Let y=y(x) be the solution of the differential equation x^ 4 ~d y+ (4 x^ 3 y+2 x) d x=0, x>0, y ((π)/(2))=0 . Then π^ 4 y((π)/(3)) is equal to :

Correct answer: D. 81 x^4\,dy + 4x^3 y\,dx + 2 x\,dx = 0 d(x^4 y) + 2 x\,dx = 0 . Integrating: x^4 y - 2 x = C . y(π/2) = 0 : (π/2)^4 · 0 - 2 (π/2) = 0 C = 0 . x^4 y = 2 x . At x = π/3 : (π/3)^4 y(π/3) = 2 (π/3) = 1 . π^4 y(π/3) = 81 · (π/3)^4 y(π/3) · (1)/((π/3)^4) · (π^4)/(81) ... Directly: y(π/3) = (81)/(π^4) , so π^4 y(π/3) = 81 .

JEE Main 202623 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation x^ 4 ~d y+...

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 :

Options

Correct answer

D. 81

Step-by-step 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

Practice Differential Equations on Quantrex Academy →

More from Differential Equations