Runge-Kutta method: Difference between revisions

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180 LET Y=Y+.1*(K1+2*(K2+K3)+K4)/6
180 LET Y=Y+.1*(K1+2*(K2+K3)+K4)/6
190 NEXT</lang>
190 NEXT</lang>

=={{header|C}}==
<lang c>#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double rk4(double(*f)(double, double), double dx, double x, double y)
{
double k1 = dx * f(x, y),
k2 = dx * f(x + dx / 2, y + k1 / 2),
k3 = dx * f(x + dx / 2, y + k2 / 2),
k4 = dx * f(x + dx, y + k3);
return y + (k1 + 2 * k2 + 2 * k3 + k4) / 6;
}

double rate(double x, double y)
{
return x * sqrt(y);
}

int main(void)
{
double *y, x, y2;
double x0 = 0, x1 = 10, dx = .1;
int i, n = 1 + (x1 - x0)/dx;
y = (double *)malloc(sizeof(double) * n);

for (y[0] = 1, i = 1; i < n; i++)
y[i] = rk4(rate, dx, x0 + dx * (i - 1), y[i-1]);

printf("x\ty\trel. err.\n------------\n");
for (i = 0; i < n; i += 10) {
x = x0 + dx * i;
y2 = pow(x * x / 4 + 1, 2);
printf("%g\t%g\t%g\n", x, y[i], y[i]/y2 - 1);
}

return 0;
}</lang>
{{out}} (errors are relative)
<pre>
x y rel. err.
------------
0 1 0
1 1.5625 -9.3262e-08
2 4 -2.2987e-07
3 10.5625 -2.75462e-07
4 25 -2.49396e-07
5 52.5625 -2.05844e-07
6 100 -1.65946e-07
7 175.562 -1.33956e-07
8 289 -1.09222e-07
9 451.562 -9.01828e-08
10 676 -7.54191e-08
</pre>


=={{header|C#}}==
=={{header|C#}}==

Revision as of 16:06, 29 October 2018

Task
Runge-Kutta method
You are encouraged to solve this task according to the task description, using any language you may know.

Given the example Differential equation:

With initial condition:

and

This equation has an exact solution:

Task

Demonstrate the commonly used explicit   fourth-order Runge–Kutta method   to solve the above differential equation.

  • Solve the given differential equation over the range with a step value of (101 total points, the first being given)
  • Print the calculated values of at whole numbered 's () along with error as compared to the exact solution.
Method summary

Starting with a given and calculate:

then:

Ada

<lang Ada>with Ada.Text_IO; use Ada.Text_IO; with Ada.Numerics.Generic_Elementary_Functions; procedure RungeKutta is 

  type Floaty is digits 15;
  type Floaty_Array is array (Natural range <>) of Floaty;
  package FIO is new Ada.Text_IO.Float_IO(Floaty); use FIO;
  type Derivative is access function(t, y : Floaty) return Floaty;
  package Math is new Ada.Numerics.Generic_Elementary_Functions (Floaty);
  function calc_err (t, calc : Floaty) return Floaty;
  
  procedure Runge (yp_func : Derivative; t, y : in out Floaty_Array;
                   dt : Floaty) is
     dy1, dy2, dy3, dy4 : Floaty;
  begin
     for n in t'First .. t'Last-1 loop
        dy1 := dt * yp_func(t(n), y(n));
        dy2 := dt * yp_func(t(n) + dt / 2.0, y(n) + dy1 / 2.0);
        dy3 := dt * yp_func(t(n) + dt / 2.0, y(n) + dy2 / 2.0);
        dy4 := dt * yp_func(t(n) + dt, y(n) + dy3);
        t(n+1) := t(n) + dt;
        y(n+1) := y(n) + (dy1 + 2.0 * (dy2 + dy3) + dy4) / 6.0;
     end loop;
  end Runge;
  
  procedure Print (t, y : Floaty_Array; modnum : Positive) is begin
     for i in t'Range loop
        if i mod modnum = 0 then
           Put("y(");   Put (t(i), Exp=>0, Fore=>0, Aft=>1);
           Put(") = "); Put (y(i), Exp=>0, Fore=>0, Aft=>8);
           Put(" Error:"); Put (calc_err(t(i),y(i)), Aft=>5);
           New_Line;
        end if;
     end loop;
  end Print;

  function yprime (t, y : Floaty) return Floaty is begin
     return t * Math.Sqrt (y);
  end yprime;
  function calc_err (t, calc : Floaty) return Floaty is
     actual : constant Floaty := (t**2 + 4.0)**2 / 16.0;
  begin return abs(actual-calc);
  end calc_err;   
  
  dt : constant Floaty := 0.10;
  N : constant Positive := 100;
  t_arr, y_arr : Floaty_Array(0 .. N);

begin 

  t_arr(0) := 0.0;
  y_arr(0) := 1.0;
  Runge (yprime'Access, t_arr, y_arr, dt);
  Print (t_arr, y_arr, 10);

end RungeKutta;</lang>

Output:
y(0.0) = 1.00000000 Error: 0.00000E+00
y(1.0) = 1.56249985 Error: 1.45722E-07
y(2.0) = 3.99999908 Error: 9.19479E-07
y(3.0) = 10.56249709 Error: 2.90956E-06
y(4.0) = 24.99999377 Error: 6.23491E-06
y(5.0) = 52.56248918 Error: 1.08197E-05
y(6.0) = 99.99998341 Error: 1.65946E-05
y(7.0) = 175.56247648 Error: 2.35177E-05
y(8.0) = 288.99996843 Error: 3.15652E-05
y(9.0) = 451.56245928 Error: 4.07232E-05
y(10.0) = 675.99994902 Error: 5.09833E-05

