JEE Main 2025 — Differential Equations Question with Solution

To solve the given problem, let's start by considering the equation:

${\int }_0^{x}g(t)dt=x-{\int }_0^x\tan g(t)dt$

Differentiate both sides with respect to $x$:

$g(x)=1-xg(x)$

Rearranging gives:

$g(x)(1+x)=1⟹g(x)=\frac{1}{1+x}$

Now, consider the differential equation:

$\frac{dy}{dx}-y\tan x=2(x+1)\sec xg(x)$

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

$\frac{dy}{dx}-y\tan x=2(x+1)\sec x\cdot \frac{1}{1+x}$

This simplifies to:

$\frac{dy}{dx}-y\tan x=2\sec x$

To solve this, we use an Integrating Factor (I.F):

$\text{I.F}=e^{\int \tan xdx}=e^{-\ln \cos x}=\cos x$

Multiply the entire differential equation by the Integrating Factor:

$\cos x\cdot \frac{dy}{dx}-y\cdot \sin x=2$

This implies:

$\frac{d}{dx}(y\cos x)=2$

Integrate both sides with respect to $x$:

$y\cos x=\int 2dx=2x+c$

Given the initial condition $y(0)=0$:

$0\cdot 1=2\cdot 0+c⟹c=0$

Thus:

$y\cos x=2x$

Evaluate at $x=\frac{\pi }{3}$:

$y\cdot \frac{1}{2}=2\cdot \frac{\pi }{3}$

Solving for $y$:

$y=\frac{4\pi }{3}$

Therefore, $y\left(\frac{\pi }{3}\right)=\frac{4\pi }{3}$.