"If ${{dy} \over {dx}} + 2y\tan x = \sin x,\,0"

Question

Accepted Answer

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${{dy} \over {dx}} + 2y\tan x = \sin x$

which is a first order linear differential equation.

Integrating factor (I. F.) $= {e^{\int {2\tan x\,dx} }}$

$= {e^{2\ln |\sec x|}} = {\sec ^2}x$

Solution of differential equation can be written as

$$y\,.\,{\sec ^2}x = \int {\sin x\,.\,{{\sec }^2}x\,dx = \int {\sec \,x\,.\,\tan x\,dx} } $$

$y~{\sec ^2}x = \sec x + C$

$$y\left( {{\pi \over 3}} \right) = 0,0 = \sec {\pi \over 3} + C \Rightarrow \,\,\,\,C = - 2$$

$y = {{\sec x - 2} \over {{{\sec }^2}x}} = \cos x - 2{\cos ^2}x$

$= {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}$

${y_{\max }} = {1 \over 8}$

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If ${{dy} \over {dx}} + 2y\tan x = \sin x,\,0 < x < {\pi \over 2}$ and $y\left( {{\pi \over 3}} \right) = 0$, then the maximum value of $y(x)$ is :