Let $y=y(x)$ be the solution of the differential equation $\left(x^{2}+1\right) y^{\prime}-2 x y=\left(x^{4}+2 x^{2}+1\right) \cos x$, $\mathrm{y}(0)=1$. Then $\int_{-3}^{3} \mathrm{y}(\mathrm{x}) \mathrm{dx}$ is :

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(x^{2} + 1) y^{'} - 2 x y = (x^{4} + 2 x^{2} + 1) cos x

y (0) = 1

. Then

\int_{- 3}^{3} y (x) dx

is :

Answer: (A)

The right side is

(x^{2} + 1)^{2} cos x

, so dividing by

x^{2} + 1

gives

y^{'} - \frac{2 x}{x ^{2} + 1} y = (x^{2} + 1) cos x

with integrating factor

\frac{1}{x ^{2} + 1}

. Then

\frac{d}{d x} (\frac{y}{x ^{2} + 1}) = cos x

, so

\frac{y}{x ^{2} + 1} = sin x + c

; the condition

y (0) = 1

gives

c = 1

, hence

y = (x^{2} + 1) (1 + sin x)

. Over

[- 3, 3]

the odd part

(x^{2} + 1) sin x

integrates to zero, leaving

\int_{- 3}^{3} (x^{2} + 1) d x = 2 [\frac{x ^{3}}{3} + x]_{0}^{3} = 24

. Thus

\int_{- 3}^{3} y d x = 24

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option

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