1. **Problem Statement:** Solve the partial differential equation (PDE) given by $$n \frac{\partial u}{\partial x} = y \frac{\partial u}{\partial y}$$ where $u = u(x,y)$. 2. **Method:** Assume a separable solution of the form $$u(x,y) = X(x)Y(y)$$ which allows us to write $$\frac{\partial u}{\partial x} = X'(x)Y(y), \quad \frac{\partial u}{\partial y} = X(x)Y'(y).$$ 3. **Substitute into PDE:** $$n X'(x) Y(y) = y X(x) Y'(y).$$ Divide both sides by $X(x)Y(y)$ (assuming nonzero): $$n \frac{X'(x)}{X(x)} = y \frac{Y'(y)}{Y(y)}.$$ 4. **Separate variables:** The left side depends only on $x$, the right side only on $y$, so both equal a constant $k$: $$n \frac{X'}{X} = k, \quad y \frac{Y'}{Y} = k.$$ 5. **Solve ODE for $X(x)$:** $$n \frac{dX}{dx} = k X \implies \frac{dX}{dx} = \frac{k}{n} X.$$ This is a first-order linear ODE with solution: $$X(x) = C_1 e^{\frac{k}{n} x}.$$ 6. **Solve ODE for $Y(y)$:** $$y \frac{dY}{dy} = k Y \implies \frac{dY}{dy} = \frac{k}{y} Y.$$ This is separable: $$\frac{dY}{Y} = \frac{k}{y} dy,$$ integrate: $$\ln |Y| = k \ln |y| + C_2 \implies Y(y) = C_3 y^k.$$ 7. **General solution:** $$u(x,y) = X(x) Y(y) = C e^{\frac{k}{n} x} y^k,$$ where $C = C_1 C_3$ is an arbitrary constant. 8. **Eigenvalues $k$:** The problem states eigenvalues $k = 0, d, -d$. - For $k=0$: $$u = C e^{0} y^{0} = C,$$ a constant solution. - For $k=d$: $$u = C e^{\frac{d}{n} x} y^{d}.$$ - For $k=-d$: $$u = C e^{-\frac{d}{n} x} y^{-d}.$$ **Summary:** The complete solution is a linear combination: $$u(x,y) = C_0 + C_1 e^{\frac{d}{n} x} y^{d} + C_2 e^{-\frac{d}{n} x} y^{-d}.$$