JEE Main 2026 Apr 2 Shift 1, Mathematics Q20: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Apr 2, Shift 1Mathematics

Let $y = y (x)$ be the solution curve of the differential equation $(1 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ , $y (0) = 0$ . If the curve $y = y (x)$ passes through the point $(α, - \frac{1}{2})$ , then a value of $α$ is :

A
$\frac{π}{6}$
B
$\frac{π}{4}$
C
$\frac{π}{3}$
D
$\frac{π}{2}$

Solution

Rearrange to separate variables: $\frac{d y}{y + 1} = - \frac{cos x}{1 + sin x} d x$ .

Integrate both sides (the right side is $- ln (1 + sin x)$ ):

$ln ∣ y + 1∣ = - ln (1 + sin x) + C$

Apply $y (0) = 0$ : $ln 1 = - ln 1 + C \Rightarrow C = 0$ .

So $y + 1 = \frac{1}{1 + sin x}$ . Setting $y = - 1/2$ : $\frac{1}{2} = \frac{1}{1 + sin α} \Rightarrow 1 + sin α = 2 \Rightarrow sin α = 1$ .

Among the options, $α = π /2$ satisfies $sin α = 1$ .

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JEE Main 2026 Apr 2 Shift 1, Mathematics Q20: Methods of Solving First Order, First Degree Differential Equation

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