If dydx+2y tanx=sinx, 0<x<π2 and yπ3=0 then the maximum value of yx is

I.F =sec2xEquation is ysec2x=secx+cC=−2⇒ysec2x=secx−2⇒y=cosx−2cos2xdydx=0⇒cosx=14∴y=14−18=18

If dydx+2y tanx=sinx, 0<x<π2 and yπ3=0 then the maximum value of yx is

I.F =sec2xEquation is ysec2x=secx+cC=−2⇒ysec2x=secx−2⇒y=cosx−2cos2xdydx=0⇒cosx=14∴y=14−18=18

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If $\frac{d y}{d x} + 2 y t a n x = \sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ then the maximum value of $y (x)$ is

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$\frac{1}{8}$

$\frac{3}{4}$

$\frac{1}{4}$

$\frac{3}{8}$

answer is A.

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Detailed Solution

I.F = $\sec^{2} x$

Equation is $y \sec^{2} x = \sec x + c$

$C = - 2$

$\Rightarrow y \sec^{2} x = \sec x - 2$

$\Rightarrow y = \cos x - 2 \cos^{2} x$

$\frac{d y}{d x} = 0 \Rightarrow \cos x = \frac{1}{4}$

$∴ y = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$

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If dydx+2y tanx=sinx, 0<x<π2 and yπ3=0 then the maximum value of yx is

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If $\frac{d y}{d x} + 2 y t a n x = \sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ then the maximum value of $y (x)$ is