Calculus – Differential Equations – JEE Main 29 Jan 2025 Shift 1

Solution:

Rearrange the given equation into the standard linear form $\frac{dy}{dx} + Py = Q$.

$\cos x(\ln(\cos x))^2 \frac{dy}{dx} – 3y \sin x \ln(\cos x) = -\sin x$

Divide the entire equation by $\cos x(\ln(\cos x))^2$:

$\frac{dy}{dx} – \frac{3\sin x}{\cos x \ln(\cos x)}y = \frac{-\sin x}{\cos x (\ln(\cos x))^2}$

$\frac{dy}{dx} – \frac{3\tan x}{\ln(\cos x)}y = \frac{-\tan x}{(\ln(\cos x))^2}$

Use the property $\ln(\cos x) = -\ln(\sec x)$ to simplify signs:

$\frac{dy}{dx} + \frac{3\tan x}{\ln(\sec x)}y = \frac{-\tan x}{(\ln(\sec x))^2}$

Now, find the Integrating Factor (I.F.):

$\text{I.F.} = e^{\int \frac{3\tan x}{\ln(\sec x)} dx}$

Let $t = \ln(\sec x) \Rightarrow dt = \tan x dx$.

$\text{I.F.} = e^{\int \frac{3}{t} dt} = e^{3\ln t} = t^3 = (\ln(\sec x))^3$

The general solution is $y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + C$:

$y (\ln(\sec x))^3 = \int \frac{-\tan x}{(\ln(\sec x))^2} \cdot (\ln(\sec x))^3 dx$

$y (\ln(\sec x))^3 = \int -\tan x \ln(\sec x) dx$

Substitute $t = \ln(\sec x)$ again ($dt = \tan x dx$):

$y (\ln(\sec x))^3 = \int -t dt = \frac{-t^2}{2} + C$

$y (\ln(\sec x))^3 = \frac{-1}{2}(\ln(\sec x))^2 + C$

Apply condition $y(\frac{\pi}{4}) = \frac{-1}{\ln 2}$. Note that $\ln(\sec \frac{\pi}{4}) = \ln(\sqrt{2}) = \frac{1}{2}\ln 2$.

$\left(\frac{-1}{\ln 2}\right) \left(\frac{1}{2}\ln 2\right)^3 = \frac{-1}{2}\left(\frac{1}{2}\ln 2\right)^2 + C$

$\frac{-1}{8}(\ln 2)^2 = \frac{-1}{8}(\ln 2)^2 + C \Rightarrow C = 0$

So, the solution simplifies to:

$y (\ln(\sec x))^3 = \frac{-1}{2}(\ln(\sec x))^2 \Rightarrow y = \frac{-1}{2\ln(\sec x)} = \frac{1}{2\ln(\cos x)}$

Now find $y(\frac{\pi}{6})$ where $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$:

$y\left(\frac{\pi}{6}\right) = \frac{1}{2 \ln(\frac{\sqrt{3}}{2})} = \frac{1}{2 (\ln \sqrt{3} – \ln 2)}$

$y\left(\frac{\pi}{6}\right) = \frac{1}{\ln 3 – 2\ln 2} = \frac{1}{\ln 3 – \ln 4}$

Ans. (4)