How to get the RadauIIA 5th order in terms of its K(i), just like the RK4 Classic?
\begin{align} k_1 &= f(t_{n},y_{n}) \\ k_2 &= f(t_{n}+ \frac{h}{2},y_{n}+h\frac{k_1}{2}) \\ k_3 &= f(t_{n}+ \frac{h}{2},y_{n}+h\frac{k_2}{2}) \\ k_4 &= f(t_{n}+ h,y_{n}+hk_3) \\\hline y_{n+1} &= y_{n} + \frac{1}{6}h(k_1+2k_2+2k_3+k_4) \end{align} Or for example the implicit Trapezium is given by the following formula:
$$ y_{n+1} = y_{n} + \frac{1}{2}h(f(t_{n},y_{n}) + f(t_{n+1},y_{n+1})) $$
Is the Radau IIA 5th order as such:
\begin{align} k_1 &= h * f(t_{n} + c1 * h, y_{n} + (a11*k1 + a12*k2 + a13*k3)) \\ k_2 &= h * f(t_{n} + c2 * h, y_{n} + (a21*k1 + a22*k2 + a23*k3)) \\ k_3 &= h * f(t_{n} + c3 * h, y_{n} + (a31*k1 + a32*k2 + a33*k3)) \\\hline y_{n+1} &= y_{n} + (b1*k1 + b2*k2 + b3*k3) \end{align}
The Butcher tableau for Radau IIA as per Hairer-Wanner: "Solving ODE II: Stiff & DAE", 2nd ed., page 74 is
\begin{array}{c|ccc} \frac{4-\sqrt{6}}{10}&\frac{88-7\sqrt6}{360}& \frac{296-169\sqrt6}{1800}& \frac{-2+3\sqrt6}{225} \\ \frac{4+\sqrt{6}}{10}& \frac{296+169\sqrt6}{1800}&\frac{88+7\sqrt6}{360}& \frac{-2-3\sqrt6}{225} \\ 1&\frac{16-\sqrt6}{36}&\frac{16+\sqrt6}{36} &\frac19 \\\hline &\frac{16-\sqrt6}{36}&\frac{16+\sqrt6}{36} &\frac19 \end{array} So it is a fully implicit method, and the point of the last stage is also the next point of the solution approximation.