Let y=y(x) be the solution of the differential equation, x y^′-y=x^2(x cos x+ sin x), x>0. If y(π)=π, then y^′ ′((π)/2)+y((π)/2) is equal to:

Let $\displaystyle y=y(x)$ be the solution of the differential equation, $\displaystyle x y^{\prime}-y=x^2(x \cos x+\sin x), x>0$. If $\displaystyle y(\pi)=\pi$, then $\displaystyle y^{\prime \prime}\left(\frac{\pi}{2}\right)+y\left(\frac{\pi}{2}\right)$ is equal to :