Let g be a differentiable function such that int 0 x g ( t ) dt = x -int 0 x tg ( t ) dt , x geq 0 and let y = y ( x ) satisfy the differential equation d y/d x-y tan x= 2(x+1) sec x g(x), x in [0, π/2 ) . If y(0)=0 , then y (π/3 ) is equal to

Mathematics · Differential Equations

JEE Main 2025 — 3 April, Morning Shift — Question 31

Let $g$ be a differentiable function such that $\int_{0}^{\mathrm{x}} \mathrm{g}(\mathrm{t}) \mathrm{dt}=\mathrm{x}-\int_{0}^{\mathrm{x}} \mathrm{tg}(\mathrm{t}) \mathrm{dt}, \mathrm{x} \geq 0$ and let $\mathrm{y}=\mathrm{y}(\mathrm{x})$

satisfy the differential equation $\frac{d y}{d x}-y \tan x=$ $2(x+1) \sec x g(x), x \in\left[0, \frac{\pi}{2}\right)$ . If $y(0)=0$ , then

$y\left(\frac{\pi}{3}\right)$ is equal to

Option A:
$\frac{2 \pi}{3 \sqrt{3}}$
Option B:
$\frac{4 \pi}{3}$
Correct
Option C:
$\frac{2 \pi}{3}$
Option D:
$\frac{4 \pi}{3 \sqrt{3}}$

Answer: B

Step-by-step solution

Diff. w.r.t. x

$\mathrm{g}(\mathrm{x})=1-\mathrm{xg}(\mathrm{x})$

$g(x)=\frac{1}{1+x}$

so $\frac{d y}{d x}-y \tan x=2 \sec x$

$\mathrm{IF}=\mathrm{e}^{-\int \tan \mathrm{dx}}=\mathrm{e}^{\log \cos \mathrm{x}}=\cos \mathrm{x}$

solution of D.E.

$\mathrm{y} \cos \mathrm{x}=\int 2 \mathrm{dx}+\mathrm{c}$

$\mathrm{y} \cos \mathrm{x}=2 \mathrm{x}+\mathrm{c}$

$y(0)=0$

$\mathrm{c}=0$

$y=\frac{2 x}{\cos x}$

$\mathrm{y}=2 \mathrm{x} \sec \mathrm{x}$

$\mathrm{y}\left(\frac{\pi}{3}\right)=2 \cdot \frac{\pi}{3} \cdot 2=\frac{4 \pi}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Mathematics
Chapter: Differential Equations
Topic: Applications of Differential Equations