(8) The solution of the equation
x
∂z
∂x + y
∂z
∂y = 0
is
(i) z = ϕ

|x|
|y|

, ϕ ∈ C
1
(R)
(ii) z = ϕ

x−1
y

, ϕ ∈ C
1
(R)
(iii) z = ϕ

x+1
y

, ϕ ∈ C
1
(R)
(iv) z = ϕ(|x| + |y|), ϕ ∈ C
1
(R)

We are given the partial differential equation:

\[
x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 0
\]

We need to determine which of the given functions is a solution to this PDE. Let's examine the general approach to solving such first-order PDEs using the method of characteristics.

### Step 1: Setting Up the Characteristic Equations

The characteristic equations corresponding to the PDE are:

\[
\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{0}
\]

### Step 2: Solving the Characteristic Equations

From the first two terms:

\[
\frac{dx}{x} = \frac{dy}{y} \implies \frac{d(x/y)}{(x/y)} = 0
\]

This implies that \( \frac{x}{y} = C_1 \), where \( C_1 \) is a constant. Thus, \( x = C_1 y \), or \( \frac{x}{y} \) is constant along the characteristic curves.

The third equation \( \frac{dz}{0} \) implies that \( z \) is constant along these characteristic curves.

Therefore, the general solution of the PDE can be written as:

\[
z = \phi\left(\frac{x}{y}\right)
\]

where \( \phi \) is an arbitrary differentiable function.

### Step 3: Analyzing the Given Options

Let's check each of the provided solutions against this form:

1. **Option (i): \( z = \phi\left(\frac{|x|}{|y|}\right) \)**

   This matches the general solution \( z = \phi\left(\frac{x}{y}\right) \) since \( \frac{|x|}{|y|} \) is still a function of \( \frac{x}{y} \). So, this is a valid solution.

2. **Option (ii): \( z = \phi\left(\frac{1}{x}y\right) \)**

   This can be rewritten as \( z = \phi\left(\frac{y}{x}\right) \), which is equivalent to \( z = \phi\left(\frac{1}{C_1}\right) \) where \( C_1 = \frac{x}{y} \). Therefore, this is also a valid solution.

3. **Option (iii): \( z = \phi\left(\frac{x+1}{y}\right) \)**

   This does **not** match the general form \( \phi\left(\frac{x}{y}\right) \). Adding 1 to \( x \) changes the characteristic form, so this is **not** a solution.

4. **Option (iv): \( z = \phi(|x| + |y|) \)**

   This solution does **not** match the form \( \phi\left(\frac{x}{y}\right) \) or any equivalent characteristic form, so this is **not** a solution.

### Conclusion

The correct solutions to the PDE \( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 0 \) are:

- **(i) \( z = \phi\left(\frac{|x|}{|y|}\right) \)**
- **(ii) \( z = \phi\left(\frac{y}{x}\right) \)**

Thus, the correct options are **(i) and (ii)**.

Would you like any further explanation or have other questions?

---

Here are some related questions to consider:
1. How would the solution change if the equation was not homogeneous?
2. What if the PDE had a source term added, like \( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = f(x, y) \)?
3. Can we generalize this method to systems of first-order PDEs?
4. How does the method of characteristics apply to higher-order PDEs?
5. What are some real-world applications where such PDEs arise?

**Tip:** The method of characteristics is a powerful tool for solving first-order PDEs. Understanding how to derive and solve characteristic equations is crucial in various fields, including fluid dynamics and wave propagation.

Learn how to solve the partial differential equation x ∂z ∂x + y ∂z ∂y = 0 using the method of characteristics. Explore detailed solutions for options (i), (ii), and analysis of options (iii) and (iv). Suitable for advanced undergraduate students studying partial differential equations.

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