Here is a simple example of comparing different numerical methods for ODEs.

We focus on the example subject to the initial condition . We know that the exact solution is .

Let’s compare Euler’s method, the trapezoid method, and the Runge-Kutta algorithm with a step size of , evolving up to time .

The result is the following picture:

Comparing numerical methods. Dashed line is exact solution; pink is Runge-Kutta algorithm, purple is trapezoid method, red is Euler’s method.

Here is the code which I used to obtain the picture:

f[x_] := x^2;
dt = .3;
n = 3;

solutioncurve = 
  Plot[(1 - t)^(-1), {t, 0, dt*n}, PlotStyle -> {Thick, Dashed}, 
   PlotRange -> All];

evalues = RecurrenceTable[{
   t[0] == 0,
   x[0] == 1,
   t[k + 1] == t[k] + dt,
   x[k + 1] == x[k] + f[x[k]]*dt},
  {t, x},
  {k, 0, n}];
eplot = ListLinePlot[evalues, PlotStyle -> Red, 
   PlotMarkers -> Automatic];

trapvalues = RecurrenceTable[
   {t[0] == 0,
    x[0] == 1,
    t[k + 1] == t[k] + dt,
    x[k + 1] == x[k] + .5*dt*(f[x[k]] + f[x[k] + f[x[k]]*dt])},
   {t, x},
   {k, 0, n}];
trapplot = 
  ListLinePlot[trapvalues, PlotStyle -> Purple, 
   PlotMarkers -> Automatic];

rkvalues = RecurrenceTable[
   {t[0] == 0,
    x[0] == 1,
    t[k + 1] == t[k] + dt,
    x[k + 1] == x[k] + dt*(
        (1/6)*(f[x[k]])
         + (1/3)*(f[x[k] + .5*dt*f[x[k]]])
         + (1/3)*f[x[k] + .5*dt*(f[x[k] + .5*dt*f[x[k]]])]
         + (1/6)*
          f[x[k] + dt*(f[x[k] + .5*dt*(f[x[k] + .5*dt*f[x[k]]])])]
        )
    },
   {t, x},
   {k, 0, n}];
rkplot = ListLinePlot[rkvalues, PlotStyle -> Pink, 
   PlotMarkers -> Automatic];

Show[solutioncurve, eplot, trapplot, rkplot]