The problem defines a harmonic function as a function of two variables that satisfies Laplace's equation, which is given by $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$. We need to show that the functions $f(x, y) = x^3y - xy^3$ and $f(x, y) = \ln(4x^2 + 4y^2)$ are harmonic functions.

1. Problem Description

The problem defines a harmonic function as a function of two variables that satisfies Laplace's equation, which is given by $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$. We need to show that the functions $f(x, y) = x^3y - xy^3$ and $f(x, y) = \ln(4x^2 + 4y^2)$ are harmonic functions.

2. Solution Steps

For function 33, $f(x, y) = x^3y - xy^3$:

First, we find the first partial derivatives with respect to $x$ and $y$:

$\frac{\partial f}{\partial x} = 3x^2y - y^3$

$\frac{\partial f}{\partial y} = x^3 - 3xy^2$

Next, we find the second partial derivatives with respect to $x$ and $y$:

$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (3x^2y - y^3) = 6xy$

$\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (x^3 - 3xy^2) = -6xy$

Now, we check if Laplace's equation holds:

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 6xy + (-6xy) = 0$

Thus, $f(x, y) = x^3y - xy^3$ is a harmonic function.

For function 34, $f(x, y) = \ln(4x^2 + 4y^2)$:

First, we find the first partial derivatives with respect to $x$ and $y$:

$\frac{\partial f}{\partial x} = \frac{8x}{4x^2 + 4y^2} = \frac{2x}{x^2 + y^2}$

$\frac{\partial f}{\partial y} = \frac{8y}{4x^2 + 4y^2} = \frac{2y}{x^2 + y^2}$

Next, we find the second partial derivatives with respect to $x$ and $y$:

$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (\frac{2x}{x^2 + y^2}) = \frac{2(x^2 + y^2) - 2x(2x)}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4x^2}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2}$

$\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (\frac{2y}{x^2 + y^2}) = \frac{2(x^2 + y^2) - 2y(2y)}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4y^2}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2}$

Now, we check if Laplace's equation holds:

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2} + \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2 + 2x^2 - 2y^2}{(x^2 + y^2)^2} = \frac{0}{(x^2 + y^2)^2} = 0$

Thus, $f(x, y) = \ln(4x^2 + 4y^2)$ is a harmonic function.

3. Final Answer

Both functions, $f(x, y) = x^3y - xy^3$ and $f(x, y) = \ln(4x^2 + 4y^2)$, satisfy Laplace's equation and are therefore harmonic functions.