Let y=y(x) be the solution of the differential equation (x^2+1) y^ -2 x y=(x^4+2 x^2+1) x , y(0)=1 . Then ∫_ -3 ^3 y(x) d x is :

Correct answer: A. 24 aligned & (x^2+1) (d y)/(d x)-2 x y=(x^4+2 x^2+1) x & (d y)/(d x)-((2 x)/(x^2+1)) y= (x^2+1)^2 x cx^2+1 =(x^2+1) x & (Linear D.E) & P=(-2 x)/(x^2+1), Q=(x^2+1) x & I.F =e^ ∫ P d x =e^ ∫ (-2 x)/(x^2+1) d x =(1)/(x^2+1) & y · (1)/(x^2+1)=∫(x^2+1) x · (1)/(x^2+1) d x & (y)/(x^2+1)= x+c y =1 c=1 & y=(x^2+1)( x+1) & ∫_ -3 ^3 y d x=∫_ -3 ^3(x^2+1)( x+1) & d x=∫_ -3 ^3 x^2 x+x^2 x+1 d x & ∫_ -3 ^3 x^2 x d x+∫_ -3 ^3 x^2 d x+∫_ -3 ^3 x d x+∫_ -3 ^3 1 d x & =0+18+0+6=24 aligned

JEE Main 20257 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation (x^2+1) y^...

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 :

Options

Correct answer

A. 24

Step-by-step solution

\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}

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