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QJEEM23Maf215: JEE Main 2023 Mathematics - Class 12 JEEM23Maf215 Solution: Let the solution curve , of the differential equation satisfy . …

Let the solution curve x=x(y), 0 < y < \frac{\pi}{2} , of the differential equation (log⁡e(cos⁡y))2cos⁡y dx−(1+3xlog⁡e(cos⁡y))sin⁡ydy=0\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(π3)=12log⁡e2x\left(\frac{\pi}{3}\right)=\frac{1}{2 \log _{e} 2} . If x(π6)=1log⁡em−log⁡enx\left(\frac{\pi}{6}\right)=\frac{1}{\log _{e} m-\log _{e} n} , where mm and nn are coprime, then mnm n is equal to __________.

Step-by-Step Verified Solution

amp;(cos⁡y)(ln⁡(cos⁡y))2dx=(1+3xln⁡cos⁡y)sin⁡ydyamp;⇒dxdy=(1+3xln⁡cos⁡y)sin⁡y(ln⁡cos⁡y)2cos⁡yamp;=tan⁡y[1(ln⁡cos⁡y)2+3xln⁡cos⁡y]amp;⇒dxdy−(3tan⁡yln⁡cos⁡y)x=tan⁡y(ln⁡cos⁡y)2\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.

 I.F. =e−∫3tan⁡yln⁡cos⁡ydy=(ln⁡cos⁡y)3\text { I.F. }=e^{-\int \frac{3 \tan y}{\ln \cos y} d y}=(\ln \cos y)^3

So, the solution is :

amp;x×(ln⁡cos⁡y)3=∫((ln⁡cos⁡y)3×tan⁡y(ln⁡cos⁡y)2)dyamp;x×(ln⁡cos⁡y)3=−(ln⁡cos⁡y)22+C\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}

 At y=π3, \text { At } y=\frac{\pi}{3} \text {, }

amp;12ln⁡2×(ln⁡(12))3=−(ln⁡(12))22+Camp;⇒C=0\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}

amp; So, x×ln⁡3cos⁡y=−ln⁡2cos⁡y2amp; At y=π6,x×(ln⁡(32))3=−12(ln⁡(32))2amp;⇒x=−12ln⁡(32)\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}

amp;=−12[ln⁡3−ln⁡2]=−12[12ln⁡3−ln⁡2]amp;=−12[ln⁡3−ln⁡42]=1ln⁡4−ln⁡3amp;⇒m=4,n=3amp;⇒mn=12\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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