Q31 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
Let $y=y(x)$ be the solution of the differential equation $x^{4} \mathrm{~d} y+\left(4 x^{3} y+2 \sin x\right) \mathrm{d} x=0, x>0, y\left(\frac{\pi}{2}\right)=0$. Then $\pi^{4} y\left(\frac{\pi}{3}\right)$ is equal to :
- A72
- B92
- C64
- Correct answerD81
Solution
$x^4\,dy + 4x^3 y\,dx + 2\sin x\,dx = 0 \Rightarrow d(x^4 y) + 2\sin x\,dx = 0$.
Integrating: $x^4 y - 2\cos x = C$.
$y(\pi/2) = 0$: $(\pi/2)^4 \cdot 0 - 2\cos(\pi/2) = 0 \Rightarrow C = 0$.
$x^4 y = 2\cos x$.
At $x = \pi/3$: $(\pi/3)^4 y(\pi/3) = 2\cos(\pi/3) = 1$.
$\pi^4 y(\pi/3) = 81 \cdot (\pi/3)^4 y(\pi/3) \cdot \frac{1}{(\pi/3)^4} \cdot \frac{\pi^4}{81}$... Directly: $y(\pi/3) = \frac{81}{\pi^4}$, so $\pi^4 y(\pi/3) = 81$.
Integrating: $x^4 y - 2\cos x = C$.
$y(\pi/2) = 0$: $(\pi/2)^4 \cdot 0 - 2\cos(\pi/2) = 0 \Rightarrow C = 0$.
$x^4 y = 2\cos x$.
At $x = \pi/3$: $(\pi/3)^4 y(\pi/3) = 2\cos(\pi/3) = 1$.
$\pi^4 y(\pi/3) = 81 \cdot (\pi/3)^4 y(\pi/3) \cdot \frac{1}{(\pi/3)^4} \cdot \frac{\pi^4}{81}$... Directly: $y(\pi/3) = \frac{81}{\pi^4}$, so $\pi^4 y(\pi/3) = 81$.
Hence, the correct answer is option: 'D'.