Editing Runge–Kutta methods (section)

==Use==
As an example, consider the two-stage second-order Runge–Kutta method with α = 2/3, also known as [[Heun's method#Runge.E2.80.93Kutta method|Ralston method]]. It is given by the tableau

{| cellspacing="0" cellpadding="3"
| width="20" |  || style="border-right:1px solid;" | 0
|-
||| style="border-right:1px solid; border-bottom:1px solid;" | 2/3 || style="border-bottom:1px solid;" | 2/3
| style="border-bottom:1px solid;" |
|-
||| style="border-right:1px solid;" | || 1/4 || 3/4
|}

with the corresponding equations

:<math> \begin{align}
k_1 &= f(t_n,\ y_n), \\
k_2 &= f(t_n + \tfrac{2}{3}h,\ y_n + \tfrac{2}{3}h k_1), \\
y_{n+1} &= y_n + h\left(\tfrac{1}{4}k_1+\tfrac{3}{4}k_2\right).
\end{align} </math>

This method is used to solve the initial-value problem
:<math> \frac{dy}{dt} = \tan(y)+1,\quad y_0=1,\ t\in [1, 1.1]</math>
with step size ''h'' = 0.025, so the method needs to take four steps.

The method proceeds as follows:

{| cellpadding="8"
|-
| <math>t_0=1 \colon</math> || ||
|-
| || <math>y_0=1</math>
|-
| <math>t_1=1.025 \colon</math>
|-
| || <math>y_0 = 1</math> || <math>k_1=2.557407725</math> || <math>k_2 = f(t_0 + \tfrac23h ,\ y_0 + \tfrac23hk_1) = 2.7138981400</math>
|-
| ||  colspan=3 |<math>y_1=y_0+h(\tfrac14k_1 + \tfrac34k_2)=\underline{1.066869388}</math>
|-
| <math>t_2=1.05 \colon</math>
|-
| || <math>y_1 = 1.066869388</math> || <math>k_1=2.813524695</math> || <math>k_2 = f(t_1 + \tfrac23h ,\ y_1 + \tfrac23hk_1)</math>
|-
| ||  colspan=3 | <math>y_2=y_1+h(\tfrac14k_1 + \tfrac34k_2)=\underline{1.141332181}</math>
|-
| <math>t_3=1.075 \colon</math>
|-
| || <math>y_2 = 1.141332181</math> || <math>k_1=3.183536647</math> || <math>k_2 = f(t_2 + \tfrac23h ,\ y_2 + \tfrac23hk_1)</math>
|-
| ||  colspan=3 | <math>y_3=y_2+h(\tfrac14k_1 + \tfrac34k_2)=\underline{1.227417567}</math>
|-
| <math>t_4=1.1 \colon</math>
|-
| || <math>y_3 = 1.227417567 </math> || <math>k_1=3.796866512</math> || <math>k_2 = f(t_3 + \tfrac23h ,\ y_3 + \tfrac23hk_1)</math>
|-
| ||  colspan=3 | <math>y_4=y_3+h(\tfrac14k_1 + \tfrac34k_2)=\underline{1.335079087}.</math>
|}

The numerical solutions correspond to the underlined values.