ALGOL 68

<lang ALGOL68> BEGIN 

  PROC rk4 = (PROC (REAL, REAL) REAL f, REAL y, x, dx) REAL :
  BEGIN  CO Fourth-order Runge-Kutta method CO
     REAL dy1 = dx * f(x, y);
     REAL dy2 = dx * f(x + dx / 2.0, y + dy1 / 2.0);
     REAL dy3 = dx * f(x + dx / 2.0, y + dy2 / 2.0);
     REAL dy4 = dx * f(x + dx, y + dy3);
     y + (dy1 + 2.0 * dy2 + 2.0 * dy3 + dy4) / 6.0
  END;
  REAL x0 = 0, x1 = 10, y0 = 1.0;			CO Boundary conditions. CO
  REAL dx = 0.1;					CO Step size. CO
  INT num points = ENTIER ((x1 - x0) / dx + 0.5);	CO Add 0.5 for rounding errors. CO
  [0:num points]REAL y;   y[0] := y0;			CO Grid and starting point.CO
  PROC dy by dx = (REAL x, y) REAL : x * sqrt(y);	CO Differential equation. CO
  FOR i TO num points
  DO 
     y[i] := rk4 (dy by dx, y[i-1], x0 + dx * (i - 1), dx)
  OD;
  print (("   x              true y         calc y       relative error", newline));
  FOR i FROM 0 BY 10 TO  num points
  DO
     REAL x = x0 + dx * i;
     REAL true y = (x * x + 4.0) ^ 2 / 16.0;
     printf (($3(-zzd.7dxxx), -d.4de-ddl$, x, true y, y[i], y[i] / true y - 1.0))
  OD

END </lang>

Output:
   x              true y         calc y       relative error
   0.0000000      1.0000000      1.0000000    0.0000e 00
   1.0000000      1.5625000      1.5624999   -9.3262e-08
   2.0000000      4.0000000      3.9999991   -2.2987e-07
   3.0000000     10.5625000     10.5624971   -2.7546e-07
   4.0000000     25.0000000     24.9999938   -2.4940e-07
   5.0000000     52.5625000     52.5624892   -2.0584e-07
   6.0000000    100.0000000     99.9999834   -1.6595e-07
   7.0000000    175.5625000    175.5624765   -1.3396e-07
   8.0000000    289.0000000    288.9999684   -1.0922e-07
   9.0000000    451.5625000    451.5624593   -9.0183e-08
  10.0000000    676.0000000    675.9999490   -7.5419e-08

APL

<lang APL> 

     ∇RK4[⎕]∇
   ∇

[0] Z←R(Y¯ RK4)Y;T;YN;TN;∆T;∆Y1;∆Y2;∆Y3;∆Y4 [1] (T R ∆T)←R [2] LOOP:→(R≤TN←¯1↑T)/EXIT [3] ∆Y1←∆T×TN Y¯ YN←¯1↑Y [4] ∆Y2←∆T×(TN+∆T÷2)Y¯ YN+∆Y1÷2 [5] ∆Y3←∆T×(TN+∆T÷2)Y¯ YN+∆Y2÷2 [6] ∆Y4←∆T×(TN+∆T)Y¯ YN+∆Y3 [7] Y←Y,YN+(∆Y1+(2×∆Y2)+(2×∆Y3)+∆Y4)÷6 [8] T←T,TN+∆T [9] →LOOP [10] EXIT:Z←T,[⎕IO+.5]Y 

   ∇

     ∇PRINT[⎕]∇
   ∇

[0] PRINT;TABLE [1] TABLE←0 10 .1({⍺×⍵*.5}RK4)1 [2] ⎕←'T' 'RK4 Y' 'ERROR'⍪TABLE,TABLE[;2]-{((4+⍵*2)*2)÷16}TABLE[;1] 

   ∇

</lang>

Output:
      PRINT
    T           RK4 Y              ERROR
  0       1               0.000000000E0
  0.1     1.005006249    ¯1.303701147E¯9
  0.2     1.020099995    ¯5.215366805E¯9
  0.3     1.045506238    ¯1.174457109E¯8
  0.4     1.081599979    ¯2.093284546E¯8
  0.5     1.128906217    ¯3.288601591E¯8
  0.6     1.188099952    ¯4.780736740E¯8
  0.7     1.260006184    ¯6.602350622E¯8
  0.8     1.345599912    ¯8.799725681E¯8
  0.9     1.446006136    ¯1.143253423E¯7
  . . .

