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

We start from the given differential equation

$$x\,y' \;-\; y \;=\; x^{2}\bigl(x\cos x + \sin x\bigr), \qquad x > 0.$$

Because a factor of $$x$$ is multiplying $$y'$$, we first divide the whole equation by $$x$$ (allowed since $$x > 0$$) so that the derivative term stands alone:

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

This is a first-order linear differential equation of the standard form

$$y' + P(x)\,y = Q(x),$$

with

$$P(x) = -\frac{1}{x}, \qquad Q(x) = x^{2}\cos x + x\sin x.$$

The integrating factor is obtained from the formula $$\mu(x)=e^{\int P(x)\,dx}.$$ Hence

$$\mu(x)=\exp\!\bigl(\int -\tfrac{1}{x}\,dx\bigr) =\exp(-\ln x)=x^{-1}.$$

Multiplying every term of the differential equation by this integrating factor $$x^{-1}$$ gives

$$x^{-1}y' \;-\;x^{-2}y \;=\;x^{-1}\!\bigl(x^{2}\cos x + x\sin x\bigr) \;=\;x\cos x + \sin x.$$

By construction the left-hand side is the exact derivative of $$\mu(x)y=x^{-1}y$$:

$$\frac{d}{dx}\!\bigl(x^{-1}y\bigr) \;=\; x\cos x + \sin x.$$

We now integrate both sides with respect to $$x$$:

$$x^{-1}y \;=\; \int \bigl(x\cos x + \sin x\bigr)\,dx + C.$$

We evaluate the integral term by term.

First, for $$\displaystyle\int x\cos x\,dx$$ we apply integration by parts:

Let $$u = x \;\Rightarrow\; du = dx,$$   and $$dv=\cos x\,dx \;\Rightarrow\; v=\sin x.$$ Then

$$\int x\cos x\,dx = x\sin x -\!\int\!\sin x\,dx = x\sin x + \cos x.$$

Second, $$\displaystyle\int \sin x\,dx = -\cos x.$$

Adding the two results, the sum of the integrals is simply

$$x\sin x + \cos x \;+\;(-\cos x)=x\sin x.$$

Thus we have

$$x^{-1}y = x\sin x + C.$$

Multiplying through by $$x$$ gives the explicit solution

$$y(x)=x^{2}\sin x + Cx.$$

We next use the initial condition $$y(\pi)=\pi$$ to find the constant $$C$$:

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

so $$C = 1.$$ Substituting this back, our solution becomes

$$y(x)=x^{2}\sin x + x.$$

Now we need $$y''\!\bigl(\tfrac{\pi}{2}\bigr)+y\!\bigl(\tfrac{\pi}{2}\bigr).$$ To do this we first differentiate $$y(x)$$ twice.

The first derivative :

$$\begin{aligned} y'(x) &= \frac{d}{dx}\bigl(x^{2}\sin x + x\bigr) \\ &= 2x\sin x + x^{2}\cos x + 1. \end{aligned}$$

The second derivative :

$$\begin{aligned} y''(x) &= \frac{d}{dx}\bigl(2x\sin x + x^{2}\cos x + 1\bigr) \\ &= 2\sin x + 2x\cos x + 2x\cos x - x^{2}\sin x \\ &= 2\sin x + 4x\cos x - x^{2}\sin x. \end{aligned}$$

We now evaluate these expressions at $$x = \dfrac{\pi}{2}.$$

Noting that $$\sin\!\bigl(\tfrac{\pi}{2}\bigr)=1$$ and $$\cos\!\bigl(\tfrac{\pi}{2}\bigr)=0,$$ we have

$$\begin{aligned} y''\!\bigl(\tfrac{\pi}{2}\bigr) &= 2\cdot 1 \;+\; 4\cdot \tfrac{\pi}{2}\cdot 0 \;-\;\bigl(\tfrac{\pi}{2}\bigr)^{2}\cdot 1 \\ &= 2 - \frac{\pi^{2}}{4}. \end{aligned}$$

Similarly,

$$\begin{aligned} y\!\bigl(\tfrac{\pi}{2}\bigr) &= \bigl(\tfrac{\pi}{2}\bigr)^{2}\!\cdot 1 \;+\;\tfrac{\pi}{2} = \frac{\pi^{2}}{4} + \frac{\pi}{2}. \end{aligned}$$

Adding these two values gives

$$\begin{aligned} y''\!\bigl(\tfrac{\pi}{2}\bigr) + y\!\bigl(\tfrac{\pi}{2}\bigr) &= \Bigl(2 - \frac{\pi^{2}}{4}\Bigr) + \Bigl(\frac{\pi^{2}}{4} + \frac{\pi}{2}\Bigr) \\ &= 2 + \frac{\pi}{2}. \end{aligned}$$

Hence, the correct answer is Option A.