Let $y=y(x)$ be the solution of the differential equation $x^{4} d y+\left(4 x^{3} y+2 \sin x\right) d x=0$, $x>0$, $y\left(\frac{\pi}{2}\right)=0$. Then $\pi^{4} \mathrm{y}\left(\frac{\pi}{3}\right)$ is equal to :

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

x^{4} d y + (4 x^{3} y + 2 sin x) d x = 0

x > 0

y (\frac{π}{2}) = 0

.

Then

π^{4} y (\frac{π}{3})

is equal to :

Answer: (A)

Rewrite the differential equation:

x^{4} d y + 4 x^{3} y d x = - 2 sin x d x

.

The left side is

d (x^{4} y)

, so integrating:

x^{4} y = 2 cos x + c

.

Using

y (\frac{π}{2}) = 0

(\frac{π}{2})^{4} \cdot 0 = 2 cos \frac{π}{2} + c

, so

c = 0

.

Thus

x^{4} y = 2 cos x

.

At

x = \frac{π}{3}

(\frac{π}{3})^{4} y (\frac{π}{3}) = 2 cos \frac{π}{3} = 2 \cdot \frac{1}{2} = 1

.

Therefore

π^{4} y (\frac{π}{3}) = 81 \cdot 1 = 81

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option

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