Numerical methods for ODEs

Runge-Kutta method
The formula for the fourth order Runge-Kutta method (RK4) is given below. Consider the
problem (
y0 = f(t; y)
y(t0) = 
Define h to be the time step size and ti = t0 + ih. Then the following formula
w0 = 
k1 = hf(ti; wi)
k2 = hf

ti +
h
2
; wi +
k1
2

k3 = hf

ti +
h
2
; wi +
k2
2

k4 = hf(ti + h; wi + k3)
wi+1 = wi +
1
6
(k1 + 2k2 + 2k3 + k4)
computes an approximate solution, that is wi  y(ti).
Let us look at an example: (
y0 = y ?? t2 + 1
y(0) = 0:5
The exact solution for this problem is y = t2 + 2t + 1 ?? 1
2et, and we are interested in the value of
y for 0  t  2.
1. We first solve this problem using RK4 with h = 0:5. From t = 0 to t = 2 with step size
h = 0:5, it takes 4 steps: t0 = 0, t1 = 0:5, t2 = 1, t3 = 1:5, t4 = 2.
Step 0 t0 = 0, w0 = 0:5.
Step 1 t1 = 0:5
k1 = hf(t0;w0) = 0:5f(0; 0:5) = 0:75
k2 = hf(t0 + h=2;w0 + k1=2) = 0:5f(0:25; 0:875) = 0:90625
K3 = hf(t0 + h=2;w0 + k2=2) = 0:5f(0:25; 0:953125) = 0:9453125
K4 = hf(t0 + h;w0 + K3) = 0:5f(0:5; 1:4453125) = 1:09765625
w1 = w0 + (k1 + 2k2 + 2k3 + k4)=6 = 1:425130208333333
Step 2 t2 = 1
k1 = hf(t1;w1) = 0:5f(0:5; 1:425130208333333) = 1:087565104166667
k2 = hf(t1 + h=2;w1 + k1=2) = 0:5f(0:75; 1:968912760416667) = 1:203206380208333
K3 = hf(t1 + h=2;w1 + k2=2) = 0:5f(0:75; 2:0267333984375) = 1:23211669921875
K4 = hf(t1 + h;w1 + K3) = 0:5f(1; 2:657246907552083) = 1:328623453776042
w2 = w1 + (k1 + 2k2 + 2k3 + k4)=6 = 2:639602661132812