The given differential equation is:

$x \frac{\partial u}{\partial x} - 2y \frac{\partial u}{\partial y} = 0$

We are asked to solve this equation using the method of separation of variables. Here's the solution:

Step 1: Rewrite the equation

Rewrite the equation as: $\frac{\frac{\partial u}{\partial x}}{2y \frac{\partial u}{\partial y}} = \frac{1}{x}$

Or equivalently: $\frac{\frac{\partial u}{\partial x}}{\frac{\partial u}{\partial y}} = \frac{2y}{x}$

Step 2: Assume separable form

Assume the solution is of the separable form $u(x, y) = X(x)Y(y)$. Substituting this into $u$, we have:

$u_x = Y(y) \frac{dX}{dx}, \quad u_y = X(x) \frac{dY}{dy}.$

Substitute these into the equation: $x \cdot Y(y) \frac{dX}{dx} - 2y \cdot X(x) \frac{dY}{dy} = 0.$

Divide through by $X(x)Y(y)$: $x \frac{\frac{dX}{dx}}{X} = 2y \frac{\frac{dY}{dy}}{Y}.$

Step 3: Separate variables

This gives: $\frac{1}{X} \frac{dX}{dx} \cdot x = \frac{2}{Y} \frac{dY}{dy} \cdot y.$

Since the left-hand side depends only on $x$, and the right-hand side depends only on $y$, they must both be equal to a constant, say $k$. So: $\frac{x}{X} \frac{dX}{dx} = k, \quad \frac{y}{Y} \frac{dY}{dy} = \frac{k}{2}.$

Step 4: Solve the equations

For $X(x)$:

$\frac{1}{X} \frac{dX}{dx} = \frac{k}{x}.$ Integrate: $\ln|X| = k \ln|x| + C_1,$ $X(x) = C_1 x^k.$

For $Y(y)$:

$\frac{1}{Y} \frac{dY}{dy} = \frac{k}{2y}.$ Integrate: $\ln|Y| = \frac{k}{2} \ln|y| + C_2,$ $Y(y) = C_2 y^{k/2}.$

Step 5: Combine the solutions

The solution is: $u(x, y) = X(x)Y(y) = C x^k y^{k/2},$ where $C = C_1C_2$ is a constant.

Let me know if you have any questions or would like further explanation! Here are related questions to explore:

1. What happens when $k = 0$ in the solution?
2. How does the method of separation of variables simplify solving partial differential equations?
3. Can this method handle non-linear equations? Why or why not?
4. What physical processes might this equation describe?
5. What other methods can be used to solve partial differential equations?

Tip: Always check if your solution satisfies the original equation by substitution