JEE Main 2025 — Differential Equations Question with Solution

$$\begin{array}{rl}& \cos x(\ln (\cos x){)}^{2}dy+(\sin x-3y(\sin x)\ln (\cos x))dx=0\\ & \cos x(\ln (\cos x){)}^2\frac{dy}{dx}-3\sin x\cdot \ln (\cos x)y=-\sin x\\ & \frac{dy}{dx}-\frac{3\tan x}{\ln (\cos x)}y=\frac{-\tan x}{(\ln (\cos x){)}^2}\\ & \frac{dy}{dx}+\frac{3\tan x}{\ln (\sec x)}y=\frac{-\tan x}{(\ln (\sec x){)}^2}\\ & \text{ I.F. }=e^{\int \frac{3\tan x}{\ln (\sec x)}dx}=(\ln (\sec x){)}^3\end{array}$$

$$\begin{array}{rl}& y\times (\ln (\sec x){)}^3=-\int \frac{\tan x}{(\ln (\sec x){)}^2}(\ln (\sec x){)}^{3}dx\\ & y\times (\ln (\sec x){)}^3=-\frac{1}{2}(\ln (\sec x){)}^2+C\\ & \text{ Given }:x=\frac{\pi }{4},y=-\frac{1}{\ln 2}\\ & \frac{-1}{\ln 2}\times (\ln \sqrt{2}{)}^3=-\frac{1}{2}\times (\ln \sqrt{2}{)}^2+C\\ & \Rightarrow \frac{-1}{8\ln 2}\times (\ln 2{)}^3=\frac{-1}{2}\times \frac{1}{4}(\ln 2{)}^2+C\\ & -\frac{1}{8}(\ln 2{)}^2=\frac{-1}{8}(\ln 2{)}^2+C\\ & \Rightarrow C=0\\ & \therefore y(\ln (\sec x){)}^3=\frac{-1}{2}(\ln (\sec x){)}^2+0\\ & y=\frac{-1}{2\ln (\sec x)}\\ & y=\frac{1}{2\ln (\cos x)}\end{array}$$

$$\begin{array}{rl}& \therefore y\left(\frac{\pi }{6}\right)=\frac{1}{2\ln \left(\cos \frac{\pi }{6}\right)}\\ & =\frac{1}{2\ln \left(\frac{\sqrt{3}}{2}\right)}\\ & =\frac{1}{2\left(\frac{1}{2}\ln 3-\ln 2\right)}\\ & =\frac{1}{\ln 3-\ln 4}\end{array}$$