AWK

<lang AWK>

  1. syntax: GAWK -f RUNGE-KUTTA_METHOD.AWK
  2. converted from BBC BASIC

BEGIN { 

   print(" t    y         error")
   y = 1
   for (i=0; i<=100; i++) {
     t = i / 10
     if (t == int(t)) {
       actual = ((t^2+4)^2) / 16
       printf("%2d %12.7f %g\n",t,y,actual-y)
     }
     k1 = t * sqrt(y)
     k2 = (t + 0.05) * sqrt(y + 0.05 * k1)
     k3 = (t + 0.05) * sqrt(y + 0.05 * k2)
     k4 = (t + 0.10) * sqrt(y + 0.10 * k3)
     y += 0.1 * (k1 + 2 * (k2 + k3) + k4) / 6
   }
   exit(0)

} </lang>

Output:
 t    y         error
 0    1.0000000 0
 1    1.5624999 1.45722e-007
 2    3.9999991 9.19479e-007
 3   10.5624971 2.90956e-006
 4   24.9999938 6.23491e-006
 5   52.5624892 1.08197e-005
 6   99.9999834 1.65946e-005
 7  175.5624765 2.35177e-005
 8  288.9999684 3.15652e-005
 9  451.5624593 4.07232e-005
10  675.9999490 5.09833e-005

BASIC

BBC BASIC

<lang bbcbasic> y = 1.0 

     FOR i% = 0 TO 100
       t = i% / 10
 
       IF t = INT(t) THEN
         actual = ((t^2 + 4)^2) / 16
         PRINT "y("; t ") = "; y TAB(20) "Error = ";  actual - y
       ENDIF
 
       k1 =  t * SQR(y)
       k2 = (t + 0.05) * SQR(y + 0.05 * k1)
       k3 = (t + 0.05) * SQR(y + 0.05 * k2)
       k4 = (t + 0.10) * SQR(y + 0.10 * k3)
       y += 0.1 * (k1 + 2 * (k2 + k3) + k4) / 6
     NEXT i%</lang>
Output:
y(0) = 1            Error = 0
y(1) = 1.56249985   Error = 1.45721892E-7
y(2) = 3.99999908   Error = 9.19479201E-7
y(3) = 10.5624971   Error = 2.90956245E-6
y(4) = 24.9999938   Error = 6.23490936E-6
y(5) = 52.5624892   Error = 1.08196974E-5
y(6) = 99.9999834   Error = 1.65945964E-5
y(7) = 175.562476   Error = 2.35177287E-5
y(8) = 288.999968   Error = 3.15652015E-5
y(9) = 451.562459   Error = 4.07231605E-5
y(10) = 675.999949  Error = 5.09832905E-5

IS-BASIC

<lang IS-BASIC>100 PROGRAM "Runge.bas" 110 LET Y=1 120 FOR T=0 TO 10 STEP .1 130 IF T=INT(T) THEN PRINT "y(";STR$(T);") =";Y;TAB(21);"Error =";((T^2+4)^2)/16-Y 140 LET K1=T*SQR(Y) 150 LET K2=(T+.05)*SQR(Y+.05*K1) 160 LET K3=(T+.05)*SQR(Y+.05*K2) 170 LET K4=(T+.1)*SQR(Y+.1*K3) 180 LET Y=Y+.1*(K1+2*(K2+K3)+K4)/6 190 NEXT</lang>

C

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <math.h>

double rk4(double(*f)(double, double), double dx, double x, double y) { double k1 = dx * f(x, y), k2 = dx * f(x + dx / 2, y + k1 / 2), k3 = dx * f(x + dx / 2, y + k2 / 2), k4 = dx * f(x + dx, y + k3); return y + (k1 + 2 * k2 + 2 * k3 + k4) / 6; }

double rate(double x, double y) { return x * sqrt(y); }

int main(void) { double *y, x, y2; double x0 = 0, x1 = 10, dx = .1; int i, n = 1 + (x1 - x0)/dx; y = (double *)malloc(sizeof(double) * n);

for (y[0] = 1, i = 1; i < n; i++) y[i] = rk4(rate, dx, x0 + dx * (i - 1), y[i-1]);

printf("x\ty\trel. err.\n------------\n"); for (i = 0; i < n; i += 10) { x = x0 + dx * i; y2 = pow(x * x / 4 + 1, 2); printf("%g\t%g\t%g\n", x, y[i], y[i]/y2 - 1); }

return 0; }</lang>

Output:

(errors are relative) 

x       y       rel. err.
------------
0       1       0
1       1.5625  -9.3262e-08
2       4       -2.2987e-07
3       10.5625 -2.75462e-07
4       25      -2.49396e-07
5       52.5625 -2.05844e-07
6       100     -1.65946e-07
7       175.562 -1.33956e-07
8       289     -1.09222e-07
9       451.562 -9.01828e-08
10      676     -7.54191e-08

C#

<lang csharp> using System;

namespace RungeKutta { 

   class Program
   {
       static void Main(string[] args)
       {
           //Incrementers to pass into the known solution
           double t = 0.0;
           double T = 10.0;
           double dt = 0.1;

           // Assign the number of elements needed for the arrays
           int n = (int)(((T - t) / dt)) + 1;

           // Initialize the arrays for the time index 's' and estimates 'y' at each index 'i'
           double[] y = new double[n];
           double[] s = new double[n];

           // RK4 Variables
           double dy1;
           double dy2;
           double dy3;
           double dy4;

