# Interpreting a Butcher Table

$y_{{i+1}} = y_{{i}} + \alpha f_i + \beta f_{i-1}$

where, if we have 3 time points that are not equally spaced: $t_{i-1}, t_i$ and $t_{i+1}$, then

$t_i - t_{i-1} = h_{prev}, \space \space t_{i+1} - t_i = h$

I am asked to calculate an alternative $y^*_{i+1}$, given a $y_i$, using the third order Runge-Kutta method defined by the Butcher table:

    0    |
(1/2)  | (1/2)
1    |  -1      2
--------------------------------
|  (1/6)  (2/3)  (1/6)


I have no idea if I wrote the explicit formula for the Runge-Kutta method correctly, however. I dont have experience working with Butcher tables which is why I'm not confident.

Here's what I got:

$y_{i+1} = y_i + h/6 (k_1 + 4k_2 +k_3)$ where

$k_1 = f(t_i,y_i)$

$k_2 = f(t_i + h/2, \space y_i + \frac{h}{2} k_1)$

$k_3 = f(t_i + h/2, \space y_i + \frac{h}{2} k_2)$

I'm hoping someone can show me if/where I went wrong. Part of what's throwing me off is that we're dealing with an adaptive step size, but I'm not sure either way. I'd really appreciate the help.

Up to $k_2$ it is correct, however, you have to read the third line as $$k_3 = f(t_i+h, y_i-h·k_1+2h·k_2).$$