Define $y : \B{R} \rightarrow \B{R}^n$ by $$y_j (t) = t^{j+1}$$ It follows that the derivative of $y(t)$ satisfies the runge_kutta_4 ODE equation where $y(0) = 0$ and $f(t, y)$ is given by $$f(t , y)_j = y_j '(t) = \left\{ \begin{array}{ll} 1 & {\; \rm if \;} j = 0 \\ (j+1) y_{j-1} (t) & {\; \rm otherwise } \end{array} \right.$$
Source Code  from pycppad import * def pycppad_test_runge_kutta_4_correct() : def fun(t , y) : n = y.size f = numpy.zeros(n) f[0] = 1. index = numpy.array( range(n-1) ) + 1 f[index] = (index + 1) * y[index-1] return f n = 5 # size of y(t) (order of method plus 1) ti = 0. # initial time dt = 2. # a very large time step size to test correctness yi = numpy.zeros(n) # initial value for y(t); i.e., y(0) # take one 4-th order Runge-Kutta integration step of size dt yf = runge_kutta_4(fun, ti, yi, dt) # check the results t_jp = 1. # t^0 at t = dt for j in range(n-1) : t_jp = t_jp * dt # t^(j+1) at t = dt assert abs( yf[j] - t_jp ) < 1e-10 # check yf[j] = t^(j+1)