Let \(y=y(x)\) be the solution of the differential equation \(\cos x\left(\log _{\mathrm{e}}(\cos…

Let \(y=y(x)\) be the solution of the differential equation \(\cos x\left(\log _{\mathrm{e}}(\cos x)\right)^2 \mathrm{dy}+\)\(\left(\sin x-3 y \sin x \log _{\mathrm{e}}(\cos x)\right) \mathrm{d} x=0, x \in\left(0, \frac{\pi}{2}\right)\). If \(y\left(\frac{\pi}{4}\right)=\frac{-1}{\log _{\mathrm{e}} 2}\), then \(y\left(\frac{\pi}{6}\right)\) is equal to :
  1. \(\frac{1}{\log _e(3)-\log _e(4)}\)
  2. \(\frac{2}{\log _e(3)-\log _e(4)}\)
  3. \(\frac{1}{\log _e(4)-\log _e(3)}\)
  4. \(-\frac{1}{\log _e(4)}\)

Solution

$\begin{aligned} & \frac{d y}{d x}-\frac{3 \sin x}{\cos x \ln \cos x} y=-\frac{\sin x}{\cos x(\ln \cos x)^2} \\ & \text { I.F. }=e^{-\int \frac{3 \tan x}{\ln \cos x} d x} \\ & \text { Let } \ln \cos x=t \\ & -\tan x d x=d t \\ & e^{3 \int^{\frac{d t}{t}}}=e^{3 \ln t}=t^3=(\ln \cos x)^3 \\ & \therefore \quad \text { Solution will be } \\ & \quad y(\ln \cos x)^3=-\int(\tan x)(\ln \cos x) d x \\ & y(\ln \cos x)^3=\frac{(\ln \cos x)^2}{2}+c \\ & \because \quad y\left(\frac{\pi}{4}\right)=-\frac{1}{\ln 2} \\ & \Rightarrow \quad c=0 \\ & \therefore \quad y=\frac{1}{2(\ln \cos x)} \\ & y\left(\frac{\pi}{6}\right)=\frac{1}{2} \times \frac{1}{\ln \left(\cos \frac{\pi}{6}\right)} \\ & =\frac{1}{2}\left[\frac{1}{\ln \left(\frac{\sqrt{3}}{2}\right)}\right] \\ & =\frac{1}{2} \times \frac{1}{\ln \sqrt{3}-\ln 2} \\ & =\frac{1}{\ln 3-\ln 4}\end{aligned}$

Asked in: JEE Main 2025 (29 Jan Shift 1)

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