           // RK4 Initializations
           int i = 0;
           s[i] = 0.0;
           y[i] = 1.0;

           Console.WriteLine(" ===================================== ");
           Console.WriteLine(" Beging 4th Order Runge Kutta Method ");
           Console.WriteLine(" ===================================== ");

           Console.WriteLine();
           Console.WriteLine(" Given the example Differential equation: \n");
           Console.WriteLine("     y' = t*sqrt(y) \n");
           Console.WriteLine(" With the initial conditions: \n");
           Console.WriteLine("     t0 = 0" + ", y(0) = 1.0 \n");
           Console.WriteLine(" Whose exact solution is known to be: \n");
           Console.WriteLine("     y(t) = 1/16*(t^2 + 4)^2 \n");
           Console.WriteLine(" Solve the given equations over the range t = 0...10 with a step value dt = 0.1 \n");
           Console.WriteLine(" Print the calculated values of y at whole numbered t's (0.0,1.0,...10.0) along with the error \n");
           Console.WriteLine();

           Console.WriteLine(" y(t) " +"RK4" + " ".PadRight(18) + "Absolute Error");
           Console.WriteLine(" -------------------------------------------------");
           Console.WriteLine(" y(0) " + y[i] + " ".PadRight(20) + (y[i] - solution(s[i])));

           // Iterate and implement the Rk4 Algorithm
           while (i < y.Length - 1)
           {

               dy1 = dt * equation(s[i], y[i]);
               dy2 = dt * equation(s[i] + dt / 2, y[i] + dy1 / 2);
               dy3 = dt * equation(s[i] + dt / 2, y[i] + dy2 / 2);
               dy4 = dt * equation(s[i] + dt, y[i] + dy3);

               s[i + 1] = s[i] + dt;
               y[i + 1] = y[i] + (dy1 + 2 * dy2 + 2 * dy3 + dy4) / 6;

               double error = Math.Abs(y[i + 1] - solution(s[i + 1]));
               double t_rounded = Math.Round(t + dt, 2);

               if (t_rounded % 1 == 0)
               {
                   Console.WriteLine(" y(" + t_rounded + ")" + " " + y[i + 1] + " ".PadRight(5) + (error));
               }

               i++;
               t += dt;

           };//End Rk4

           Console.ReadLine();
       }

       // Differential Equation
       public static double equation(double t, double y)
       {
           double y_prime;
           return y_prime = t*Math.Sqrt(y);
       }

       // Exact Solution
       public static double solution(double t)
       {
           double actual;
           actual = Math.Pow((Math.Pow(t, 2) + 4), 2)/16;
           return actual;
       }
   }

}</lang>

C++

Using Lambdas <lang cpp>/* 

* compiled with gcc 5.4:
* g++-mp-5 -std=c++14 -o rk4 rk4.cc
*
*/
  1. include <iostream>
  2. include <math.h>

using namespace std;

auto rk4(double f(double, double)) { 

       return
       [       f            ](double t, double y, double dt ) -> double{ return
       [t,y,dt,f            ](                    double dy1) -> double{ return
       [t,y,dt,f,dy1        ](                    double dy2) -> double{ return
       [t,y,dt,f,dy1,dy2    ](                    double dy3) -> double{ return
       [t,y,dt,f,dy1,dy2,dy3](                    double dy4) -> double{ return
       ( dy1 + 2*dy2 + 2*dy3 + dy4 ) / 6   ;} (
       dt * f( t+dt  , y+dy3   )          );} (
       dt * f( t+dt/2, y+dy2/2 )          );} (
       dt * f( t+dt/2, y+dy1/2 )          );} (
       dt * f( t     , y       )          );} ;

}

int main(void) { 

       const double TIME_MAXIMUM = 10.0, WHOLE_TOLERANCE = 1e-12 ;
       const double T_START = 0.0, Y_START = 1.0, DT = 0.10;

       auto eval_diff_eqn = [               ](double t, double y)->double{ return t*sqrt(y)                         ; } ;
       auto eval_solution = [               ](double t          )->double{ return pow(t*t+4,2)/16                   ; } ;
       auto find_error    = [eval_solution  ](double t, double y)->double{ return fabs(y-eval_solution(t))          ; } ;
       auto is_whole      = [WHOLE_TOLERANCE](double t          )->bool  { return fabs(t-round(t)) < WHOLE_TOLERANCE; } ;

       auto dy = rk4( eval_diff_eqn ) ;

       double y = Y_START, t = T_START ;

       while(t <= TIME_MAXIMUM) {
         if (is_whole(t)) { printf("y(%4.1f)\t=%12.6f \t error: %12.6e\n",t,y,find_error(t,y)); }
         y += dy(t,y,DT) ; t += DT;
       }
       return 0;

}</lang>

Common Lisp

<lang lisp>(defun runge-kutta (f x y x-end n) 

   (let ((h (float (/ (- x-end x) n) 1d0))
         k1 k2 k3 k4)
       (setf x (float x 1d0)
             y (float y 1d0))
       (cons (cons x y)
           (loop for i below n do
               (setf k1 (* h (funcall f x y))
                     k2 (* h (funcall f (+ x (* 0.5d0 h)) (+ y (* 0.5d0 k1))))
                     k3 (* h (funcall f (+ x (* 0.5d0 h)) (+ y (* 0.5d0 k2))))
                     k4 (* h (funcall f (+ x h) (+ y k3)))
                     x (+ x h)
                     y (+ y (/ (+ k1 k2 k2 k3 k3 k4) 6)))
               collect (cons x y)))))

