Let y = y ( x ) be the solution of the differential equation, x y ^...

Given $x\dfrac{dy}{dx} - y = {x}^{2}\left(xcos + \sin x\right)$

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

$∴$ I.F $= {e}^{- \ln x} = \dfrac{1}{x}$

$∴$ Solution is $y . \dfrac{1}{x} = ∫ \dfrac{1}{x} . x\left(xcosx + \sin x\right)dx$

$\dfrac{y}{x} = ∫ \left(xcosx + \sin x\right)dx$

$\dfrac{y}{x} = xsinx + C$

$∵ y\left(\pi\right) = \pi ⇒ C = 1$

$y = {x}^{2}\sin x + x$

$\dfrac{dy}{dx} = {x}^{2}\cos x + 2xsinx + 1$

$\dfrac{{d}^{2}y}{d{x}^{2}} = - {x}^{2}\sin x + 2xcosx + 2\sin x + 2xcosx$

$= - {x}^{2}\sin x + 4xcosx + 2\sin x$

$∴$ ${y}^{''}\left(\dfrac{\pi}{2}\right) + y\left(\dfrac{\pi}{2}\right) = \left(- \dfrac{{\pi}^{2}}{4} + 0 + 2\right) + \left(\dfrac{{\pi}^{2}}{4} + \dfrac{\pi}{2}\right)$

$= - \dfrac{{\pi}^{2}}{4} + 2 + \dfrac{{\pi}^{2}}{4} + \dfrac{\pi}{2}$

$= \dfrac{\pi}{2} + 2$.