Start free

Mathematics · Differential Equations

JEE Main 2026 — 21 January, Evening Shift — Question 16

Let y=y(x)\mathrm{y}=\mathrm{y}(\mathrm{x}) be the solution of the differential equation sec⁡xdydx−2y=2+3sin⁡x,x∈(−π2,π2)\sec x \frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{y}=2+3 \sin \mathrm{x}, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), y(0)=−74y(0)=-\frac{7}{4}. Then y(π6)y\left(\frac{\pi}{6}\right) is equal to:

  1. Option A:

    −52-\frac{5}{2}

    Correct
  2. Option B:

    −54-\frac{5}{4}

  3. Option C:

    −33−7-3 \sqrt{3}-7

  4. Option D:

    −32−7-3 \sqrt{2}-7

Answer: A

Step-by-step solution

dydx−2ycos⁡x=2cos⁡x+3sin⁡x⋅cos⁡x\frac{d y}{d x}-2 y \cos x=2 \cos x+3 \sin x \cdot \cos x

I.F. =e−2sin⁡x=e^{-2 \sin x}

e−2sin⁡x⋅y=∫e−2sin⁡x(3sin⁡xcos⁡x+2cos⁡x)dxe^{-2 \sin x} \cdot y=\int e^{-2 \sin x}(3 \sin x \cos x+2 \cos x) d x

y.e−2sin⁡x=e−2sin⁡x(−32sin⁡x−74)+C\mathrm{y} . \mathrm{e}^{-2 \sin \mathrm{x}}=\mathrm{e}^{-2 \sin \mathrm{x}}\left(-\frac{3}{2} \sin \mathrm{x}-\frac{7}{4}\right)+\mathrm{C}

⇒y=−32sin⁡x−74+C⋅e2sin⁡x\Rightarrow \mathrm{y}=-\frac{3}{2} \sin \mathrm{x}-\frac{7}{4}+\mathrm{C} \cdot \mathrm{e}^{2 \sin \mathrm{x}}

∵y(0)=−74⇒C=0\because y(0)=-\frac{7}{4} \Rightarrow C=0

y(π6)=−32⋅12−74=−52y\left(\frac{\pi}{6}\right)=\frac{-3}{2} \cdot \frac{1}{2}-\frac{7}{4}=\frac{-5}{2}

Answer key and solution verified before publishing.

Practise Differential Equations

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2026
Paper
21 January, Evening Shift
Subject
Mathematics
Chapter
Differential Equations
Topic
Methods of solving a First Order,First Degree Differential