JEE Main 2025 — Differential Equations Question with Solution
JEE Main 2025 (7 Apr Shift 2)
Question
Let be the solution of the differential equation , . Then is :
Choose an option
Show full solutionCorrect option: A
Correct answer
A24
Step-by-step explanation
$\begin{aligned}
& \left(x^2+1\right) \frac{d y}{d x}-2 x y=\left(x^4+2 x^2+1\right) \cos x \\ & \frac{d y}{d x}-\left(\frac{2 x}{x^2+1}\right) y=\frac{\left(x^2+1\right)^2 \cos x}{\mathrm{cx}^2+1}=\left(\mathrm{x}^2+1\right) \cos x \\ & \text { (Linear D.E) } \\ & P=\frac{-2 x}{x^2+1}, Q=\left(x^2+1\right) \cos x \\ & \text { I.F }=e^{\int P d x}=e^{\int \frac{-2 x}{x^2+1} d x}=\frac{1}{x^2+1} \\ & y \cdot \frac{1}{x^2+1}=\int\left(x^2+1\right) \cos x \cdot \frac{1}{x^2+1} d x \\ & \frac{y}{x^2+1}=\sin x+c \Rightarrow y \cos =1 \Rightarrow c=1 \\ & y=\left(x^2+1\right)(\sin x+1) \\ & \int_{-3}^3 y d x=\int_{-3}^3\left(x^2+1\right)(\sin x+1) \\ & d x=\int_{-3}^3 x^2 \sin x+x^2 \sin x+1 d x \\ & \Rightarrow \int_{-3}^3 x^2 \sin x d x+\int_{-3}^3 x^2 d x+\int_{-3}^3 \sin x d x+\int_{-3}^3 1 d x \\ & =0+18+0+6=24
\end{aligned}$
& \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}{\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
\end{aligned}$
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