To form a partial differential equation (PDE) by eliminating the arbitrary functions $f$ and $g$ from the given equation $z = f(ax + by) + g(\alpha x + \beta y)$, we need to differentiate the equation with respect to $x$ and $y$ and then eliminate the functions $f$ and $g$ using the resulting equations.

Given:

$z = f(ax + by) + g(\alpha x + \beta y)$

Step 1: Differentiate with respect to $x$

First, we differentiate with respect to $x$:

$\frac{\partial z}{\partial x} = f'(ax + by) \cdot (a) + g'(\alpha x + \beta y) \cdot (\alpha)$

$\frac{\partial z}{\partial x} = a f'(ax + by) + \alpha g'(\alpha x + \beta y)$

Step 2: Differentiate with respect to $y$

Next, we differentiate with respect to $y$:

$\frac{\partial z}{\partial y} = f'(ax + by) \cdot (b) + g'(\alpha x + \beta y) \cdot (\beta)$

$\frac{\partial z}{\partial y} = b f'(ax + by) + \beta g'(\alpha x + \beta y)$

Step 3: Differentiate with respect to $x$ again

Differentiate $\frac{\partial z}{\partial x}$ with respect to $x$:

$\frac{\partial^2 z}{\partial x^2} = a^2 f''(ax + by) + \alpha^2 g''(\alpha x + \beta y)$

Step 4: Differentiate with respect to $y$ again

Differentiate $\frac{\partial z}{\partial y}$ with respect to $y$:

$\frac{\partial^2 z}{\partial y^2} = b^2 f''(ax + by) + \beta^2 g''(\alpha x + \beta y)$

Step 5: Mixed Partial Derivative with respect to $x$ and $y$

Differentiate $\frac{\partial z}{\partial x}$ with respect to $y$:

$\frac{\partial^2 z}{\partial x \partial y} = ab f''(ax + by) + \alpha\beta g''(\alpha x + \beta y)$

Step 6: Eliminate $f''$ and $g''$

To eliminate the arbitrary functions, consider the following:

• $\frac{\partial^2 z}{\partial x^2}$ multiplied by $\beta^2$:

  $\beta^2 \frac{\partial^2 z}{\partial x^2} = \beta^2 a^2 f''(ax + by) + \alpha^2 \beta^2 g''(\alpha x + \beta y)$

• $\frac{\partial^2 z}{\partial y^2}$ multiplied by $\alpha^2$:

  $\alpha^2 \frac{\partial^2 z}{\partial y^2} = \alpha^2 b^2 f''(ax + by) + \alpha^2 \beta^2 g''(\alpha x + \beta y)$

• Subtracting these two equations:

  $\beta^2 \frac{\partial^2 z}{\partial x^2} - \alpha^2 \frac{\partial^2 z}{\partial y^2} = (\beta^2 a^2 - \alpha^2 b^2) f''(ax + by)$

Since the right side must be zero to eliminate $f''(ax + by)$, we have:

$\boxed{\beta^2 \frac{\partial^2 z}{\partial x^2} - \alpha^2 \frac{\partial^2 z}{\partial y^2} = 0}$

This is the required PDE after eliminating the arbitrary functions $f$ and $g$.

Would you like further details or have any other questions?

Here are 8 related questions to explore:

1. How would you form the PDE if the functions were $z = f(x + y) + g(x - y)$?
2. What are the characteristics of the PDE obtained?
3. How does the choice of arbitrary functions affect the type of PDE formed?
4. What methods can be used to solve the derived PDE?
5. What role do the coefficients $\alpha$ and $\beta$ play in the PDE formation?
6. How can this approach be extended to multiple arbitrary functions?
7. What are the boundary conditions necessary to solve the PDE?
8. How would the PDE change if the original function was $z = f(a^2x + b^2y) + g(\alpha^2 x + \beta^2 y)$?

Tip: Always check if your derived PDE satisfies the initial conditions or any symmetries that might simplify the problem.