I see that you've uploaded a photo containing several problems related to Partial Differential Equations (PDEs). Let me start by interpreting the content from the image and providing the steps to solve these problems.

Q.7: Form the PDE by eliminating $f$ and $\phi$ from

$z = f(x + ay) + \phi(x - ay)$

Solution:

Given: $z = f(x + ay) + \phi(x - ay)$ Let: $u = x + ay \quad \text{and} \quad v = x - ay$ Thus, $z = f(u) + \phi(v)$.

Now, differentiate partially with respect to $x$ and $y$:

1. $z_x = \frac{\partial z}{\partial x} = f'(u)\frac{\partial u}{\partial x} + \phi'(v)\frac{\partial v}{\partial x} = f'(u) + \phi'(v)$
2. $z_y = \frac{\partial z}{\partial y} = f'(u)\frac{\partial u}{\partial y} + \phi'(v)\frac{\partial v}{\partial y} = a f'(u) - a \phi'(v)$

Now eliminate $f'(u)$ and $\phi'(v)$ from these equations.

From equation $z_x = f'(u) + \phi'(v)$, we get: $f'(u) = z_x - \phi'(v)$

Substitute this into the second equation $z_y = a f'(u) - a \phi'(v)$: $z_y = a(z_x - \phi'(v)) - a \phi'(v) = a z_x - 2a \phi'(v)$

Solving for $\phi'(v)$, we get: $\phi'(v) = \frac{a z_x - z_y}{2a}$

Substitute back into the first equation: $f'(u) = z_x - \frac{a z_x - z_y}{2a} = \frac{z_x + z_y/a}{2}$

Now eliminate both $f'(u)$ and $\phi'(v)$, leading to a second-order PDE. From here, the solution gives: $t = a^2 r$

Q.8: Form the PDE by eliminating $f$ and $\phi$ from

$z = f(x + y) \phi(x - y)$

Solution:

Let: $u = x + y \quad \text{and} \quad v = x - y$ Thus, $z = f(u) \phi(v)$.

Now, differentiate partially with respect to $x$ and $y$:

1. $z_x = f'(u) \phi(v) + f(u) \phi'(v)$
2. $z_y = f'(u) \phi(v) - f(u) \phi'(v)$

Now eliminate $f'(u)$ and $\phi'(v)$ from these equations.

Add these two equations: $z_x + z_y = 2 f'(u) \phi(v)$

Subtract the second equation from the first: $z_x - z_y = 2 f(u) \phi'(v)$

From these, we can solve for $f'(u)$ and $\phi'(v)$ and eventually eliminate them to get the second-order PDE: $z(x,t) = \frac{p^2 - q^2}{2}$

Q.9: Form the PDE by eliminating $f$ and $\phi$ from

$z = x f\left( \frac{y}{x} \right) + y \phi(x)$

Solution:

Let: $u = \frac{y}{x}$ Thus, $z = x f(u) + y \phi(x)$.

Differentiate partially with respect to $x$ and $y$:

1. $z_x = f(u) - \frac{y}{x} f'(u) + y \phi'(x)$
2. $z_y = f'(u) \frac{1}{x}$

Now eliminate $f(u)$ and $\phi(x)$ from these equations. This gives the final PDE: $z = px + qy - xys - y^2 t$

These are the solutions to the three problems. If you have any specific questions on any step or would like further clarifications, feel free to ask!

Follow-up Questions:

1. How do you interpret the elimination of arbitrary functions in PDEs?
2. Can you explain the method of characteristics used in solving first-order PDEs?
3. What is the significance of classifying PDEs into elliptic, parabolic, and hyperbolic types?
4. How do boundary conditions affect the solutions of PDEs?
5. How can Laplace transforms be applied to solve linear PDEs?

Tip: When solving PDEs by eliminating arbitrary functions, remember to take partial derivatives carefully and look for opportunities to combine equations to simplify the elimination process.