Let y = y(x) be the solution of the differential equation , . If , then is equal to: [2025]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $\cos x (\log_{e} {(\cos x))}^{2} d y + (\sin x - 3 y \sin x \log_{e} (\cos x)) d x = 0$ , $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{4}) = \frac{- 1}{\log_{e} 2}$ , then $y (\frac{π}{6})$ is equal to: [2025]

1 $- \frac{1}{\log_{e} (4)}$

2 $\frac{1}{\log_{e} (3) - \log_{e} (4)}$

3 $\frac{2}{\log_{e} (3) - \log_{e} (4)}$

4 $\frac{1}{\log_{e} (4) - \log_{e} (3)}$

Ans.
(2)

We have, $\cos x (\ln {(\cos x))}^{2} d y + (\sin x - 3 y (\sin x) \ln (\cos x)) d x = 0$

$\Rightarrow \cos x (\ln {(\cos x))}^{2} \frac{d y}{d x} - 3 \sin x \cdot \ln (\cos x) y = - \sin x$

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

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

$I . F . = e^{\int \frac{3 \tan x}{\ln (\sec x)} d x} = \ln {(\sec x))}^{3}$

$∴$ Solution is $y \times (\ln {(\sec x))}^{3} = - \int \frac{\tan x}{(\ln {(\sec x))}^{2}} (\ln {(\sec x))}^{3} d x + C$

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

At $x = \frac{π}{4}, y = - \frac{1}{\ln 2}$

$∴ \frac{- 1}{\ln 2} \times {(\ln \sqrt{2})}^{3} = - \frac{1}{2} {(\ln \sqrt{2})}^{2} + C$

$\Rightarrow \frac{- 1}{8 \ln 2} \times {(\ln 2)}^{3} = \frac{- 1}{2} \times \frac{1}{4} {(\ln 2)}^{2} + C$

$\Rightarrow - \frac{1}{8} {(\ln 2)}^{2} = \frac{- 1}{8} {(\ln 2)}^{2} + C \Rightarrow C = 0$

$∴ y (\ln {(\sec x))}^{3} = \frac{- 1}{2} (\ln {(\sec x))}^{2} + 0$

$\Rightarrow y = \frac{- 1}{2 \ln (\sec x)} \Rightarrow y = \frac{1}{2 \ln (\cos x)}$

$∴ y (\frac{π}{6}) = \frac{1}{2 \ln (\cos \frac{π}{6})} = \frac{1}{2 \ln (\frac{\sqrt{3}}{2})}$

$= \frac{1}{2 (\frac{1}{2} \ln 3 - \ln 2)} = \frac{1}{\ln 3 - \ln 4}$

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