Let the solution curve x = x ( y ) , 0 < y < π 2 , of the differential equation log e cos y 2…

Given,

${\left({\log }_e\left(\cos y\right)\right)}^2\cos y dx-\left(1+3x{\log }_e\left(\cos y\right)\right)\sin y dy=0$

$\Rightarrow \frac{dx}{dy}+\left(\frac{-3\tan y}{\mathrm{lncos}y}\right)x=\frac{\sin y}{{\left(In\cos y\right)}^2·\cos y}$

Which is a linear differential equation,

So, Integrating factor will be, $IF=e^{3\int \frac{-\tan y}{\mathrm{lncos}y}dy}={\left(\mathrm{lncos}y\right)}^3$

So, solution of differential equation is given by,
$x{\left(\mathrm{lncos}y\right)}^3=\int \frac{\sin y}{\cos y}\left(\mathrm{lncos}y\right)dy$
$\Rightarrow x{\left(\mathrm{lncos}y\right)}^3=\frac{-{\left(\mathrm{lncos}y\right)}^2}{2}+c$

Now using the given value of $x\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2\ln 2}$
We get, $c=0$
Hence, $x=\frac{-1}{2\ln \cos y}$

Now finding, $x\left(\frac{\mathrm{\pi }}{6}\right)=-\frac{1}{2\ln {\displaystyle \left(\frac{\sqrt{3}}{2}\right)}}=\frac{1}{\ln 4-\ln 3}$

Now on comparing with $\frac{1}{{\log }_{e}m-{\log }_{e}n}$ we get, $m=4, n=3$

Hence, $mn=12$