I need to apply a Runge-Kutta method, with a 3 line Butcher Tableau (h=0.5).
I've rewritten the problem in standard form (as required in part(a)) and got the following:
$$\begin{eqnarray} u_1' &=& u_2\\ u_2' &=& xu_2 - \frac1{1+u_1} \end{eqnarray}$$
If that's wrong, please correct me.
Thanks!
The method to use for the task is a third order method introduced by Karl Heun in 1900. He proposed to use chains of Euler-like steps to probe the vector field close to the solution curve over the next step and combine the final values of several chains. This construction gives an easy structure in the analysis of the method via Taylor expansion.
As we do not solve homework problems here, I will demonstrate how to read a Butcher tableau and apply the method to a system of ODE using the less accurate explicit midpoint method (RK2 or modified Euler), $$ \begin{array}{c|cc} 0\\ \frac12&\frac12\\ \hline &0&1 \end{array} \implies \left\{\begin{align} \vec k_1&=F(x_n,\vec u_n)\\ \vec k_2&=F(x_n+\tfrac12h,\vec u_n+\tfrac12h\vec k_1)\\ \vec u_{n+1}&=\vec u_n+h\vec k_2 \end{align}\right. $$ Splitting the system into its components you would have to compute
\begin{align} k_{11}&=f_1(x_0,u_0)&&=u_{02}&&=1,\\ k_{12}&=f_2(x_0,u_0)&&=x_0u_{02}-\frac1{1+u_{01}}&&=0.5, \\[0.8em] k_{21}&=f_1(x_0+0.5h, u_0+0.5hk_1)&&=u_{02}+0.25k_{12}&&=1.125,\\ k_{22}&=f_2(x_0+0.5h, u_0+0.5hk_1)&&=~...&&=0.961805555...,\\[0.5em] \text{so that at }x_1&=x_0+h=1.5& \\[0.8em] u_{11}&=u_{01}+hk_{21}&&&&=1.5625, \\ u_{12}&=u_{02}+hk_{22}&&&&=1.480902777... \end{align}
For your task you have just to apply this same computation to the given 3-stage method 3rd order method.
Going back to the midpoint method, to get better accuracy one can reduce the step size. Thus compute 5 steps with step size $h=0.1$, the intermediate values are \begin{align} x_0= 1.000, ~~ u_0&=[1.0, 1.0]\\[1em] k_1 &= [1.0, 0.5]\\ k_2&=[1.025, 0.5884451219512195]\\ x_1= 1.100, ~~ u_1&=[1.1025, 1.058844512195122]\\[1em] k_1 &= [1.058844512195122, 0.6891047065775355]\\ k_2&=[1.0932997475239987, 0.7933527917860186]\\ x_2= 1.200, ~~ u_2&=[1.2118299747524, 1.138179791373724]\\[1em] k_1 &= [1.138179791373724, 0.9137014319048803]\\ k_2&=[1.1838648629689679, 1.0390575848336334]\\ x_3= 1.300, ~~ u_3&=[1.3302164610492968, 1.2420855498570873]\\[1em] k_1 &= [1.2420855498570873, 1.1855665337446706]\\ k_2&=[1.3013638765443207, 1.3388370820146442]\\ x_4= 1.400, ~~ u_4&=[1.460352848703729, 1.3759692580585516]\\[1em] k_1 &= [1.3759692580585516, 1.519911194559178]\\ k_2&=[1.4519648177865105, 1.709959435386379]\\ x_5= 1.500, ~~ u_5&=[1.60554933048238, 1.5469652015971895] \end{align} The result you compute with the 3rd order method should be somewhere in-between these, but more to the side of the second computation.
