We are given that both functions $u(x,y) = (x - y)^{100}$ and $u(x,y) = (x - 2y)^{200}$ are solutions to the following partial differential equation (PDE):

$a \frac{\partial^2 u}{\partial x^2} + b \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} = 0$

We are tasked with finding the constants $a$ and $b$.

Step-by-Step Solution:

1. First Function: $u(x, y) = (x - y)^{100}$

   • Compute the second partial derivatives: $\frac{\partial u}{\partial x} = 100 (x - y)^{99}$ $\frac{\partial^2 u}{\partial x^2} = 100 \cdot 99 (x - y)^{98}$ $\frac{\partial u}{\partial y} = -100 (x - y)^{99}$ $\frac{\partial^2 u}{\partial y^2} = 100 \cdot 99 (x - y)^{98}$ $\frac{\partial^2 u}{\partial x \partial y} = -100 \cdot 99 (x - y)^{98}$

   • Substitute into the PDE: $a \cdot 100 \cdot 99 (x - y)^{98} + b \cdot (-100 \cdot 99 (x - y)^{98}) + 100 \cdot 99 (x - y)^{98} = 0$ Factor out $100 \cdot 99 (x - y)^{98}$: $(a - b + 1) \cdot 100 \cdot 99 (x - y)^{98} = 0$ Since $(x - y)^{98} \neq 0$, we have: a - b + 1 = 0 \quad \Rightarrow \quad a = b - 1 \tag{1}

2. Second Function: $u(x, y) = (x - 2y)^{200}$

   • Compute the second partial derivatives: $\frac{\partial u}{\partial x} = 200 (x - 2y)^{199}$ $\frac{\partial^2 u}{\partial x^2} = 200 \cdot 199 (x - 2y)^{198}$ $\frac{\partial u}{\partial y} = -400 (x - 2y)^{199}$ $\frac{\partial^2 u}{\partial y^2} = 4 \cdot 200 \cdot 199 (x - 2y)^{198}$ $\frac{\partial^2 u}{\partial x \partial y} = -2 \cdot 200 \cdot 199 (x - 2y)^{198}$

   • Substitute into the PDE: $a \cdot 200 \cdot 199 (x - 2y)^{198} + b \cdot (-2 \cdot 200 \cdot 199 (x - 2y)^{198}) + 4 \cdot 200 \cdot 199 (x - 2y)^{198} = 0$ Factor out $200 \cdot 199 (x - 2y)^{198}$: $(a - 2b + 4) \cdot 200 \cdot 199 (x - 2y)^{198} = 0$ Since $(x - 2y)^{198} \neq 0$, we have: a - 2b + 4 = 0 \quad \Rightarrow \quad a = 2b - 4 \tag{2}

3. Solve the System of Equations:

   From equation (1): $a = b - 1$

   Substitute this into equation (2): $b - 1 = 2b - 4$ Solving for $b$: $-1 + 4 = 2b - b \quad \Rightarrow \quad b = 3$

   Now substitute $b = 3$ into equation (1): $a = 3 - 1 = 2$

Thus, the constants are $a = 2$ and $b = 3$.

The correct answer is: $\boxed{a = 2, b = 3}$

Would you like further details on any step or have additional questions?

Here are some related questions:

1. How do partial derivatives change for different types of functions?
2. What are the different methods to solve second-order PDEs?
3. How do boundary conditions affect the solution of a PDE?
4. Can you explain the method of separation of variables for PDEs?
5. What are the applications of PDEs in physics and engineering?

Tip: When solving PDEs, it's essential to verify if the given functions satisfy the boundary conditions to ensure the solution's validity.