(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}(uv) = u'v + uv'\).

Let \(u = x\) and \(v = \cos \frac{1}{2}x\). Then \(u' = 1\) and \(v' = -\frac{1}{2} \sin \frac{1}{2}x\).

Thus, \(\frac{dy}{dx} = 1 \cdot \cos \frac{1}{2}x + x \cdot \left(-\frac{1}{2} \sin \frac{1}{2}x\right) = \cos \frac{1}{2}x - \frac{1}{2}x \sin \frac{1}{2}x\).

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\) again using the product rule.

\(\frac{d^2y}{dx^2} = -\frac{1}{2} \sin \frac{1}{2}x - \frac{1}{2} \left( \cos \frac{1}{2}x + \frac{1}{2}x \sin \frac{1}{2}x \right)\).

Simplify to get \(\frac{d^2y}{dx^2} = -\frac{1}{2} \sin \frac{1}{2}x - \frac{1}{4}x \cos \frac{1}{2}x - \frac{1}{2} \sin \frac{1}{2}x\).

Substitute into \(4 \frac{d^2y}{dx^2} + y + 4 \sin \frac{1}{2}x = 0\) to verify.

(ii) To find the area, integrate \(y = x \cos \frac{1}{2}x\) from \(0\) to \(\pi\).

\(\int_0^\pi x \cos \frac{1}{2}x \, dx = \left[ 2x \sin \frac{1}{2}x - 4 \cos \frac{1}{2}x \right]_0^\pi\).

Evaluate: \(2\pi - 4\).