## 5 Numerical Results  
In this section, the methods discussed on Sections 4.1 and 4.2 were tested on six problems. The numerical results for the proposed methods are compared with other existing implicit RK methods of the same order. The methods chosen in the numerical experiments are as follows:  
* DIRKT6: The new sixth-order four-stage diagonally implicit Runge-Kutta type method which was derived in this paper.
* DIRKT5: The new fifth-order three-stage diagonally implicit Runge-Kutta type method which was derived in this paper.
* RKRI5: The fifth-order three-stage implicit Runge-Kutta Radau I method given by [PERSON] [18].
* RKRIIA5: The fifth-order three-stage implicit Runge-Kutta Radau IIA method as given by [PERSON] [19].
* DIRK5: The five-stage diagonally implicit Runge-Kutta method of order five given by [PERSON] et al. [20].
* RKLIMIC6: The sixth-order four-stage implicit Runge-Kutta Lobatto IIIC method as given by [PERSON] [18].  
**Problem 1**.: _Consider the homogeneous linear problem given in [PERSON] et al. [17]_  
\[u^{(4)}(t)=-4\,u(t),\quad u(0)=0,\quad u^{\prime}(0)=1,\quad u^{\prime\prime}(0 )=2,\quad u^{\prime\prime\prime}(0)=2.\]  
The exact solution is \(u(t)=e^{t}\,\sin(t),\quad 0\leq t\leq 5\).  
**Problem 2**.: _Consider the homogeneous linear problem with non constant coefficients given in [PERSON] et al. [17]_  
\[u^{(4)}(t)=(16\,t^{4}-48\,t^{2}+12)u(t),\quad u(0)=1,\quad u^{\prime}(0)=0, \quad u^{\prime\prime}(0)=-2,\quad u^{\prime\prime\prime}(0)=0.\]  
The exact solution is \(u(t)=\,e^{-t^{2}},\quad 0\leq t\leq 3\).  
**Problem 3**.: _Consider non linear problem given in [PERSON] et al. [17]_  
\[u^{(4)}(t)=\frac{3\sin(u)(3+2\sin^{2}(u))}{\cos^{7}(u)},\quad u(0)=0,\quad u^{ \prime}(0)=1,\quad u^{\prime\prime}(0)=0,\quad u^{\prime\prime\prime}(0)=1.\]  
The exact solution is \(u(t)=\arcsin(t),\quad 0\leq t\leq\frac{\pi}{4}\).  
\begin{table}
\begin{tabular}{c|c c c} \hline \(\frac{1}{2}-\frac{\sqrt{5}}{10}\) & \(\frac{1}{225}\) & & \\ \(\frac{1}{2}+\frac{\sqrt{5}}{10}\) & 0 & \(\frac{1}{225}\) & \\
0 & 0 & \(-\frac{1}{100}\) & \(\frac{1}{225}\) & \\
1 & 0 & \(-\frac{1}{30}\) & \(\frac{17}{300}-\frac{\sqrt{5}}{100}\) & \(\frac{1}{225}\) \\ \hline  & \(\frac{1}{72}+\frac{\sqrt{5}}{180}\) & \(\frac{1}{72}-\frac{\sqrt{5}}{180}\) & \(\frac{1}{72}\) & 0 \\  & \(\frac{1}{16}+\frac{\sqrt{5}}{48}\) & \(\frac{1}{16}-\frac{\sqrt{5}}{48}\) & \(\frac{1}{24}\) & 0 \\  & \(\frac{5}{24}+\frac{\sqrt{5}}{24}\) & \(\frac{5}{24}-\frac{\sqrt{5}}{24}\) & \(\frac{1}{12}\) & 0 \\  & \(\frac{5}{12}\) & \(\frac{5}{12}\) & \(\frac{1}{12}\) & \(\frac{1}{12}\) \\ \hline \end{tabular}
\end{table}
Table 3: The DIRKT6 Method.  
**Problem 4**.: _Consider the linear system homogeneous given in [PERSON] et al. [17]_  
\[u_{1}^{(4)}(t) =\,e^{3\,t}u_{4}(t), u_{1}(0) =1, u_{1}^{\prime}(0) =-1, u_{1}^{\prime\prime}(0) =1, u_{1}^{\prime\prime\prime}(0) =-1,\] \[u_{2}^{(4)}(t) =16\,e^{-t}u_{1}(t), u_{2}(0) =1, u_{2}^{\prime}(0) =-2, u_{2}^{\prime\prime}(0) =4, u_{2}^{\prime\prime\prime}(0) =-8,\] \[u_{3}^{(4)}(t) =81\,e^{-t}u_{2}(t), u_{3}(0) =1, u_{3}^{\prime}(0) =-3, u_{3}^{\prime\prime}(0) =9, u_{3}^{\prime\prime\prime}(0) =-27,\] \[u_{4}^{(4)}(t) =256\,e^{-t}u_{3}(t), u_{4}(0) =1, u_{4}^{\prime}(0) =-4, u_{4}^{\prime\prime}(0) =16, u_{4}^{\prime\prime\prime}(0) =-64,\]  
The exact solution are  
\[u_{1}(t) =\,e^{-t}, u_{2}(t) =\,e^{-2\,t}, u_{3}(t) =\,e^{-3\,t}, u_{4}(t) =\,e^{-4\,t},0\leq t\leq 1.\]  
**Problem 5**.: _Application to Problem the Ill-Posed Problem of a Beam on Elastic Foundation given in [PERSON] et al. [21]_  
\[u^{(4)}(t) =-u(t)+1, u(0) =1, u^{\prime}(0) =0, u^{\prime\prime}(0) =0, u^{\prime\prime\prime}(0) =0, u^{\prime\prime\prime}(0) =0,\]  
The exact solution is \(u(t) =1-\frac{1}{2}\,e^{-\frac{t}{\sqrt{2}}}\,(1+\,e^{\sqrt{2}t})\,\cos (\frac{t}{\sqrt{2}}), \quad 0<t<1.\)