Question

Let \(y = y(x)\) satisfy 

\[ \cos x\,(\log(\cos x))^2\,dy + \left( \sin x - 3y\sin x\log(\cos x) \right)dx = 0, \quad x \in \left(0,\frac{\pi}{2}\right). \]

If \[ y\!\left(\frac{\pi}{4}\right) = -\frac{1}{\log 2}, \] then \[ y\!\left(\frac{\pi}{6}\right) \] equals:

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

Hint:

When logarithms of trig functions appear, substitution \(t=\log(\cos x)\) is the standard CUET trick to reduce complexity.

Answer 1

The Correct Option is A

Concept: The given equation is a first-order linear differential equation in disguise. The key idea is to reduce it using substitution: \[ t=\log(\cos x) \] This converts trigonometric-logarithmic structure into a rational differential equation.

Step 1: Rewrite the equation in standard form. Given: \[ \cos x(\log(\cos x))^2 dy + (\sin x - 3y\sin x\log(\cos x))dx = 0 \] Divide by \(\cos x(\log(\cos x))^2\): \[ \frac{dy}{dx} -\frac{3\tan x}{\log(\cos x)}y = -\frac{\tan x}{(\log(\cos x))^2} \] So it becomes: \[ \frac{dy}{dx}+P(x)y=Q(x) \] where: \[ P(x)=-\frac{3\tan x}{\log(\cos x)} \]

Step 2: Substitute \(t=\log(\cos x)\). \[ t=\log(\cos x) \Rightarrow \frac{dt}{dx}=-\tan x \] So: \[ \frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx} =-\tan x \frac{dy}{dt} \] Substitute into equation: \[ -\tan x \frac{dy}{dt} -\frac{3\tan x}{t}y = -\frac{\tan x}{t^2} \] Divide by \(-\tan x\): \[ \frac{dy}{dt}+\frac{3}{t}y=\frac{1}{t^2} \]

Step 3: Find integrating factor. \[ I.F.=e^{\int \frac{3}{t}dt}=t^3 \] Multiply: \[ t^3\frac{dy}{dt}+3t^2 y=t \] \[ \frac{d}{dt}(t^3 y)=t \]

Step 4: Integrate. \[ t^3 y=\frac{t^2}{2}+C \] \[ y=\frac{1}{2t}+\frac{C}{t^3} \]

Step 5: Use initial condition. At \(x=\pi/4\): \[ t=\log(\cos\pi/4)=\log\left(\frac{1}{\sqrt2}\right)=-\frac12\log2 \] \[ y=-\frac{1}{\log2} \] Substitute to find \(C\).

Step 6: Evaluate at \(x=\pi/6\). \[ t=\log(\cos\pi/6)=\log\left(\frac{\sqrt3}{2}\right) =\log3-\log4 \] Substitute into general solution: \[ y(\pi/6)=\frac{1}{\log3-\log4} \] \[ \boxed{\frac{1}{\log3-\log4}} \]