Aspire Faculty ID #17992 · Topic: JEE Main 2026 (23 January Morning Shift) · Just now
JEE Main 2026 (23 January Morning Shift)

Let $y = y(x)$ be the solution of the differential equation

$x^4dy + (4x^3y + 2\sin x)dx = 0,; x > 0,; y\left(\frac{\pi}{2}\right) = 0.$

Then $\pi^4 y\left(\frac{\pi}{3}\right)$ is equal to:

Solution

$(x^4dy + 4x^3y,dx) = -2\sin x,dx$

$\Rightarrow \int d(x^4y) = \int -2\sin x,dx$

$\Rightarrow x^4y = 2\cos x + c$

$\Rightarrow x^4f(x) = 2\cos x + c$

As $f\left(\frac{\pi}{2}\right) = 0$

So, $c = 0$

$\left(\frac{\pi}{3}\right)^4 f\left(\frac{\pi}{3}\right) = 2\cos \frac{\pi}{3}$

$\Rightarrow \pi^4 f\left(\frac{\pi}{3}\right) = 81$

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