Recursion relation of fourth order Runge-Kutta method applied on system I'm trying to apply the Gauss-Legendre method of fourth order (as Runge-Kutta method) on the following system of equations
$$\left\{ \begin{matrix}
              \dot{a} =& -b \\ 
              \dot{b} =& a \\
         \end{matrix} 
\right.$$
and I want the result to be a recurrence relation, i.e. of the form
$$\begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix}= \begin{pmatrix} d_1 & d_2 \\ d_3 & d_4 \end{pmatrix} \cdot \begin{pmatrix} a_{n} \\ b_{n} \end{pmatrix}$$
where I still have to find the $d$ elements. The problem I'm struggling with is that the $c$-elements of the butcher tableau (from the Gauss-Legendre method of order 4), are $c_1 = \frac{1}{2}-\frac{\sqrt{3}}{6}$ and $c_2=\frac{1}{2}+\frac{\sqrt{3}}{6}$, which will result in a different discretization step from the $a_{n+1} = a_n + h$ and $b_{n+1} = b_n + h$ of the recurrence relation. Any tips and help are welcome!
Edit: the butcher tableau for the 2-steps Gauss-Legendre method (fourth order) is given by
\begin{array}{c|ccc}
\frac{1}{2}-\frac{\sqrt{3}}{6} & \frac{1}{4} & \frac{1}{4}-\frac{\sqrt{3}}{6}\\
\frac{1}{2}+\frac{\sqrt{3}}{6} & \frac{1}{4}+\frac{\sqrt{3}}{6} & \frac{1}{4} \\
\hline
& \frac{1}{2} & \frac{1}{2}
\end{array}
Added: these are the equations I've got at the moment
$$
a_{n+1} = a_n + \frac{1}{2}hk_1 + \frac{1}{2}hk_2 \\
\, \, k_1 = f(t_n + (\frac{1}{2} - \frac{\sqrt{3}}{6})h, a_n + \frac{1}{4}k_1 + (\frac{1}{4}-\frac{\sqrt{3}}{6})k_2) \\
\, \, k_2 = f(t_n + (\frac{1}{2} + \frac{\sqrt{3}}{6})h, a_n + (\frac{1}{4}+\frac{\sqrt{3}}{6})k_1 + \frac{1}{4}k_2)
$$
The same holds for the $b$ part (with $g$ instead of $f$).
 A: For those who are interested, I've found the solution by making use of an alternative definition of a Runge-Kutta method. 
We have two functions
$$
 f_1(a,b) = -b \\
 f_2(a,b) = \,\,\,\,  a \\
$$
and the Runge-Kutta relation
$$
 Y_i = y_n + h \sum_{i=1}^s a_{i,j} f(Y_i)
$$
By making use of this relation, we obtain
$$
\begin{cases}
A_1 = a_0 + h\left(-\frac{1}{4}B_1 -\left(\frac{1}{4}-\frac{\sqrt{3}}{6}\right)B_2\right) \\
B_1 = b_0 + h \left( \frac{1}{4}A_1 + \left( \frac{1}{4}-\frac{\sqrt{3}}{6}\right)A_2 \right) \\
A_2 = a_0 + h\left(-\left(\frac{1}{4}+\frac{\sqrt{3}}{6}\right)B_1 -\frac{1}{4}B_2\right) \\
B_2 = b_0 + h \left( \left(\frac{1}{4} + \frac{\sqrt{3}}{6}\right)A_1 + \frac{1}{4}A_2 \right)
\end{cases}
$$
This is a set of equations in function of $a_0$ and $b_0$, which we can solve by making use of  computer software. Assume we have found $A_1$, $B_1$, $A_2$ and $B_2$ in terms of $a_0$ and $b_0$, then we can use the next equations
$$
a_{n+1} = a_n + h \left( b_1 \cdot f_1(A_1,B_1) + b_2 \cdot f_1(A_2, B_2) \right) \\
b_{n+1} = b_n + h \left( b_1 \cdot f_2(A_1,B_1) + b_2 \cdot f_2(A_2, B_2) \right) \\
$$
where the $b_1$'s on the right hand side are the Runge-Kutta coefficients ($b_1 = b_2 = \frac{1}{2}$)
Substitution the relations for $A_1$, $B_1$, $A_2$, $B_2$ and writing this out in matrix notation results in 
$$
\begin{bmatrix}
a_{n+1} \\
b_{n+1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{h^4-60h^2+144}{h^4+12h^2+144} & \frac{12h(h^2-12)}{h^4+12h^2+144} \\
-\frac{12h(h^2-12)}{h^4+12h^2+144} & \frac{h^4-60h^2+144}{h^4+12h^2+144}
\end{bmatrix}
\cdot
\begin{bmatrix}
a_n \\
b_n
\end{bmatrix}
$$
