A function y = f ( x ) satisfies f(x) 2x + x - (1 + ^ 2 x) f ^ '...

Given:

$f\left(x\right)\sin 2x + \sin x - \left(1 + \cos^{2}x\right){f}^{'}\left(x\right) = 0$

Now, let $f\left(x\right) = y$ we get,

$⇒ ysin2x + \sin x - \left(1 + \cos^{2}x\right)\dfrac{dy}{dx} = 0$

$⇒ \dfrac{ysin2x}{\left(1 + \cos^{2}x\right)} + \dfrac{\sin x}{\left(1 + \cos^{2}x\right)} - \dfrac{dy}{dx} = 0$

$⇒ \dfrac{dy}{dx} - \left(\dfrac{\sin 2x}{1 + \cos^{2}x}\right)y = \dfrac{\sin x}{1 + \cos^{2}x}$

$⇒ I . F . = {e}^{∫ - \left(\dfrac{\sin 2x}{1 + \cos^{2}x}\right)dx}$

$⇒ I . F . = {e}^{\log \left(1 + \cos^{2}x\right)}$

$⇒ I . F . = 1 + \cos^{2}x$

So, solution of the equation is given by,

$y\left(1 + \cos^{2}x\right) = ∫ \dfrac{\sin x}{\left(1 + \cos^{2}x\right)}\left(1 + \cos^{2}x\right)dx$

$⇒ y\left(1 + \cos^{2}x\right) = ∫ sinxdx$

$⇒ y\left(1 + \cos^{2}x\right) = - \cos x + C$

It is given that, $f\left(0\right) = 0$.

$⇒ 0\left(1 + 1\right) = - \cos 0 + C$

$⇒ C = 1$

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

$⇒ y = \dfrac{1 - \cos x}{1 + \cos^{2}x}$

$⇒ y\left(\dfrac{\pi}{2}\right) = \dfrac{1 - cos\dfrac{\pi}{2}}{1 + \cos^{2}\dfrac{\pi}{2}}$

$⇒ y\left(\dfrac{\pi}{2}\right) = 1$