The prompt is to solve the differential equation using the RADAU IA $3^{rd}$ order using the butcher table.
0 | 1/4 -1/4
(2/3) | (1/4) 5/12
--------------------------------
| (1/4) (3/4)
$$4y^{''}+4y'+5y=0$$
$$t\in[0,10]$$ and $$y(0)=4 $$ $$y'(0)=-2$$
how do I proceed with arriving at the matrices for solving the equation.
$$y_n = y_{n-1}+h(\frac{3}{4}f_1 + \frac{1}{4}f_2)$$ with
$$f_1=f(t_{n-1},y_{n-1}+h(\frac{1}{4}f_1-\frac{1}{4}f_2))$$ $$f_2=f(t_{n-1}+\frac{2}{3}h,y_{n-1}+h(\frac{1}{4}f_1+\frac{5}{12}f_2))$$
having arrived at those equations i'm a little lost on how to proceed further. Could anyone guide me please?
Let's separate the problem and the method by calling the state of the method $u$.
In general, no direct solution is available, so one needs to compute iteratively improving approximations. In any case, that needs initial guesses for the variables $k_1,k_2$, either extrapolated from the previous step or just $k_1=k_2=f(t,u)$.
The most simple, but also slowest and most restricted method is then to directly iterate \begin{align} k_1^{m+1}&=f(t, u+\tfrac14h(k_1^m-k_2^m))\\ k_2^{m+1}&=f(t+\tfrac23h, u+\tfrac1{12}h(3k_1^m+5k_2^m)) \end{align} Using a Lipschitz constant $L$ of $f$ one can find out for what values of $Lh$ this is a contraction.
Having some approximation $J$ of the Jacobian of $f$ at $(t,u)$ one can establish a Newton-like method in solving \begin{align} k_1+Δk_1&=f(t, u+\tfrac14h(k_1-k_2))+\tfrac14hJ(Δk_1-Δk_2)\\ k_2+Δk_2&=f(t+\tfrac23h, u+\tfrac1{12}h(3k_1+5k_2))+\tfrac1{12}hJ(3Δk_1+5Δk_2) \end{align} This now is a linear system for $(Δk_1,Δk_2)$. The iteration converges in general faster and for a larger range of values $Lh$.
However, for the given equation you get a linear first-order system $u'=Au$. When inserted into the step equations, this gives a linear system \begin{align} k_1&=Au+\tfrac14hA(k_1-k_2))\\ k_2&=Au+\tfrac1{12}hA(3k_1+5k_2)) \end{align} which can be directly solved for $k_1,k_2$.