JEE Main 2025 — Differential Equations Question with Solution
JEE Main 2025 (3 Apr Shift 1)
Question
Let be a differentiable function such that and let satisfy the differential equation . If , then is equal to
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
Diff. w.r.t. x
$\begin{aligned}
& g(x)=1-x g(x) \\ & g(x)=\frac{1}{1+x} \\ & \text { so } \frac{d y}{d x}-y \tan x=2 \sec x \\ & I F=e^{-\int \tan d x}=e^{\log \cos x}=\cos x
\end{aligned}\begin{aligned}
& y \cos x=\int 2 d x+c \\ & y \cos x=2 x+c \\ & y(0)=0
\end{aligned}\mathrm{c}=0\begin{aligned} & y=\frac{2 x}{\cos x} \\ & y=2 x \sec x \\ & y\left(\frac{\pi}{3}\right)=2 \cdot \frac{\pi}{3} \cdot 2=\frac{4 \pi}{3}\end{aligned}$
$\begin{aligned}
& g(x)=1-x g(x) \\ & g(x)=\frac{1}{1+x} \\ & \text { so } \frac{d y}{d x}-y \tan x=2 \sec x \\ & I F=e^{-\int \tan d x}=e^{\log \cos x}=\cos x
\end{aligned}\begin{aligned}
& y \cos x=\int 2 d x+c \\ & y \cos x=2 x+c \\ & y(0)=0
\end{aligned}\mathrm{c}=0\begin{aligned} & y=\frac{2 x}{\cos x} \\ & y=2 x \sec x \\ & y\left(\frac{\pi}{3}\right)=2 \cdot \frac{\pi}{3} \cdot 2=\frac{4 \pi}{3}\end{aligned}$
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Differential Equations chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2025, covering the Differential Equations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.