(let ((sol (runge-kutta (lambda (x y) (* x (sqrt y))) 0 1 10 100))) 

   (loop for n from 0
         for (x . y) in sol
         when (zerop (mod n 10))
         collect (list x y (- y (/ (expt (+ 4 (* x x)) 2) 16)))))

((0.0d0 1.0d0 0.0d0) 

(0.9999999999999999d0 1.562499854278108d0 -1.4572189210859676d-7)
(2.0000000000000004d0 3.9999990805207988d0 -9.194792029987298d-7)
(3.0000000000000013d0 10.562497090437557d0 -2.9095624576314094d-6)
(4.000000000000002d0 24.999993765090643d0 -6.234909392333066d-6)
(4.999999999999998d0 52.56248918030259d0 -1.081969734428867d-5)
(5.999999999999995d0 99.9999834054036d0 -1.659459609015812d-5)
(6.999999999999991d0 175.56247648227117d0 -2.3517728038768837d-5)
(7.999999999999988d0 288.9999684347983d0 -3.156520000402452d-5)
(8.999999999999984d0 451.56245927683887d0 -4.072315812209126d-5)
(9.99999999999998d0 675.9999490167083d0 -5.0983286655537086d-5))</lang>

D

Translation of: Ada

<lang d>import std.stdio, std.math, std.typecons;

alias FP = real; alias FPs = Typedef!(FP[101]);

void runge(in FP function(in FP, in FP) 

          pure nothrow @safe @nogc yp_func,
          ref FPs t, ref FPs y, in FP dt) pure nothrow @safe @nogc {
   foreach (immutable n; 0 .. t.length - 1) {
       immutable FP
           dy1 = dt * yp_func(t[n], y[n]),
           dy2 = dt * yp_func(t[n] + dt / 2.0, y[n] + dy1 / 2.0),
           dy3 = dt * yp_func(t[n] + dt / 2.0, y[n] + dy2 / 2.0),
           dy4 = dt * yp_func(t[n] + dt, y[n] + dy3);
       t[n + 1] = t[n] + dt;
       y[n + 1] = y[n] + (dy1 + 2.0 * (dy2 + dy3) + dy4) / 6.0;
   }

}

FP calc_err(in FP t, in FP calc) pure nothrow @safe @nogc { 

   immutable FP actual = (t ^^ 2 + 4.0) ^^ 2 / 16.0;
   return abs(actual - calc);

}

void main() { 

   enum FP dt = 0.10;
   FPs t_arr, y_arr;

   t_arr[0] = 0.0;
   y_arr[0] = 1.0;
   runge((t, y) => t * y.sqrt, t_arr, y_arr, dt);

   foreach (immutable i; 0 .. t_arr.length)
       if (i % 10 == 0)
           writefln("y(%.1f) = %.8f Error: %.6g",
                    t_arr[i], y_arr[i],
                    calc_err(t_arr[i], y_arr[i]));

}</lang>

Output:
y(0.0) = 1.00000000 Error: 0
y(1.0) = 1.56249985 Error: 1.45722e-07
y(2.0) = 3.99999908 Error: 9.19479e-07
y(3.0) = 10.56249709 Error: 2.90956e-06
y(4.0) = 24.99999377 Error: 6.23491e-06
y(5.0) = 52.56248918 Error: 1.08197e-05
y(6.0) = 99.99998341 Error: 1.65946e-05
y(7.0) = 175.56247648 Error: 2.35177e-05
y(8.0) = 288.99996843 Error: 3.15652e-05
y(9.0) = 451.56245928 Error: 4.07232e-05
y(10.0) = 675.99994902 Error: 5.09833e-05

Dart

<lang dart>import 'dart:math' as Math;

num RungeKutta4(Function f, num t, num y, num dt){ 

 num k1 = dt * f(t,y);
 num k2 = dt * f(t+0.5*dt, y + 0.5*k1);
 num k3 = dt * f(t+0.5*dt, y + 0.5*k2);
 num k4 = dt * f(t + dt, y + k3);
 return y + (1/6) * (k1 + 2*k2 + 2*k3 + k4);

}

void main(){ 

 num t  = 0;
 num dt = 0.1;
 num tf = 10;
 num totalPoints = ((tf-t)/dt).floor()+1;
 num y  = 1;
 Function f  = (num t, num y) => t * Math.sqrt(y);
 Function actual = (num t) => (1/16) * (t*t+4)*(t*t+4);
 for (num i = 0; i <= totalPoints; i++){
   num relativeError = (actual(t) - y)/actual(t);
   if (i%10 == 0){
     print('y(${t.round().toStringAsPrecision(3)}) = ${y.toStringAsPrecision(11)}  Error = ${relativeError.toStringAsPrecision(11)}');
   }
   y  = RungeKutta4(f, t, y, dt);
   t += dt;
 }

