1. Given differential equation

$xy' - y = x^2(x\cos x + \sin x), \qquad x>0$

We rewrite it in linear form by dividing by $x$:

$y' - \frac{1}{x}y = x(x\cos x + \sin x)=x^2\cos x + x\sin x$

2. Identify integrating factor

For $y' + P(x)y = Q(x),$ we have $P(x)=-\frac{1}{x}$ so the integrating factor is

$\text{I.F.}=e^{\int -\frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$ (since $x>0$).

3. Solve the differential equation

Multiply throughout by $\frac{1}{x}$:

$\frac{1}{x}y' - \frac{1}{x^2}y = x\cos x + \sin x$

The left side is:

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

Integrate:

$\frac{y}{x}=\int (x\cos x + \sin x)\,dx + C$

Now,

$\int x\cos x\,dx = x\sin x + \cos x$ by integration by parts, and

$\int \sin x\,dx = -\cos x$

Therefore,

$\int (x\cos x + \sin x)dx = x\sin x + \cos x - \cos x = x\sin x$

So,

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

Hence,

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

4. Use the condition $y(\pi)=\pi$

Substitute $x=\pi$:

$y(\pi)=\pi^2\sin \pi + C\pi = 0 + C\pi = \pi$

Thus,

$C=1$

So the solution is

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

5. Find $y''(\pi/2)$ and $y(\pi/2)$

First derivative:

$y' = \frac{d}{dx}(x^2\sin x + x)=2x\sin x + x^2\cos x + 1$

Second derivative:

$y'' = \frac{d}{dx}(2x\sin x + x^2\cos x + 1)$

Compute termwise:

$\frac{d}{dx}(2x\sin x)=2\sin x + 2x\cos x$

$\frac{d}{dx}(x^2\cos x)=2x\cos x - x^2\sin x$

Therefore,

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

At $x=\frac{\pi}{2}$, we use $\sin \frac{\pi}{2}=1, \qquad \cos \frac{\pi}{2}=0$

So,

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

Also,

y\left(\frac{\pi}{2}\right)=\left(\frac{\pi}{2}\right)^2\sin\frac{\pi}{2}+\frac{\pi}{2}=rac{\pi^2}{4}+\frac{\pi}{2}

6. Add them

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

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

7. Match with options

$2+\frac{\pi}{2}$ corresponds to Option C.