Let y = y(x) be the solution of the differential equation : x( _e( x))^2dy+( x-3y x _e( x)…

Question

Let y = y(x) be the solution of the differential equation :

$\cos x{({\log }_{e}(\cos x))}^{2}dy+(\sin x-3y\sin x{\log }_{e}(\cos x))dx=0$, x \in  (0, $\frac{\pi }{2}$ ). If $y(\frac{\pi }{4})$ = $-\frac{1}{{\log }_{e}2}$, then $y(\frac{\pi }{6})$ is equal to :

Accepted Answer

$\begin{gathered} \cos x(\ln (\cos x){)}^{2}dy+(\sin x-3y(\sin x)\ln (\cos x))dx=0 \ \cos x(\ln (\cos x){)}^{2}\frac{dy}{dx}-3\sin x\cdot \ln (\cos x)y=-\sin x \ \frac{dy}{dx}-\frac{3	an x}{\ln (\cos x)}y=\frac{-	an x}{(\ln (\cos x){)}^{2}} \ \frac{dy}{dx}+\frac{3	an x}{\ln (\sec x)}y=\frac{-	an x}{(\ln (\sec x){)}^{2}} \ ={e}^{\int \frac{3	an x}{\ln (\sec x)}dx}=(\ln (\sec x){)}^{3} \end{gathered}$

$\begin{gathered} y	imes (\ln (\sec x){)}^{3}=-\int \frac{	an x}{(\ln (\sec x){)}^{2}}(\ln (\sec x){)}^{3}dx \ y	imes (\ln (\sec x){)}^{3}=-\frac{1}{2}(\ln (\sec x){)}^{2}+C \ :x=\frac{\pi }{4},y=-\frac{1}{\ln 2} \ \frac{-1}{\ln 2}	imes (\ln \sqrt{2}{)}^{3}=-\frac{1}{2}	imes (\ln \sqrt{2}{)}^{2}+C \ \Rightarrow \frac{-1}{8\ln 2}	imes (\ln 2{)}^{3}=\frac{-1}{2}	imes \frac{1}{4}(\ln 2{)}^{2}+C \ -\frac{1}{8}(\ln 2{)}^{2}=\frac{-1}{8}(\ln 2{)}^{2}+C \ \Rightarrow C=0 \ 	herefore y(\ln (\sec x){)}^{3}=\frac{-1}{2}(\ln (\sec x){)}^{2}+0 \ y=\frac{-1}{2\ln (\sec x)} \ y=\frac{1}{2\ln (\cos x)} \end{gathered}$

$\begin{gathered} 	herefore y(\frac{\pi }{6})=\frac{1}{2\ln (\cos \frac{\pi }{6})} \ =\frac{1}{2\ln (\frac{\sqrt{3}}{2})} \ =\frac{1}{2(\frac{1}{2}\ln 3-\ln 2)} \ =\frac{1}{\ln 3-\ln 4} \end{gathered}$

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