}</lang>

Output:
y(0.00) = 1.0000000000  Error = 0.0000000000
y(1.00) = 1.5624998543  Error = 9.3262010950e-8
y(2.00) = 3.9999990805  Error = 2.2986980086e-7
y(3.00) = 10.562497090  Error = 2.7546153479e-7
y(4.00) = 24.999993765  Error = 2.4939637555e-7
y(5.00) = 52.562489180  Error = 2.0584442034e-7
y(6.00) = 99.999983405  Error = 1.6594596090e-7
y(7.00) = 175.56247648  Error = 1.3395644308e-7
y(8.00) = 288.99996843  Error = 1.0922214534e-7
y(9.00) = 451.56245928  Error = 9.0182772312e-8
y(10.0) = 675.99994902  Error = 7.5419063100e-8

ERRE

<lang ERRE> PROGRAM RUNGE_KUTTA

CONST DELTA_T=0.1

FUNCTION Y1(T,Y) 

    Y1=T*SQR(Y)

END FUNCTION

BEGIN 

  Y=1.0
  FOR I%=0 TO 100 DO
     T=I%*DELTA_T

     IF T=INT(T) THEN           ! print every tenth
         ACTUAL=((T^2+4)^2)/16  ! exact solution
         PRINT("Y(";T;")=";Y;TAB(20);"Error=";ACTUAL-Y)
     END IF

     K1=Y1(T,Y)
     K2=Y1(T+DELTA_T/2,Y+DELTA_T/2*K1)
     K3=Y1(T+DELTA_T/2,Y+DELTA_T/2*K2)
     K4=Y1(T+DELTA_T,Y+DELTA_T*K3)
     Y+=DELTA_T*(K1+2*(K2+K3)+K4)/6
  END FOR

END PROGRAM</lang>

Output:
Y( 0 )= 1          Error= 0
Y( 1 )= 1.5625     Error= 2.384186E-07
Y( 2 )= 3.999999   Error= 7.152558E-07
Y( 3 )= 10.5625    Error= 1.907349E-06
Y( 4 )= 25         Error= 3.814697E-06
Y( 5 )= 52.56249   Error= 7.629395E-06
Y( 6 )= 100        Error= 0
Y( 7 )= 175.5625   Error= 0
Y( 8 )= 289        Error= 0
Y( 9 )= 451.5625   Error= 0
Y( 10 )= 676.0001  Error=-6.103516E-05

F#

Works with: F# interactive (fsi.exe)

<lang F Sharp> open System

let y'(t,y) = t * sqrt(y)

let RungeKutta4 t0 y0 t_max dt = 

   let dy1(t,y) = dt * y'(t,y)
   let dy2(t,y) = dt * y'(t+dt/2.0, y+dy1(t,y)/2.0)
   let dy3(t,y) = dt * y'(t+dt/2.0, y+dy2(t,y)/2.0)
   let dy4(t,y) = dt * y'(t+dt, y+dy3(t,y))

   (t0,y0) |> Seq.unfold (fun (t,y) ->
       if ( t <= t_max) then Some((t,y), (Math.Round(t+dt, 6), y + ( dy1(t,y) + 2.0*dy2(t,y) + 2.0*dy3(t,y) + dy4(t,y))/6.0)) 
       else None
       )

let y_exact t = (pown (pown t 2 + 4.0) 2)/16.0

RungeKutta4 0.0 1.0 10.0 0.1 

   |> Seq.filter (fun (t,y) -> t % 1.0 = 0.0 )
   |> Seq.iter (fun (t,y) -> Console.WriteLine("y({0})={1}\t(relative error:{2})", t, y, (y / y_exact(t))-1.0) )</lang>
Output:
y(0)=1			(relative error:0)
y(1)=1.56249985427811	(relative error:-9.32620110027926E-08)
y(2)=3.9999990805208	(relative error:-2.29869800194571E-07)
y(3)=10.5624970904376	(relative error:-2.75461533583155E-07)
y(4)=24.9999937650906	(relative error:-2.49396374552013E-07)
y(5)=52.5624891803026	(relative error:-2.05844421730106E-07)
y(6)=99.9999834054036	(relative error:-1.65945964192282E-07)
y(7)=175.562476482271	(relative error:-1.33956447156969E-07)
y(8)=288.999968434799	(relative error:-1.09222150213029E-07)
y(9)=451.56245927684	(relative error:-9.01827772459285E-08)
y(10)=675.99994901671	(relative error:-7.54190684348899E-08)

Fortran

<lang fortran>program rungekutta implicit none real(kind=kind(1.0D0)) :: t,dt,tstart,tstop real(kind=kind(1.0D0)) :: y,k1,k2,k3,k4 tstart =0.0D0 ; tstop =10.0D0 ; dt = 0.1D0 y = 1.0D0 t = tstart write(6,'(A,f4.1,A,f12.8,A,es13.6)') 'y(',t,') = ',y,' Error = '& 

