Step 1: The given equation is a linear differential equation of the type

$$\frac{dy}{dx} + Py = Q$$

where $P = \cot x$ and $Q = 4x \text{ cosec } x$.

Step 2: Find the Integrating Factor (I.F.):

$$I.F. = e^{\int \cot x \, dx} = e^{\log \sin x} = \sin x$$

Step 3: The solution is given by

$$y \cdot \sin x = \int (4x \text{ cosec } x) \sin x \, dx + C$$

Since $\text{cosec } x = \frac{1}{\sin x}$, we get

$$y \sin x = \int 4x \, dx + C = 2x^2 + C$$

Step 4: Using the condition $y = 0$ when $x = \frac{\pi}{2}$:

$$0 \cdot \sin \frac{\pi}{2} = 2\left(\frac{\pi}{2}\right)^2 + C$$

which gives

$$C = -\frac{\pi^2}{2}$$

Step 5: Substituting $C$ back:

$$y \sin x = 2x^2 - \frac{\pi^2}{2}$$

or

$$y = \frac{2x^2}{\sin x} - \frac{\pi^2}{2\sin x}$$

∴ The particular solution is $\mathbf{y = \frac{2x^2}{\sin x} - \frac{\pi^2}{2\sin x}}$.