Let $y=y(x)$ be the solution of the differential equation $x d y=\left(y+x^{3} \cos x\right) d x$ with $y(\pi)=0$, then $y\left(\frac{\pi}{2}\right)$ is equal to :

(B) $\frac{\pi^{2}}{4}+\frac{\pi}{2}$

(A) $\frac{\pi^{2}}{4}-\frac{\pi}{2}$

(C) $\frac{\pi^{2}}{2}-\frac{\pi}{4}$

(D) $\frac{\pi^{2}}{2}+\frac{\pi}{4}$

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

x d y = (y + x^{3} cos x) d x

with

y (π) = 0

, then

y (\frac{π}{2})

is equal to :

Answer: (B)

Writing the equation as

\frac{d y}{d x} - \frac{y}{x} = x^{2} cos x

, the integrating factor is

\frac{1}{x}

, so

\frac{d}{d x} (\frac{y}{x}) = x cos x ⟹ \frac{y}{x} = x sin x + cos x + C

From

y (π) = 0

0 = π sin π + cos π + C = - 1 + C

, so

C = 1

. At

x = \frac{π}{2}

\frac{y}{π /2} = \frac{π}{2} + 0 + 1

y (\frac{π}{2}) = \frac{π}{2} (\frac{π}{2} + 1) = \frac{π ^{2}}{4} + \frac{π}{2}

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option

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