JEE Main 2025 (Online) 3rd April Morning ShiftQuestion 1
Let $g$ be a differentiable function such that $\int_{0}^{x}g(t)\,dt=x-\int_{0}^{x}tg(t)\,dt$, $x\ge0$ and let $y=y(x)$ satisfy the differential equation $\frac{dy}{dx}-y\tan x=2(x+1)\sec x\,g(x)$, $x\in\left[0,\frac{\pi}{2}\right)$. If $y(0)=0$, then $y\left(\frac{\pi}{3}\right)$ is equal to
Differential Equation
Solution
Concept. Differentiate the integral equation to find $g$; then the DE is linear with IF $\cos x$. Formulas used. (1) $\frac{d}{dx}\int_0^xh(t)\,dt=h(x)$(2) $\text{IF}=e^{-\int\tan x\,dx}=\cos x$Step 1 - Find $g$. By (1), $g=1-xg$, so $g(x)=\dfrac{1}{1+x}$. Step 2 - The DE. $2(x+1)\sec x\cdot\frac{1}{1+x}=2\sec x$, so $$y'-y\tan x=2\sec x.$$ Step 3 - Solve. By (2), $(y\cos x)'=2$, so $y\cos x=2x+C$. $y(0)=0$ gives $C=0$. Step 4 - At $x=\frac\pi3$. $y=\dfrac{2\pi/3}{1/2}=\dfrac{4\pi}{3}$. Answer: (A)