To solve for the constants $a$ and $b$, we must check whether both given solutions $u(x, y) = (x - y)^{100}$ and $u(x, y) = (x - 2y)^{200}$ satisfy the 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 will calculate the second partial derivatives of both functions and substitute them into the PDE.

First solution: $u(x, y) = (x - y)^{100}$

Let’s compute the partial derivatives for $u_1(x, y) = (x - y)^{100}$:

1. First derivatives: $\frac{\partial u_1}{\partial x} = 100(x - y)^{99}, \quad \frac{\partial u_1}{\partial y} = -100(x - y)^{99}.$

2. Second derivatives: $\frac{\partial^2 u_1}{\partial x^2} = 100 \cdot 99(x - y)^{98} = 9900(x - y)^{98},$ $\frac{\partial^2 u_1}{\partial x \partial y} = -100 \cdot 99(x - y)^{98} = -9900(x - y)^{98},$ $\frac{\partial^2 u_1}{\partial y^2} = 100 \cdot 99(x - y)^{98} = 9900(x - y)^{98}.$

Substitute these into the PDE: $a \cdot 9900(x - y)^{98} + b \cdot (-9900(x - y)^{98}) + 9900(x - y)^{98} = 0,$ $9900(x - y)^{98} [a - b + 1] = 0.$ Since $(x - y)^{98} \neq 0$, we get: a - b + 1 = 0 \quad \text{or} \quad a = b - 1. \tag{1}

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

Now, we compute the partial derivatives for $u_2(x, y) = (x - 2y)^{200}$:

1. First derivatives: $\frac{\partial u_2}{\partial x} = 200(x - 2y)^{199}, \quad \frac{\partial u_2}{\partial y} = -400(x - 2y)^{199}.$

2. Second derivatives: $\frac{\partial^2 u_2}{\partial x^2} = 200 \cdot 199(x - 2y)^{198} = 39800(x - 2y)^{198},$ $\frac{\partial^2 u_2}{\partial x \partial y} = -400 \cdot 199(x - 2y)^{198} = -79600(x - 2y)^{198},$ $\frac{\partial^2 u_2}{\partial y^2} = 4 \cdot 200 \cdot 199(x - 2y)^{198} = 159200(x - 2y)^{198}.$

Substitute these into the PDE: $a \cdot 39800(x - 2y)^{198} + b \cdot (-79600(x - 2y)^{198}) + 159200(x - 2y)^{198} = 0,$ $39800(x - 2y)^{198} [a - 2b + 4] = 0.$ Since $(x - 2y)^{198} \neq 0$, we get: a - 2b + 4 = 0 \quad \text{or} \quad a = 2b - 4. \tag{2}

Solving the system of equations

From equations (1) and (2), we have: $a = b - 1 \quad \text{and} \quad a = 2b - 4.$ Equating both expressions for $a$: $b - 1 = 2b - 4,$ $-b = -3 \quad \Rightarrow \quad b = 3.$ Substitute $b = 3$ into $a = b - 1$: $a = 3 - 1 = 2.$

Final answer:

The constants are $a = 2$ and $b = 3$. Therefore, the correct option is: $\boxed{a = 2, \, b = 3}.$

Would you like further explanation, or have any questions?

Here are 5 questions to explore further:

1. How do partial derivatives behave in a general PDE?
2. What other methods can be used to solve systems of linear equations?
3. What is the significance of boundary conditions in solving PDEs?
4. How does the chain rule apply to finding higher-order partial derivatives?
5. What are some common types of second-order PDEs in physics?

Tip: Always check that your solutions to PDEs satisfy both the equation and any boundary or initial conditions.