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Mathematics · Differential Equations

JEE Main 2026 — 2 April, Morning Shift — Question 36

Let y = y(x) be the solution curve of the differential equation (1 + sin x) dydx\frac{dy}{dx} + (y + 1) cos x = 0, y(0) = 0. If the curve y = y(x) passes through the point (α,−12)\left(\alpha, -\frac{1}{2}\right), then a value of α is :

  1. Option A:

    π6\frac{\pi}{6}

  2. Option B:

    π4\frac{\pi}{4}

  3. Option C:

    π3\frac{\pi}{3}

  4. Option D:

    π2\frac{\pi}{2}

    Correct

Answer: D

Step-by-step solution

∫dyy+1=−∫cos⁡x1+sin⁡xdx\int \frac{d y}{y+1}=-\int \frac{\cos x}{1+\sin x} d x ⇒ln⁡(y+1)=−ln⁡(1+sin⁡x)+C\Rightarrow \ln (\mathrm{y}+1)=-\ln (1+\sin \mathrm{x})+\mathrm{C} ∵y(0)=0⇒C=0\because y(0)=0 \Rightarrow C=0 ⇒y+1=11+sin⁡x\Rightarrow y+1=\frac{1}{1+\sin x} passes through (α,−12)\left(\alpha,-\frac{1}{2}\right) ⇒sin⁡x=1\Rightarrow \sin \mathrm{x}=1 ⇒x=π2\Rightarrow \mathrm{x}=\frac{\pi}{2} (as per options)

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Morning Shift
Subject
Mathematics
Chapter
Differential Equations
Topic
Methods of solving a First Order,First Degree Differential

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