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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.

If u(x, y) = (x − y)^100 and u(x, y) = (x − 2y)^200 are both solutions of the PDE a ∂^2u/∂x^2 + b ∂^2u/∂x∂y + ∂^2u/∂y^2 = 0 at all points (x, y), find the constants a and b.

This problem involves finding the constants a and b such that both u(x, y) = (x − y)^100 and u(x, y) = (x − 2y)^200 are solutions to the given second-order partial differential equation. The solution process involves computing second partial derivatives and solving a system of equations for a and b.

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