JEE Main 2026 Jan 28 Shift 2, Mathematics Q14: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Jan 28, Shift 2Mathematics

Let $y = y (x)$ be the solution of the differential equation $x \frac{d y}{d x} - y = x^{2} cot x, x \in (0, π)$ . If $y (\frac{π}{2}) = \frac{π}{2}$ , then $6 y (\frac{π}{6}) - 8 y (\frac{π}{4})$ is equal to :

A
$3 π$
B
$- 3 π$
C
$- π$
D
$π$

Solution

Rewrite: $x d y - y d x = x^{2} cot x d x$ . Divide both sides by $x^{2}$ to recognise the exact form:

$\frac{x d y - y d x}{x ^{2}} = cot x d x \Rightarrow d (\frac{y}{x}) = cot x d x$ .

Integrating: $\frac{y}{x} = lo g_{e} ∣ sin x ∣ + C$ .

Apply $y (π /2) = π /2$ : $1 = lo g_{e} 1 + C \Rightarrow C = 1$ . So $y = x (lo g_{e} sin x + 1)$ .

$y (\frac{π}{6}) = \frac{π}{6} (lo g_{e} (1/2) + 1) = \frac{π}{6} (1 - lo g_{e} 2)$ .

$y (\frac{π}{4}) = \frac{π}{4} (lo g_{e} (1/ 2) + 1) = \frac{π}{4} (1 - \frac{1}{2} lo g_{e} 2)$ .

$6 y (π /6) - 8 y (π /4) = π (1 - lo g_{e} 2) - 2 π (1 - \frac{1}{2} lo g_{e} 2) = π - π lo g_{e} 2 - 2 π + π lo g_{e} 2 = - π$ .

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JEE Main 2026 Jan 28 Shift 2, Mathematics Q14: Methods of Solving First Order, First Degree Differential Equation

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