JEE Main 2025 Jan 29 Shift 1, Mathematics Q10: Methods of Solving First Order, First Degree Differential Equation

Rearrange the ODE by dividing by $\cos x(\ln \cos x{)}^2$:

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

Since $\ln \cos x=-\ln \sec x$, this becomes

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

Integrating factor: with $u=\ln \sec x$, $du=\tan xdx$, so $\int \frac{3\tan x}{\ln \sec x}dx=3\ln (\ln \sec x)$ and IF $=(\ln \sec x{)}^3.$

Hence $y(\ln \sec x{)}^3=-\int \frac{\tan x}{(\ln \sec x{)}^2}(\ln \sec x{)}^{3}dx+C=-\int udu+C=-\frac{u^2}{2}+C.$

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

Boundary condition $y(\pi /4)=-1/\ln 2$: here $\ln \sec (\pi /4)=\ln \sqrt{2}=\frac{1}{2}\ln 2.$ Substituting gives $C=0.$

Therefore $y=\frac{-1}{2\ln \sec x}=\frac{1}{2\ln \cos x}.$

At $x=\pi /6$: $\cos (\pi /6)=\sqrt{3}/2$, so $\ln \cos (\pi /6)=\frac{1}{2}\ln 3-\ln 2.$

$y(\pi /6)=\frac{1}{2(\frac{1}{2}\ln 3-\ln 2)}=\frac{1}{\ln 3-\ln 4}.$