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

Given differential equation,

$$\frac{dy}{dx}+2y\tan x=\sin x,\qquad 0<x<\frac{\pi}{2}$$

This is a linear differential equation.

Integrating factor is

$$IF=e^{\int2\tan x\,dx}$$

$$=e^{-2\ln(\cos x)}$$

$$=\sec^2x$$

Multiplying throughout by $$\sec^2x,$$

$$\sec^2x\frac{dy}{dx}+2y\sec^2x\tan x=\sin x\sec^2x$$

Hence,

$$\frac{d}{dx}(y\sec^2x)=\tan x\sec x$$

Integrating,

$$y\sec^2x=\sec x+C$$

$$y=\cos x+C\cos^2x$$

Using

$$y\left(\frac{\pi}{3}\right)=0,$$

$$0=\cos\frac{\pi}{3}+C\cos^2\frac{\pi}{3}$$

$$0=\frac12+\frac C4$$

$$C=-2$$

Therefore,

$$y=\cos x-2\cos^2x$$

Now find maximum value.

Differentiate:

$$\frac{dy}{dx}=-\sin x+4\sin x\cos x$$

$$=\sin x(4\cos x-1)$$

Since

$$0<x<\frac{\pi}{2},$$

$$\sin x>0$$

Hence,

$$\frac{dy}{dx}=0\iff4\cos x-1=0$$

$$\cos x=\frac14$$

Now,

$$y=\cos x-2\cos^2x$$

Put

$$t=\cos x,\qquad 0<t<1$$

Then,

$$y=t-2t^2$$

This is a downward parabola.

Maximum occurs at

$$t=\frac{-1}{2(-2)}=\frac14$$

Therefore,

$$y_{\max}=\frac14-2\left(\frac1{16}\right)$$

$$=\frac14-\frac18$$

$$=\frac18$$

Hence, the maximum value of $$y(x)$$ is

$$\boxed{\frac18}$$.