If (dy)/(dx) + 2ytanx = x, 0 < x < (π)/(2) and y((π)/(3)) = 0 ,...

We know that,

$\dfrac{dy}{dx} + y\left(2\tan x\right) = \sin x$ is a linear differential equation so,

$I . F = {e}^{∫ 2tanxdx} = {e}^{2\ln \left|\sec x\right|} = \sec^{2}x$

The general solution will be

$y{sec}^{2}x = ∫ \sin x{sec}^{2}xdx + C$

$⇒ y{sec}^{2}x = \sec x + C$

$∵ y\left(\dfrac{\pi}{3}\right) = 0 ⇒ C = - 2$

Hence the particular solution is

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

$⇒ y = \cos x - 2\cos^{2}x$

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

So, the maximum value of $y\left(x\right)$ is $\dfrac{1}{8}$