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

Given,   $\frac{dy}{dx}+2y\tan x=\sin x,0<x<\frac{\pi }{2}$
Here  $p=2\tan xandQ=\sin x$
$$\begin{gathered} IF=e^{\int pdx}=e^{\int 2\tan xdx} \\ {e}^{2\log \left|\sec x\right|}={\sec }^{2}x \\ y\times IF=\int Q\times IFdx \\ y\times {\sec }^{2}x=\int \sin x\times {\sec }^{2}xdx \\ =\int \tan x\sec xdx \\ \therefore y\times {\sec }^{2}x=\sec x+c \\ 0\times {\sec }^2\frac{\pi }{3}=\sec \frac{\pi }{3}+C \\ O=2+c\Rightarrow c=-2 \\ \therefore y{\sec }^{2}x=\sec x-2 \\ \therefore y=\cos x-2{\cos }^{2}x \\ \Rightarrow y=-2({\cos }^{2}x-\frac{1}{2}\cos x) \\ \Rightarrow y=-2[{\cos }^{2}x-\frac{1}{2}\cos x+{\left(\frac{1}{4}\right)}^2-{\left(\frac{1}{4}\right)}^2] \\ \Rightarrow y=-2[{(\cos x-\frac{1}{4})}^2-\frac{1}{16}] \\ \Rightarrow y=\frac{1}{8}-2{(\cos x-\frac{1}{4})}^2 \end{gathered}$$
Minimum value of  $\cos x-\frac{1}{4}$  is  0
Maximum value of  $y=\frac{1}{8}-0=\frac{1}{8}$