1. **State the problem:** Solve the partial differential equation (PDE) given by $$x \frac{\partial u}{\partial x} = y \frac{\partial u}{\partial y}$$ where $u = u(x,y)$. 2. **Assume a separable solution:** Let $$u(x,y) = X(x) Y(y)$$ which implies $$\frac{\partial u}{\partial x} = X'(x) Y(y), \quad \frac{\partial u}{\partial y} = X(x) Y'(y).$$ 3. **Substitute into PDE:** $$x X'(x) Y(y) = y X(x) Y'(y).$$ 4. **Divide both sides by $X(x) Y(y)$ (assuming nonzero):** $$x \frac{X'(x)}{X(x)} = y \frac{Y'(y)}{Y(y)}.$$ 5. **Separate variables:** The left side depends only on $x$, the right side only on $y$, so both equal a constant $k$: $$x \frac{X'}{X} = k, \quad y \frac{Y'}{Y} = k.$$ 6. **Solve ODE for $X(x)$:** $$x \frac{dX}{dx} = k X \implies \frac{dX}{dx} = \frac{k}{x} X.$$ This is separable: $$\frac{dX}{X} = \frac{k}{x} dx,$$ integrate: $$\ln|X| = k \ln|x| + C_1 \implies X(x) = C_1 x^k.$$ 7. **Solve ODE for $Y(y)$:** $$y \frac{dY}{dy} = k Y \implies \frac{dY}{dy} = \frac{k}{y} Y.$$ Similarly separable: $$\frac{dY}{Y} = \frac{k}{y} dy,$$ integrate: $$\ln|Y| = k \ln|y| + C_2 \implies Y(y) = C_2 y^k.$$ 8. **General solution:** $$u(x,y) = X(x) Y(y) = C x^k y^k = C (xy)^k,$$ where $C = C_1 C_2$ is an arbitrary constant. 9. **Interpretation:** The solution family is $$u(x,y) = C (xy)^k$$ for any constant $k$. This satisfies the PDE for all $k$.