To solve the differential equation

$y' \sin(2\pi x) = \pi y \cos(2\pi x),$

we can rewrite it in a standard form:

$\frac{dy}{dx} = \frac{\pi y \cos(2\pi x)}{\sin(2\pi x)}.$

This can be simplified to:

$\frac{dy}{dx} = \pi y \cot(2\pi x).$

This is a first-order linear ordinary differential equation (ODE) of the form:

$\frac{dy}{dx} - \pi y \cot(2\pi x) = 0.$

To solve it, we can use an integrating factor. The integrating factor $\mu(x)$ is given by:

$\mu(x) = e^{\int -\pi \cot(2\pi x) \, dx}.$

Calculating the integral:

$\int -\pi \cot(2\pi x) \, dx = -\frac{1}{2} \ln(\sin(2\pi x)) + C.$

Thus, the integrating factor is:

$\mu(x) = \sin(2\pi x)^{-1/2}.$

Multiplying through the differential equation by this integrating factor:

$\sin(2\pi x)^{-1/2} \frac{dy}{dx} - \frac{\pi y}{2} \sin(2\pi x)^{-3/2} \cos(2\pi x) = 0.$

This can be solved, and ultimately we can write the general solution in implicit or explicit form.

If you want to proceed with finding the exact solution or need any specific details, let me know!

Follow-up Questions:

1. Do you need a step-by-step solution for finding the integrating factor?
2. Would you like to see the general solution explicitly?
3. Are you interested in the behavior of the solution for specific values of $x$?
4. Would you like to discuss the significance of this differential equation in a particular context?
5. Do you have any additional questions about solving first-order ODEs?

Tip: Always check for special solutions or initial conditions to fully define the solution of a differential equation.