This problem requires us to determine the increment in \(x\) for a given ordinary differential equation using the Runge-Kutta fourth-order method. We are provided with the differential equation, an initial condition, and the step size.

Differential Equation and Initial Conditions

The ordinary differential equation provided is:

\[ \frac{{{\rm{dx}}}}{{{\rm{dt}}}} = 4{\rm{t}} + 4 \]

We can define the function \( f(t, x) \) as \( f(t, x) = 4t + 4 \). It's important to note that this specific function \(f\) depends only on \(t\) and not on \(x\).

The initial condition states that \(x = x_0\) at \(t = 0\), which means our starting time is \(t_0 = 0\).

The given step size is \( \Delta t = h = 0.2 \).

Our goal is to calculate the increment in \(x\) over one step, which is \( x(t_0 + h) - x(t_0) \) or simply \( \Delta x \), by applying the Runge-Kutta fourth-order method.

Runge-Kutta Fourth Order Method

The Runge-Kutta fourth-order (RK4) method is a powerful numerical technique widely used to approximate solutions to ordinary differential equations. For an initial value problem stated as \( \frac{dx}{dt} = f(t, x) \) with an initial value \( x(t_n) = x_n \), the value of \( x_{n+1} \) at the next time step \( t_{n+1} = t_n + h \) is calculated using the formula:

\[ x_{n+1} = x_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \]

The increment in \(x\) over this single step is given by \( \Delta x = x_{n+1} - x_n \), which simplifies to:

\[ \Delta x = \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \]

The coefficients \( k_1, k_2, k_3, \) and \( k_4 \) are computed as follows:

• \( k_1 = h \cdot f(t_n, x_n) \)
• \( k_2 = h \cdot f(t_n + \frac{h}{2}, x_n + \frac{k_1}{2}) \)
• \( k_3 = h \cdot f(t_n + \frac{h}{2}, x_n + \frac{k_2}{2}) \)
• \( k_4 = h \cdot f(t_n + h, x_n + k_3) \)

Calculating RK4 Coefficients

Let's apply these formulas using our specific values: \( t_n = t_0 = 0 \), step size \( h = 0.2 \), and the function \( f(t, x) = 4t + 4 \).

1. Calculate \(k_1\):

We use the initial values \(t_0 = 0\) and \(x_0\).

\[ k_1 = h \cdot f(t_0, x_0) \]

Since our function \( f(t, x) \) only depends on \( t \), the value of \( x_0 \) does not influence the evaluation of \( f \).
\[ k_1 = 0.2 \cdot f(0) = 0.2 \cdot (4 \cdot 0 + 4) = 0.2 \cdot 4 = 0.8 \]

2. Calculate \(k_2\):

For \(k_2\), we evaluate \(f\) at \(t_0 + h/2\) and \(x_0 + k_1/2\).

\[ k_2 = h \cdot f(t_0 + \frac{h}{2}, x_0 + \frac{k_1}{2}) \]

First, calculate the time argument: \( t_0 + \frac{h}{2} = 0 + \frac{0.2}{2} = 0.1 \).
As \( f(t, x) \) does not depend on \( x \), the term \( x_0 + \frac{k_1}{2} \) is not needed for evaluating \( f \).
\[ k_2 = 0.2 \cdot f(0.1) = 0.2 \cdot (4 \cdot 0.1 + 4) = 0.2 \cdot (0.4 + 4) = 0.2 \cdot 4.4 = 0.88 \]

3. Calculate \(k_3\):

For \(k_3\), we evaluate \(f\) at \(t_0 + h/2\) and \(x_0 + k_2/2\).

\[ k_3 = h \cdot f(t_0 + \frac{h}{2}, x_0 + \frac{k_2}{2}) \]

The time argument remains the same as for \(k_2\): \( t_0 + \frac{h}{2} = 0.1 \).
Again, since \( f \) is independent of \( x \), the term \( x_0 + \frac{k_2}{2} \) is irrelevant for the evaluation of \( f \).
\[ k_3 = 0.2 \cdot f(0.1) = 0.2 \cdot (4 \cdot 0.1 + 4) = 0.2 \cdot (0.4 + 4) = 0.2 \cdot 4.4 = 0.88 \]

4. Calculate \(k_4\):

For \(k_4\), we evaluate \(f\) at \(t_0 + h\) and \(x_0 + k_3\).

\[ k_4 = h \cdot f(t_0 + h, x_0 + k_3) \]

First, calculate the time argument: \( t_0 + h = 0 + 0.2 = 0.2 \).
The term \( x_0 + k_3 \) is not needed for evaluating \( f \) because \( f \) is independent of \( x \).
\[ k_4 = 0.2 \cdot f(0.2) = 0.2 \cdot (4 \cdot 0.2 + 4) = 0.2 \cdot (0.8 + 4) = 0.2 \cdot 4.8 = 0.96 \]

Increment in x using RK4

Now, we use the main RK4 formula to find the increment in \(x\):

\[ \Delta x = \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \]

Substitute the calculated values of \( k_1, k_2, k_3, \) and \( k_4 \) into the formula:

\[ \Delta x = \frac{1}{6}(0.8 + 2 \cdot 0.88 + 2 \cdot 0.88 + 0.96) \]

Perform the multiplications:

\[ \Delta x = \frac{1}{6}(0.8 + 1.76 + 1.76 + 0.96) \]

Sum the terms inside the parenthesis:

\[ 0.8 + 1.76 + 1.76 + 0.96 = 5.28 \]

Finally, calculate the increment in \(x\):

\[ \Delta x = \frac{1}{6}(5.28) = 0.88 \]

Therefore, the increment in \(x\) calculated using the Runge-Kutta fourth-order method with a step size of \( \Delta t = 0.2 \) is \(0.88\).

Runge-Kutta Fourth Order Exactness

It's worth noting that for differential equations of the form \( \frac{dx}{dt} = f(t) \), where \(f(t)\) is a polynomial, the Runge-Kutta fourth-order method gives an exact solution if \(f(t)\) is a polynomial of degree up to 4. In this problem, \(f(t) = 4t + 4\) is a polynomial of degree 1. Let's briefly check the exact increment by direct integration:

\[ x(t) = \int (4t + 4) \, dt = 2t^2 + 4t + C \]

The increment in \(x\) from \(t=0\) to \(t=0.2\) is:

\[ \Delta x = x(0.2) - x(0) \]

\[ \Delta x = (2(0.2)^2 + 4(0.2) + C) - (2(0)^2 + 4(0) + C) \]

\[ \Delta x = (2(0.04) + 0.8 + C) - (0 + 0 + C) \]

\[ \Delta x = 0.08 + 0.8 \]

\[ \Delta x = 0.88 \]

This confirms that our Runge-Kutta calculation matches the exact increment for this specific type of differential equation.