Let y = y ( x ) be the solution of the differential equation, x y ' - y = x 2 ( x cos x + sin x ) , x…

Given $x\frac{dy}{dx}-y=x^2\left(\mathrm{xcos}+\mathrm{sinx}\right)$

$\Rightarrow$$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{x}}\mathrm{y}=\mathrm{x}\left(\mathrm{xcosx}+\mathrm{sinx}\right)$

$\therefore$ I.F $={\mathrm{e}}^{-l\mathrm{nx}}=\frac{1}{\mathrm{x}}$

$\therefore$ Solution is $y.\frac{1}{x}=\int \frac{1}{x}.x\left(x\cos x+\sin x\right)\mathrm{d}x$

$\frac{y}{x}=\int \left(x\cos x+\sin x\right)\mathrm{dx}$

$\frac{y}{x}=x\sin x+\mathrm{C}$

$\because \mathrm{y}\left(\mathrm{\pi }\right)=\mathrm{\pi } \Rightarrow \mathrm{C}=1$

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

$\frac{\mathrm{d}y}{\mathrm{d}x}=x^2\cos x+2x\sin x+1$

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=-x^2\sin x+2x\cos x+2\sin x+2x\cos x$

$=-x^2\sin x+4x\cos x+2\sin x$

$\therefore$ $y^{\text{'}\text{'}}\left(\frac{\pi }{2}\right)+y\left(\frac{\pi }{2}\right)=\left(-\frac{{\pi }^2}{4}+0+2\right)+\left(\frac{{\pi }^2}{4}+\frac{\pi }{2}\right)$

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

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