          &,abs(y-(t**2+4.0d0)**2/16.0d0)

do; if ( t .ge. tstop ) exit 

  k1 = f (t           , y                 )
  k2 = f (t+0.5D0 * dt, y +0.5D0 * dt * k1)
  k3 = f (t+0.5D0 * dt, y +0.5D0 * dt * k2)
  k4 = f (t+        dt, y +        dt * k3)
  y = y + dt *( k1 + 2.0D0 *( k2 + k3 ) + k4 )/6.0D0
  t = t + dt
  if(abs(real(nint(t))-t) .le. 1.0D-12) then
     write(6,'(A,f4.1,A,f12.8,A,es13.6)') 'y(',t,') = ',y,' Error = '&
          &,abs(y-(t**2+4.0d0)**2/16.0d0)
  end if

end do contains 

 function f (t,y)
   implicit none
   real(kind=kind(1.0D0)),intent(in) :: y,t
   real(kind=kind(1.0D0)) :: f
   f = t*sqrt(y)
 end function f

end program rungekutta </lang>

Output:
y( 0.0) =   1.00000000 Error =  0.000000E+00
y( 1.0) =   1.56249985 Error =  1.457219E-07
y( 2.0) =   3.99999908 Error =  9.194792E-07
y( 3.0) =  10.56249709 Error =  2.909562E-06
y( 4.0) =  24.99999377 Error =  6.234909E-06
y( 5.0) =  52.56248918 Error =  1.081970E-05
y( 6.0) =  99.99998341 Error =  1.659460E-05
y( 7.0) = 175.56247648 Error =  2.351773E-05
y( 8.0) = 288.99996843 Error =  3.156520E-05
y( 9.0) = 451.56245928 Error =  4.072316E-05
y(10.0) = 675.99994902 Error =  5.098329E-05

FreeBASIC

Translation of: BBC BASIC

<lang freebasic>' version 03-10-2015 ' compile with: fbc -s console ' translation of BBC BASIC

Dim As Double y = 1, t, actual, k1, k2, k3, k4

Print

For i As Integer = 0 To 100 

   t = i / 10

   If t = Int(t) Then
       actual = ((t ^ 2 + 4) ^ 2) / 16
       Print  "y("; Str(t); ") ="; y ; Tab(27); "Error = "; actual - y
   End If

   k1 = t * Sqr(y)
   k2 = (t + 0.05) * Sqr(y + 0.05 * k1)
   k3 = (t + 0.05) * Sqr(y + 0.05 * k2)
   k4 = (t + 0.10) * Sqr(y + 0.10 * k3)
   y += 0.1 * (k1 + 2 * (k2 + k3) + k4) / 6

Next i


' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>

Output:
y(0) = 1                  Error =  0
y(1) = 1.562499854278108  Error =  1.457218921085968e-007
y(2) = 3.999999080520799  Error =  9.194792012223729e-007
y(3) = 10.56249709043755  Error =  2.909562448749625e-006
y(4) = 24.99999376509064  Error =  6.234909363911356e-006
y(5) = 52.56248918030259  Error =  1.081969741534294e-005
y(6) = 99.99998340540358  Error =  1.659459641700778e-005
y(7) = 175.5624764822713  Error =  2.351772874931157e-005
y(8) = 288.9999684347985  Error =  3.156520148195341e-005
y(9) = 451.5624592768396  Error =  4.072316039582802e-005
y(10) = 675.9999490167097 Error =  5.098329029351589e-005

FutureBasic

<lang futurebasic> include "ConsoleWindow"

def tab 9

local fn dydx( x as double, y as double ) as double end fn = x * sqr(y)

local fn exactY( x as long ) as double end fn = ( x ^2 + 4 ) ^2 / 16

dim as long i dim as double h, k1, k2, k3, k4, x, y, result

h = 0.1 y = 1 for i = 0 to 100 x = i * h if x == int(x) result = fn exactY( x ) print "y("; mid$( str$(x), 2, len(str$(x) )); ") = "; y, "Error = "; result - y end if

k1 = h * fn dydx( x, y ) k2 = h * fn dydx( x + h / 2, y + k1 / 2 ) k3 = h * fn dydx( x + h / 2, y + k2 / 2 ) k4 = h * fn dydx( x + h, y + k3 )

y = y + 1 / 6 * ( k1 + 2 * k2 + 2 * k3 + k4 ) next </lang> Output: 

y(0) =  1         Error =  0
y(1) =  1.5624998543       Error =  1.45721892e-7
y(2) =  3.9999990805       Error =  9.19479201e-7
y(3) =  10.5624970904      Error =  2.90956245e-6
y(4) =  24.9999937651      Error =  6.23490936e-6
y(5) =  52.56248918        Error =  1.08196974e-5
y(6) =  99.999983405       Error =  1.65945964e-5
y(7) =  175.562476482      Error =  2.35177287e-5
y(8) =  288.99996843       Error =  3.15652014e-5
y(9) =  451.56245928       Error =  4.07231603e-5
y(10) =  675.99994902      Error =  5.09832903e-5

Go

Works with: Go1

<lang go>package main

import ( 

   "fmt"
   "math"

)

type ypFunc func(t, y float64) float64 type ypStepFunc func(t, y, dt float64) float64

// newRKStep takes a function representing a differential equation // and returns a function that performs a single step of the forth-order // Runge-Kutta method. func newRK4Step(yp ypFunc) ypStepFunc { 

   return func(t, y, dt float64) float64 {
       dy1 := dt * yp(t, y)
       dy2 := dt * yp(t+dt/2, y+dy1/2)
       dy3 := dt * yp(t+dt/2, y+dy2/2)
       dy4 := dt * yp(t+dt, y+dy3)
       return y + (dy1+2*(dy2+dy3)+dy4)/6
   }

