Let y = y(x) be the solution of the differential equation cosx(loge(cosx))2dy + (sinx –3ysinx loge(cosx))dx = 0, x∈0,π2. If yπ4=−1loge⁡2, then yπ6 is:

cos⁡x(ln⁡(cos⁡x))2dy+(sin⁡x−3y(sin⁡x)ln⁡(cos⁡x))dx=0cos⁡x(ln⁡(cos⁡x))2dydx−3sin⁡x⋅ln⁡(cos⁡x)y=−sin⁡xdydx−3tan⁡xln⁡(cos⁡x)y=−tan⁡x(ln⁡(cos⁡x))2dydx+3tan⁡xln⁡(sec⁡x)y=−tan⁡x(ln⁡(sec⁡x))2I.F =e∫3tan⁡xln⁡(sec⁡x)dx=(ln⁡(sec⁡x))3 y×(ln⁡(sec⁡x))3=−∫tan⁡x(ln⁡(sec⁡x))2(ln⁡(sec⁡x))3dx+Cy×(ln⁡(sec⁡x))3=−12(ln⁡(sec⁡x))2+CGiven :x=π4,y=−1ln⁡2−1ln⁡2×(ln⁡2)3=−12×(ln⁡2)2+C⇒−18ln⁡2×(ln⁡2)3=−12×14(ln⁡2)2+C−18(ln⁡2)2=−18(ln⁡2)2+C⇒C=0∴y(ln⁡(sec⁡x))3=−12(ln⁡(sec⁡x))2+0y=−12ln⁡(sec⁡x)y=12ln⁡(cos⁡x)∴yπ6=12ln⁡cos⁡π6=12ln⁡32=1212ln⁡3−ln⁡2=1ln⁡3−ln⁡4

Let y = y(x) be the solution of the differential equation cosx(loge(cosx))2dy + (sinx –3ysinx loge(cosx))dx = 0, x∈0,π2. If yπ4=−1loge⁡2, then yπ6 is:

cos⁡x(ln⁡(cos⁡x))2dy+(sin⁡x−3y(sin⁡x)ln⁡(cos⁡x))dx=0cos⁡x(ln⁡(cos⁡x))2dydx−3sin⁡x⋅ln⁡(cos⁡x)y=−sin⁡xdydx−3tan⁡xln⁡(cos⁡x)y=−tan⁡x(ln⁡(cos⁡x))2dydx+3tan⁡xln⁡(sec⁡x)y=−tan⁡x(ln⁡(sec⁡x))2I.F =e∫3tan⁡xln⁡(sec⁡x)dx=(ln⁡(sec⁡x))3 y×(ln⁡(sec⁡x))3=−∫tan⁡x(ln⁡(sec⁡x))2(ln⁡(sec⁡x))3dx+Cy×(ln⁡(sec⁡x))3=−12(ln⁡(sec⁡x))2+CGiven :x=π4,y=−1ln⁡2−1ln⁡2×(ln⁡2)3=−12×(ln⁡2)2+C⇒−18ln⁡2×(ln⁡2)3=−12×14(ln⁡2)2+C−18(ln⁡2)2=−18(ln⁡2)2+C⇒C=0∴y(ln⁡(sec⁡x))3=−12(ln⁡(sec⁡x))2+0y=−12ln⁡(sec⁡x)y=12ln⁡(cos⁡x)∴yπ6=12ln⁡cos⁡π6=12ln⁡32=1212ln⁡3−ln⁡2=1ln⁡3−ln⁡4

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Let y = y(x) be the solution of the differential equation cosx(log_e(cosx))²dy + (sinx –3ysinx log_e(cosx))dx = 0, $x \in (0, \frac{π}{2}) . If y (\frac{π}{4}) = \frac{- 1}{\log_{e} 2}, then y (\frac{π}{6})$ is:

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

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

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

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

answer is D.

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Detailed Solution

$\begin{array}{l} \cos x (\ln (\cos x))^{2} dy + (\sin x - 3 y (\sin x) \ln (\cos x)) dx = 0 \\ \cos x (\ln (\cos x))^{2} \frac{dy}{dx} - 3 \sin x \cdot \ln (\cos x) y = - \sin x \\ \frac{dy}{dx} - \frac{3 \tan x}{\ln (\cos x)} y = \frac{- \tan x}{(\ln (\cos x))^{2}} \\ \frac{dy}{dx} + \frac{3 \tan x}{\ln (\sec x)} y = \frac{- \tan x}{(\ln (\sec x))^{2}} \\ I.F = e^{\int \frac{3 \tan x}{\ln (\sec x)} dx} = (\ln (\sec x))^{3} \end{array} \begin{array}{l} y \times (\ln (\sec x))^{3} = - \int \frac{\tan x}{(\ln (\sec x))^{2}} (\ln (\sec x))^{3} dx + C \\ y \times (\ln (\sec x))^{3} = - \frac{1}{2} (\ln (\sec x))^{2} + C \\ Given : x = \frac{π}{4}, y = - \frac{1}{\ln 2} \\ \frac{- 1}{\ln 2} \times (\ln \sqrt{2})^{3} = - \frac{1}{2} \times (\ln \sqrt{2})^{2} + C \end{array}$
$\begin{array}{l} \Rightarrow \frac{- 1}{8 \ln 2} \times (\ln 2)^{3} = \frac{- 1}{2} \times \frac{1}{4} (\ln 2)^{2} + C \\ - \frac{1}{8} (\ln 2)^{2} = \frac{- 1}{8} (\ln 2)^{2} + C \\ \Rightarrow C = 0 \\ ∴ y (\ln (\sec x))^{3} = \frac{- 1}{2} (\ln (\sec x))^{2} + 0 \\ y = \frac{- 1}{2 \ln (\sec x)} \\ y = \frac{1}{2 \ln (\cos x)} \\ ∴ y (\frac{π}{6}) = \frac{1}{2 \ln (\cos \frac{π}{6})} \end{array}$
$\begin{array}{l} = \frac{1}{2 \ln (\frac{\sqrt{3}}{2})} \\ = \frac{1}{2 (\frac{1}{2} \ln 3 - \ln 2)} \\ = \frac{1}{\ln 3 - \ln 4} \end{array}$