Form the PDE by eliminating the arbitrary function from z= f(x2 + y2).

To form the partial differential equation (PDE) by eliminating the arbitrary function from \( z = f(x^2 + y^2) \), follow these steps:

### Step 1: Define a new variable
Let \( u = x^2 + y^2 \), so that \( z = f(u) \), where \( f(u) \) is an arbitrary function of \( u \).

### Step 2: Find the first-order partial derivatives
We need to compute the partial derivatives of \( z \) with respect to \( x \) and \( y \).

#### Partial derivative of \( z \) with respect to \( x \):
Using the chain rule:
\[
\frac{\partial z}{\partial x} = \frac{d f}{d u} \cdot \frac{\partial u}{\partial x}
\]
Since \( u = x^2 + y^2 \), we have:
\[
\frac{\partial u}{\partial x} = 2x
\]
Therefore:
\[
\frac{\partial z}{\partial x} = \frac{d f}{d u} \cdot 2x
\]

#### Partial derivative of \( z \) with respect to \( y \):
Similarly, using the chain rule:
\[
\frac{\partial z}{\partial y} = \frac{d f}{d u} \cdot \frac{\partial u}{\partial y}
\]
Since \( u = x^2 + y^2 \), we have:
\[
\frac{\partial u}{\partial y} = 2y
\]
Therefore:
\[
\frac{\partial z}{\partial y} = \frac{d f}{d u} \cdot 2y
\]

### Step 3: Find the second-order partial derivatives
Now, compute the second-order partial derivatives of \( z \) with respect to \( x \) and \( y \).

#### Second-order partial derivative with respect to \( x \):
\[
\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{d f}{d u} \cdot 2x \right)
\]
Using the product rule:
\[
\frac{\partial^2 z}{\partial x^2} = \frac{d^2 f}{d u^2} \cdot 4x^2 + \frac{d f}{d u} \cdot 2
\]

#### Second-order partial derivative with respect to \( y \):
Similarly:
\[
\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{d f}{d u} \cdot 2y \right)
\]
Using the product rule:
\[
\frac{\partial^2 z}{\partial y^2} = \frac{d^2 f}{d u^2} \cdot 4y^2 + \frac{d f}{d u} \cdot 2
\]

### Step 4: Form the PDE
The required PDE involves the sum of the second-order partial derivatives:
\[
\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2}
\]
Substitute the expressions for \( \frac{\partial^2 z}{\partial x^2} \) and \( \frac{\partial^2 z}{\partial y^2} \):
\[
\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \left( \frac{d^2 f}{d u^2} \cdot 4x^2 + \frac{d f}{d u} \cdot 2 \right) + \left( \frac{d^2 f}{d u^2} \cdot 4y^2 + \frac{d f}{d u} \cdot 2 \right)
\]
Simplify the expression:
\[
\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \frac{d^2 f}{d u^2} \cdot 4(x^2 + y^2) + 4 \frac{d f}{d u}
\]
Since \( u = x^2 + y^2 \), this becomes:
\[
\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 4u \frac{d^2 f}{d u^2} + 4 \frac{d f}{d u}
\]

Thus, the PDE is:
\[
\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 4 \left( u \frac{d^2 f}{d u^2} + \frac{d f}{d u} \right)
\]

This is the partial differential equation after eliminating the arbitrary function \( f \).

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Do you want more details or have any questions? Here are 5 related questions for further exploration:

1. How can this PDE be solved for a specific form of the function \( f(u) \)?
2. What is the physical significance of \( z = f(x^2 + y^2) \) in various applications like wave equations?
3. How do you derive second-order PDEs from other types of arbitrary functions?
4. Can this method be extended to functions of more variables, like \( f(x^2 + y^2 + z^2) \)?
5. What are the boundary conditions that could be applied to this PDE?

**Tip**: When dealing with arbitrary functions, substitution with new variables often simplifies differentiation steps.

Form the PDE by eliminating the arbitrary function from z = f(x^2 + y^2).

Explore how to form a partial differential equation (PDE) by eliminating the arbitrary function from z = f(x² + y²). This detailed guide explains the use of the chain rule and second-order derivatives in solving PDEs, suitable for undergraduate students.

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