I am trying to solve this exercise and have a few questions.

It is given the initial value problem $y^{\prime \prime \prime} (t) + y^{\prime}(t) = e^{-t}y(t)$ with initial values $ y(0) = y_0 \in \mathbb{R}, y^{\prime}(0) = v_0 \in \mathbb{R} \, \text{and} \, y^{\prime \prime}(0) = w0 \in \mathbb{R}. $

Furthermore, the following Runge-Kutta method as a butcher tableau is given:

$$\begin{array} {c|cccc} 0\\ \frac{1}{2} & \frac{1}{4} &\frac{1}{4}\\ 1& 0& 1& 0\\ \hline & \frac{1}{6} &\frac{2}{3} &\frac{1}{6} \end{array} $$

For the numerical solution of the initial value problem, the step width $h>0$ with $n = 0, 1, 2,\dots ,N = \frac{T}{h}$ should be specified. The time steps $t_n$ are derived from the decomposition of the interval $[0, T]$. Here $y_n$ approximates the solution of the initial value problem at time $t_n$.

(i) Rewrite the Butcher tableau in step form and outline the derivation of the Runge Kutta method using known square formulas

(ii) Convert the initial value problem to a system of ordinary first-order differential equations:

$$ z^{\prime} = A z \, \text{with} \, z := \bigg( \begin{array}{c} y \\ v \\w \end{array} \bigg) = \bigg( \begin{array}{c} y \\ y^{\prime} \\y^{\prime \prime} \end{array} \bigg) \in \mathbb{R^3} $$

The matrix $A:= A (t) \in \mathbb{R}^{3 \times 3}$ depends on the time $t$.

Also write a pseudo-code, which generates the matrix $A (t)$ for the time $t$. The routine should only be receive the parameter $t$ and return the final result $A$.

(iii) Apply the step form of the Runga Kutta method to the differential equation.

My solution/idea to the tasks: i) At first I rewrite the Butcher tableau in step form and I get

$k_1=f(t,y(t)) \\ k_2=f(t+\frac{h}{2}, y+\frac{h}{4} (k_1 + k_2)) \\ k_3=f(t+h, y+hk_2) \\ y_{j+1} = y_{j} + \frac{1}{6}(k_1+4k_2+k_3)$

Now i try to compute this Runge Kutta Method using the Quadrature Formula

$l_i(\alpha) = \prod_{j=1}^{s} \frac{\alpha - \alpha_j}{\alpha_i - \alpha_j} \, \text{with} \, i \neq j \\ \beta_{ij} = \int_{0}^{\alpha_i} l_j(\alpha) d\alpha \, \text{and} \, \gamma_j = \int_{0}^{1} l_j(\alpha) d\alpha$

I calculated the values but i get not the same values for $\beta_{ij}$.

$l_1(\alpha) = \frac{\alpha - 0.5}{0-0.5} \frac{\alpha - 1}{0-1} = (1-2\alpha)(1-\alpha) = (1-3\alpha 2 \alpha^2) \\ l_2(\alpha) = ... = 4\alpha - 4\alpha^2 \\ l_3(\alpha)=...= 2\alpha^2 -\alpha$ Then

$\gamma_1 = \int_{0}^{1} l_1(\alpha) d\alpha = \alpha - 1.5 \alpha^2 + 2/3 \alpha^3 \bigg \vert_{0}^{1} = 1/6 \\ \gamma_2 =...= 2 - 4/3 = 2/3 \\ \gamma_3 =...=4/6 - 3/6 = 1/6$

$\beta_{1j} = \int_{0}^{0} l_j(\alpha) d\alpha = 0 \\ \beta_{21} = \int_{0}{0.5} l_1(\alpha) d\alpha = \alpha - 3/2 \alpha^2 + 2/3 \alpha^3 \bigg \vert_{0}^{0.5}=..= 5/24$

This is not equal to the value in the Butcher Tableau.

ii) Converting to a system of ordinary first order differntial equations yields to:

Define $y_1 := y, y_2 := y^{\prime}, y_3 := y^{\prime \prime} $ with $ y_1(0) = y_0, y_2(0) =v_0, y_3(0)=w_0 $. Then we get:

$ y_{3}^{\prime} = e^{-t}y_1(t) - y_2(t) $ and $y_{1}^{\prime} = y_2, y_{2}^{\prime} =y_3$

$ \bigg( \begin{array}{c} y_{1}^{\prime} \\ y_{2}^{\prime} \\ y_{3}^{\prime} \end{array} \bigg ) = \bigg ( \begin{array}{A} 0 &1 &0 \\ 0 &0 &1 \\ e^{-t} &-1 &0 \end{array} \bigg ) \bigg( \begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \end{array} \bigg )$

I have problems writing Pseudo Code, can someone help me?

iii) At the task I have absolutely no idea what to do. Can someone help me with that?

I am trying to solve this exercise

the following Runge-Kutta method as a butcher tableau is given:

$$\begin{array} {c|cccc} 0\\ \frac{1}{2} & \frac{1}{4} &\frac{1}{4}\\ 1& 0& 1& 0\\ \hline & \frac{1}{6} &\frac{2}{3} &\frac{1}{6} \end{array} $$

(i) Rewrite the Butcher tableau in step form and outline the derivation of the Runge Kutta method using known square formulas

(iii) Apply the step form of the Runga Kutta method to the differential equation.