+\int_{-3}^3 \sin x d x+\int_{-3}^3 1 d x \\
& =0+18+0+6=24<br/>\end{aligned}$
Hence, the correct answer is option: 'A'.