JEE Main 2026 Jan 21 Shift 2, Mathematics Q16: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Jan 21, Shift 2Mathematics

Let $y = y (x)$ be the solution of the differential equation $sec x \frac{d y}{d x} - 2 y = 2 + 3 sin x, x \in (- \frac{π}{2}, \frac{π}{2})$ , $y (0) = - \frac{7}{4}$ . Then $y (\frac{π}{6})$ is equal to:

A
$- \frac{5}{2}$
B
$- \frac{5}{4}$
C
$- 33 - 7$
D
$- 32 - 7$

Solution

Multiply through by $cos x$ to put it in linear form: $\frac{d y}{d x} - 2 cos x \cdot y = 2 cos x + 3 sin x cos x$ .

Integrating factor: $μ (x) = e^{- 2 s i n x}$ . Then $(e^{- 2 s i n x} y)^{'} = e^{- 2 s i n x} (2 cos x + 3 sin x cos x)$ .

Integrate the right side. With $u = - 2 sin x$ , $d u = - 2 cos x d x$ , partial work gives $e^{- 2 s i n x} y = e^{- 2 s i n x} (- \frac{3}{2} sin x - \frac{7}{4}) + C$ , i.e. $y = - \frac{3}{2} sin x - \frac{7}{4} + C e^{2 s i n x}$ .

Apply $y (0) = - 7/4$ : $- \frac{7}{4} = 0 - \frac{7}{4} + C \Rightarrow C = 0$ .

So $y (x) = - \frac{3}{2} sin x - \frac{7}{4}$ . Evaluate at $x = π /6$ : $y (π /6) = - \frac{3}{2} \cdot \frac{1}{2} - \frac{7}{4} = - \frac{3}{4} - \frac{7}{4} = - \frac{10}{4} = - \frac{5}{2}$ .

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JEE Main 2026 Jan 21 Shift 2, Mathematics Q16: Methods of Solving First Order, First Degree Differential Equation

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