Let $$y = y(x)$$ satisfies the equation $$\frac{dy}{dx} - |A| = 0$$, for all $$x > 0$$, where $$A = \begin{bmatrix} y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x} \end{bmatrix}$$. If $$y(\pi) = \pi + 2$$, then the value of $$y\left(\frac{\pi}{2}\right)$$ is:

We first compute $$|A| = \det(A)$$ where $$A = \begin{bmatrix} y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x} \end{bmatrix}$$.

Expanding along the first row: $$|A| = y\left(-\frac{1}{x} - 0\right) - \sin x\left(0 - 2\right) + 1\left(0 + 2\right) = -\frac{y}{x} + 2\sin x + 2.$$

The ODE becomes $$\frac{dy}{dx} = -\frac{y}{x} + 2\sin x + 2$$, i.e., $$\frac{dy}{dx} + \frac{y}{x} = 2\sin x + 2.$$

The integrating factor is $$e^{\int \frac{1}{x}dx} = x$$. Multiplying through: $$\frac{d(xy)}{dx} = 2x\sin x + 2x.$$

Integrating: $$xy = \int 2x\sin x\,dx + \int 2x\,dx$$. Using integration by parts, $$\int 2x\sin x\,dx = -2x\cos x + 2\sin x$$, so: $$xy = -2x\cos x + 2\sin x + x^2 + C.$$

Applying the condition $$y(\pi) = \pi + 2$$: $$\pi(\pi + 2) = -2\pi\cos\pi + 2\sin\pi + \pi^2 + C = 2\pi + \pi^2 + C.$$ $$\pi^2 + 2\pi = 2\pi + \pi^2 + C \implies C = 0.$$

Thus $$y = -2\cos x + \frac{2\sin x}{x} + x$$. Evaluating at $$x = \frac{\pi}{2}$$: $$y\!\left(\frac{\pi}{2}\right) = -2\cos\frac{\pi}{2} + \frac{2\sin\frac{\pi}{2}}{\frac{\pi}{2}} + \frac{\pi}{2} = 0 + \frac{4}{\pi} + \frac{\pi}{2} = \frac{\pi}{2} + \frac{4}{\pi}.$$