If y = y x is the solution of the differential equation, d y d x + 2 y tan x = sin x , y π 3 = 0 , then…

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

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

$\mathrm{I}.\mathrm{F}.={\sec }^{2}x$

$y.\left({\sec }^{2}x\right)=\int \sin x.{\sec }^{2}xdx+C$

$y.\left({\sec }^{2}x\right)=\int \sec x\tan xdx+C$

$y.\left({\sec }^{2}x\right)=\sec x+C$

$x=\frac{\pi }{3};y=0$

$\Rightarrow C=-2$

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

Let $\cos x=t, -1\le t\le 1$

$\Rightarrow y=t-2t^2\Rightarrow \frac{dy}{dt}=1-4t=0\Rightarrow t=\frac{1}{4}$

Second-order derivative is negative

$\therefore \max =\frac{1}{4}-\frac{1}{8}=\frac{2-1}{8}=\frac{1}{8}$