Let y=y(x) be the solution of the differential equation ( x( _ e ( x))^2 dy+( x-3 y x _ e ( x)) d x=0, x ∈(0, (π)/(2))). If (y((π)/(4))= -1 _ e 2 ), then (y((π)/(6))) is equal to :

Correct answer: A. ((1)/( _e(3)- _e(4))) aligned & (d y)/(d x)-(3 x)/( x x) y=-( x)/( x( x)^2) & I.F. =e^ -∫ (3 x)/( x) d x & Let x=t & - x d x=d t & e^ 3 ∫^ (d t)/(t) =e^ 3 t =t^3=( x)^3 & Solution will be & y( x)^3=-∫( x)( x) d x & y( x)^3=(( x)^2)/(2)+c & y((π)/(4))=-(1)/( 2) & c=0 & y=(1)/(2( x)) & y((π)/(6))=(1)/(2) × (1)/( ( (π)/(6))) & =(1)/(2)[ 1 ( 3 2 ) ] & =(1)/(2) × 1 3 - 2 & =(1)/( 3- 4) aligned

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation ( x( _ e (...

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 :

Options

A(\frac{1}{\log _e(3)-\log _e(4)})
B(\frac{2}{\log _e(3)-\log _e(4)})
C(\frac{1}{\log _e(4)-\log _e(3)})
D(-\frac{1}{\log _e(4)})

Correct answer

A. (\frac{1}{\log _e(3)-\log _e(4)})

Step-by-step 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}

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