JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 19)
Let the solution curve $$x=x(y), 0 < y < \frac{\pi}{2}$$, of the differential equation $$\left(\log _{e}(\cos y)\right)^{2} \cos y \mathrm{~d} x-\left(1+3 x \log _{e}(\cos y)\right) \sin \mathrm{y} d y=0$$ satisfy $$x\left(\frac{\pi}{3}\right)=\frac{1}{2 \log _{e} 2}$$. If $$x\left(\frac{\pi}{6}\right)=\frac{1}{\log _{e} m-\log _{e} n}$$, where $$m$$ and $$n$$ are coprime, then $$m n$$ is equal to __________.
Answer
12
Explanation
$$
\begin{aligned}
& (\cos y)(\ln (\cos y))^2 d x=(1+3 x \ln \cos y) \sin y d y \\\\
& \Rightarrow \frac{d x}{d y}=\frac{(1+3 x \ln \cos y) \sin y}{(\ln \cos y)^2 \cos y} \\\\
& =\tan y\left[\frac{1}{(\ln \cos y)^2}+\frac{3 x}{\ln \cos y}\right] \\\\
& \Rightarrow \frac{d x}{d y}-\left(\frac{3 \tan y}{\ln \cos y}\right) x=\frac{\tan y}{(\ln \cos y)^2}
\end{aligned}
$$
Which is a linear differential equation.
$$ \text { I.F. }=e^{-\int \frac{3 \tan y}{\ln \cos y} d y}=(\ln \cos y)^3 $$
So, the solution is :
$$ \begin{aligned} & x \times(\ln \cos y)^3=\int\left((\ln \cos y)^3 \times \frac{\tan y}{(\ln \cos y)^2}\right) d y \\\\ & x \times(\ln \cos y)^3=\frac{-(\ln \cos y)^2}{2}+C \end{aligned} $$
$$ \text { At } y=\frac{\pi}{3} \text {, } $$
$$ \begin{aligned} & \frac{1}{2 \ln 2} \times\left(\ln \left(\frac{1}{2}\right)\right)^3=-\frac{\left(\ln \left(\frac{1}{2}\right)\right)^2}{2}+C \\\\ & \Rightarrow C=0 \end{aligned} $$
$$ \begin{aligned} & \text { So, } x \times \ln ^3 \cos y=\frac{-\ln ^2 \cos y}{2} \\\\ & \text { At } y=\frac{\pi}{6}, x \times\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^3=-\frac{1}{2}\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^2 \\\\ & \Rightarrow x=-\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)} \end{aligned} $$
$$ \begin{aligned} & =-\frac{1}{2[\ln \sqrt{3}-\ln 2]}=\frac{-1}{2\left[\frac{1}{2} \ln 3-\ln 2\right]} \\\\ & =\frac{-1}{2\left[\frac{\ln 3-\ln 4}{2}\right]}=\frac{1}{\ln 4-\ln 3} \\\\ & \Rightarrow m=4, n=3 \\\\ & \Rightarrow m n=12 \end{aligned} $$
Which is a linear differential equation.
$$ \text { I.F. }=e^{-\int \frac{3 \tan y}{\ln \cos y} d y}=(\ln \cos y)^3 $$
So, the solution is :
$$ \begin{aligned} & x \times(\ln \cos y)^3=\int\left((\ln \cos y)^3 \times \frac{\tan y}{(\ln \cos y)^2}\right) d y \\\\ & x \times(\ln \cos y)^3=\frac{-(\ln \cos y)^2}{2}+C \end{aligned} $$
$$ \text { At } y=\frac{\pi}{3} \text {, } $$
$$ \begin{aligned} & \frac{1}{2 \ln 2} \times\left(\ln \left(\frac{1}{2}\right)\right)^3=-\frac{\left(\ln \left(\frac{1}{2}\right)\right)^2}{2}+C \\\\ & \Rightarrow C=0 \end{aligned} $$
$$ \begin{aligned} & \text { So, } x \times \ln ^3 \cos y=\frac{-\ln ^2 \cos y}{2} \\\\ & \text { At } y=\frac{\pi}{6}, x \times\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^3=-\frac{1}{2}\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^2 \\\\ & \Rightarrow x=-\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)} \end{aligned} $$
$$ \begin{aligned} & =-\frac{1}{2[\ln \sqrt{3}-\ln 2]}=\frac{-1}{2\left[\frac{1}{2} \ln 3-\ln 2\right]} \\\\ & =\frac{-1}{2\left[\frac{\ln 3-\ln 4}{2}\right]}=\frac{1}{\ln 4-\ln 3} \\\\ & \Rightarrow m=4, n=3 \\\\ & \Rightarrow m n=12 \end{aligned} $$
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