JEE Main 2026 Jan 23 Shift 1, Mathematics Q7: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Jan 23, Shift 1Mathematics

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 :

A
$81$
B
$92$
C
$64$
D
$72$

Solution

Spot the exact form. $x^{4} d y + 4 x^{3} y d x = d (x^{4} y)$ , so:

$d (x^{4} y) = - 2 sin x d x$

Integrating: $x^{4} y = 2 cos x + C$ .

Apply $y (π /2) = 0$ : $0 = 2 cos (π /2) + C = C$ , so $C = 0$ .

At $x = π /3$ : $(\frac{π}{3})^{4} y (\frac{π}{3}) = 2 cos \frac{π}{3} = 1$ .

$∴ π^{4} y (\frac{π}{3}) = 81$ .

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JEE Main 2026 Jan 23 Shift 1, Mathematics Q7: Methods of Solving First Order, First Degree Differential Equation

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