Q33 / 75 · JEE Main 2025 (3 Apr Shift 1) · Mathematics
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
- A$\frac{2 \pi}{3 \sqrt{3}}$
- Correct answerB$\frac{4 \pi}{3}$
- C$\frac{2 \pi}{3}$
- D$\frac{4 \pi}{3 \sqrt{3}}$
Solution
Diff. w.r.t. x
$\begin{aligned}<br/>& g(x)=1-x g(x) \\ & g(x)=\frac{1}{1+x} \\ & \text { so } \frac{d y}{d x}-y \tan x=2 \sec x \\ & I F=e^{-\int \tan d x}=e^{\log \cos x}=\cos x<br/>\end{aligned}$
solution of D.E.
$\begin{aligned}<br/>& y \cos x=\int 2 d x+c \\ & y \cos x=2 x+c \\ & y(0)=0<br/>\end{aligned}$
$\mathrm{c}=0$
$\begin{aligned} & y=\frac{2 x}{\cos x} \\ & y=2 x \sec x \\ & y\left(\frac{\pi}{3}\right)=2 \cdot \frac{\pi}{3} \cdot 2=\frac{4 \pi}{3}\end{aligned}$
$\begin{aligned}<br/>& g(x)=1-x g(x) \\ & g(x)=\frac{1}{1+x} \\ & \text { so } \frac{d y}{d x}-y \tan x=2 \sec x \\ & I F=e^{-\int \tan d x}=e^{\log \cos x}=\cos x<br/>\end{aligned}$
solution of D.E.
$\begin{aligned}<br/>& y \cos x=\int 2 d x+c \\ & y \cos x=2 x+c \\ & y(0)=0<br/>\end{aligned}$
$\mathrm{c}=0$
$\begin{aligned} & y=\frac{2 x}{\cos x} \\ & y=2 x \sec x \\ & y\left(\frac{\pi}{3}\right)=2 \cdot \frac{\pi}{3} \cdot 2=\frac{4 \pi}{3}\end{aligned}$
Hence, the correct answer is option: 'B'.