Let g be a differentiable function such that ∫_0^x g(t) d t=x-∫_0^x tg(t) d t, x ≥ 0 and let y=y(x) satisfy the differential equation (d y)/(d x)-y x= 2(x+1) x g(x), x ∈[0, (π)/(2)) . If y(0)=0 , then y((π)/(3)) is

Correct answer: B. (4 π)/(3) Diff. w.r.t. x aligned & g(x)=1-x g(x) & g(x)=(1)/(1+x) & so (d y)/(d x)-y x=2 x & I F=e^ -∫ d x =e^ x = x aligned solution of D.E. aligned & y x=∫ 2 d x+c & y x=2 x+c & y(0)=0 aligned c=0 aligned & y=(2 x)/( x) & y=2 x x & y((π)/(3))=2 · (π)/(3) · 2=(4 π)/(3) aligned

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Differential Equations

Let g be a differentiable function such that ∫_0^x g(t) d t=x-∫_0^x...

Let

g

be a differentiable function such that

\int_0^x g(t) d t=x-\int_0^x \operatorname{tg}(t) d t, x \geq 0

and let

y=y(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

Options

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

Correct answer

B. $\frac{4 \pi}{3}$

Step-by-step solution

Diff. w.r.t. x

\begin{aligned}& g(x)=1-x g(x) \\ & g(x)=\frac{1}{1+x} \\ & \text{ so } \frac{d y}{d x}-y \tan x=2 \sec x \\ & I F=e^{-\int \tan d x}=e^{\log \cos x}=\cos x\end{aligned}

solution of D.E.

\begin{aligned}& y \cos x=\int 2 d x+c \\ & y \cos x=2 x+c \\ & y(0)=0\end{aligned}

\mathrm{c}=0

\begin{aligned} & y=\frac{2 x}{\cos x} \\ & y=2 x \sec x \\ & y\left(\frac{\pi}{3}\right)=2 \cdot \frac{\pi}{3} \cdot 2=\frac{4 \pi}{3}\end{aligned}

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