Runge Kutta Method using quadrature formula 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. 

*

*I do not see what you are computing. Obviously the quadrature method underlying the method is the Simpson method, this fixes both the $b_k$ as also the $c_k$. The computation of $k_2$ is by the implicit trapezoidal method, $k_2=y'(x+\frac12h)+O(h^2)$. $k_3$ is computed by the midpoint method, so that $k_3=y'(x+h)+O(h^2)$. Up to this point this gives second order. The remaining condition for order $3$ is (see for instance the first and second set of slides from J. Butcher's tutorials)
$$
\frac16=\sum b_ia_{ij}c_j=b_2a_{22}c_2+b_3a_{32}c_2
\\
\implies \frac23\frac14\frac12+\frac16 a_{32}\frac12=\frac16
\\
\implies a_{32}=1\implies a_{31}=0
$$
Now one could also check the 4th order conditions
$$
\frac18=\sum b_ic_ia_{ij}c_j=b_2c_2a_{22}c_2+b_3c_3a_{32}c_2=\frac1{24}+\frac1{12}
\\
\frac1{12}=\sum b_ia_{ij}c_j^2=(b_2a_{22}+b_3a_{32})c_2^2=\frac13\frac14
\\
\frac1{24}=\sum b_ia_{ij}a_{jm}c_m=(b_2a_{22}+b_3a_{32})a_{22}c_2=\frac13\frac18
$$
Which confirms that this method is a 4th order method.





*For a linear system $\dot z=f(t,z)=A(t)z+b(t)$ one can implement the method as
\begin{align}
k_1&=A(t)z+b(t)\\
k_2&=(I-\tfrac14hA(t+\tfrac12h))^{-1}\Bigl(A(t+\tfrac12h)(z+\tfrac14hk_1)+b(t+\tfrac12h)\Bigr)\\
k_3&=A(t+h)(z+hk_2)+b(t+h)\\
\hline
z_{+1}&=z+\frac{h}6(k_1+4k_2+k_3)
\end{align}
In the current case, $b(t)=0$, as the right side of the equation is another term that is linear in $y$.

