Let y = y(x) be the solution curve of the differential equation (1 + x)(dy)/(dx) + (y+1) x = 0 , y(0) = 0 . If the curve y = y(x) passes through the point (α, (-1)/(2)) , then a value of α is :

Correct answer: D. (π)/(2) The given differential equation can be written as: (1 + x)dy + (y+1) x dx = 0 d((y+1)(1 + x)) = 0 Integrating both sides: (y+1)(1 + x) = C Given y(0) = 0 , substituting x = 0 and y = 0 : (0+1)(1 + 0) = C C = 1 The equation of the curve is (y+1)(1 + x) = 1 Since the curve passes through (α, (-1)/(2)) , substituting x = α and y = (-1)/(2) : ((-1)/(2) + 1)(1 + α) = 1 (1)/(2)(1 + α) = 1 1 + α = 2 α = 1 α = (π)/(2) Answer: (π)/(2)

JEE Main 20262 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2026 · 2 April · Morning Shift · Mathematics · Differential Equations

Let y = y(x) be the solution curve of the differential equation (1...

Let

y = y(x)

be the solution curve of the differential equation

(1 + \sin x)\dfrac{dy}{dx} + (y+1)\cos x = 0

y(0) = 0

. If the curve

y = y(x)

passes through the point \left(\alpha, \dfrac{-1}{2}\right)

, then a value of

\alpha is :

Options

A $\dfrac{\pi}{6}$
B $\dfrac{\pi}{4}$
C $\dfrac{\pi}{3}$
D $\dfrac{\pi}{2}$

Correct answer

D. $\dfrac{\pi}{2}$

Step-by-step solution

The given differential equation can be written as:

(1 + \sin x)dy + (y+1)\cos x dx = 0

d((y+1)(1 + \sin x)) = 0

Integrating both sides:

(y+1)(1 + \sin x) = C

Given

y(0) = 0

, substituting

x = 0 and y = 0

(0+1)(1 + \sin 0) = C \Rightarrow C = 1

The equation of the curve is

(y+1)(1 + \sin x) = 1

Since the curve passes through \left(\alpha, \dfrac{-1}{2}\right)

, substituting

x = \alpha

and

y = \dfrac{-1}{2}

:<br><br>\left(\dfrac{-1}{2} + 1\right)(1 + \sin \alpha) = 1

\dfrac{1}{2}(1 + \sin \alpha) = 1

1 + \sin \alpha = 2

\sin \alpha = 1

\alpha = \dfrac{\pi}{2}

Answer:

\dfrac{\pi}{2}

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