If d y d x + 2 y tan x = sin x , 0 < x < π 2 and y π 3 = 0 , then the maximum value of y x…

We know that,

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

$I.F=e^{\int 2\tan xdx}=e^{2\ln \left|\sec x\right|}={\sec }^{2}x$

The general solution will be 

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

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

$\because y\left(\frac{\pi }{3}\right)=0\Rightarrow C=-2$

Hence the particular solution is

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

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

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

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