Let the solution curve x = x ( y ) , 0 y 2 , of the differential equation ( log e ( cos y ) ) 2 cos y d x - ( 1 + 3 x log e ( cos y ) ) sin y ? d y = 0 satisfy x ( 3 ) = 1 2 log e 2 . If x ( 6 ) = 1 log e m - log e n , where m and n are coprime

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Solution

Q.
Let the solution curve $x = x (y), 0 < y < \frac{π}{2}$ , of the differential equation $(\log_{e} {(\cos y))}^{2} \cos y d x - (1 + 3 x \log_{e} (\cos y)) \sin y d y = 0$ satisfy $x (\frac{π}{3}) = \frac{1}{2 \log_{e} 2} .$ If $x (\frac{π}{6}) = \frac{1}{\log_{e} m - \log_{e} n},$ where $m$ and $n$ are coprime, then $m n$ is equal to _______ . [2023]

Ans.
(12)

Given differential equation is, $\cos y {(logcos y)}^{2} d x = (1 + 3 x (logcos y)) \sin y d y$

$\Rightarrow \frac{d x}{d y} = \tan y (\frac{3 x}{logcos y} + \frac{1}{{(logcos y)}^{2}})$

$\Rightarrow \frac{d x}{d y} - (\frac{3 \tan y}{logcos y}) x = \frac{\tan y}{{(logcos y)}^{2}}$

$I.F. = e^{\int - \frac{3 \tan y}{logcos y} d y} = (\log {(\cos y))}^{3}$

Solution is given by,

$x \cdot {(logcos y)}^{3} = \int \frac{\tan y}{{(logcos y)}^{2}} \times {(logcos y)}^{3} d y + C$

$= - \frac{{(logcos y)}^{2}}{2} + C$

$Given, x (\frac{π}{3}) = \frac{1}{2 \log 2}$

So, $\frac{1}{2 \log 2} (\log {(\frac{1}{2}))}^{3} = - \frac{(\log {(\frac{1}{2}))}^{2}}{2} + C \Rightarrow C = 0$

For $y = \frac{π}{6}$ , we have $x {(\log \frac{\sqrt{3}}{2})}^{3} = - \frac{1}{2} {(\log \frac{\sqrt{3}}{2})}^{2} + 0$

$\Rightarrow x = - \frac{1}{2 \log (\frac{\sqrt{3}}{2})} = \frac{1}{\log (\frac{4}{3})}$

$= \frac{1}{\log 4 - \log 3} = \frac{1}{\log_{e} m - \log_{e} n} \Rightarrow m = 4, n = 3$

$∴$ $m n = 4 \times 3 = 12$

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