}

// example differential equation func yprime(t, y float64) float64 { 

   return t * math.Sqrt(y)

}

// exact solution of example func actual(t float64) float64 { 

   t = t*t + 4
   return t * t / 16

}

func main() { 

   t0, tFinal := 0, 10 // task specifies times as integers,
   dtPrint := 1        // and to print at whole numbers.
   y0 := 1.            // initial y.
   dtStep := .1        // step value.

   t, y := float64(t0), y0
   ypStep := newRK4Step(yprime)
   for t1 := t0 + dtPrint; t1 <= tFinal; t1 += dtPrint {
       printErr(t, y) // print intermediate result
       for steps := int(float64(dtPrint)/dtStep + .5); steps > 1; steps-- {
           y = ypStep(t, y, dtStep)
           t += dtStep
       }
       y = ypStep(t, y, float64(t1)-t) // adjust step to integer time
       t = float64(t1)
   }
   printErr(t, y) // print final result

}

func printErr(t, y float64) { 

   fmt.Printf("y(%.1f) = %f Error: %e\n", t, y, math.Abs(actual(t)-y))

}</lang>

Output:
y(0.0) = 1.000000 Error: 0.000000e+00
y(1.0) = 1.562500 Error: 1.457219e-07
y(2.0) = 3.999999 Error: 9.194792e-07
y(3.0) = 10.562497 Error: 2.909562e-06
y(4.0) = 24.999994 Error: 6.234909e-06
y(5.0) = 52.562489 Error: 1.081970e-05
y(6.0) = 99.999983 Error: 1.659460e-05
y(7.0) = 175.562476 Error: 2.351773e-05
y(8.0) = 288.999968 Error: 3.156520e-05
y(9.0) = 451.562459 Error: 4.072316e-05
y(10.0) = 675.999949 Error: 5.098329e-05

Groovy

<lang Groovy> class Runge_Kutta{ static void main(String[] args){ def y=1.0,t=0.0,counter=0; def dy1,dy2,dy3,dy4; def real; while(t<=10) {if(counter%10==0) {real=(t*t+4)*(t*t+4)/16; println("y("+t+")="+ y+ " Error:"+ (real-y)); }

dy1=dy(dery(y,t)); dy2=dy(dery(y+dy1/2,t+0.05)); dy3=dy(dery(y+dy2/2,t+0.05)); dy4=dy(dery(y+dy3,t+0.1));

y=y+(dy1+2*dy2+2*dy3+dy4)/6; t=t+0.1; counter++; } } static def dery(def y,def t){return t*(Math.sqrt(y));} static def dy(def x){return x*0.1;} } </lang>

Output:
y(0.0)=1.0 Error:0.0000
y(1.0)=1.562499854278108 Error:1.4572189210859676E-7
y(2.0)=3.999999080520799 Error:9.194792007782837E-7
y(3.0)=10.562497090437551 Error:2.9095624487496252E-6
y(4.0)=24.999993765090636 Error:6.234909363911356E-6
y(5.0)=52.562489180302585 Error:1.0819697415342944E-5
y(6.0)=99.99998340540358 Error:1.659459641700778E-5
y(7.0)=175.56247648227125 Error:2.3517728749311573E-5
y(8.0)=288.9999684347986 Error:3.156520142510999E-5
y(9.0)=451.56245927683966 Error:4.07231603389846E-5
y(10.0)=675.9999490167097 Error:5.098329029351589E-5

Haskell

Using GHC 7.4.1.

<lang haskell>dv 

 :: Floating a
 => a -> a -> a

dv = (. sqrt) . (*)

fy t = 1 / 16 * (4 + t ^ 2) ^ 2

rk4 

 :: (Enum a, Fractional a)
 => (a -> a -> a) -> a -> a -> a -> [(a, a)]

rk4 fd y0 a h = zip ts $ scanl (flip fc) y0 ts 

 where
   ts = [a,h ..]
   fc t y =
     sum . (y :) . zipWith (*) [1 / 6, 1 / 3, 1 / 3, 1 / 6] $
     scanl
       (\k f -> h * fd (t + f * h) (y + f * k))
       (h * fd t y)
       [1 / 2, 1 / 2, 1]

task = 

 mapM_
   (print . (\(x, y) -> (truncate x, y, fy x - y)))
   (filter (\(x, _) -> 0 == mod (truncate $ 10 * x) 10) $
    take 101 $ rk4 dv 1.0 0 0.1)</lang>

Example executed in GHCi: <lang haskell>*Main> task (0,1.0,0.0) (1,1.5624998542781088,1.4572189122041834e-7) (2,3.9999990805208006,9.194792029987298e-7) (3,10.562497090437557,2.909562461184123e-6) (4,24.999993765090654,6.234909399438493e-6) (5,52.56248918030265,1.0819697635611192e-5) (6,99.99998340540378,1.6594596999652822e-5) (7,175.56247648227165,2.3517730085131916e-5) (8,288.99996843479926,3.1565204153594095e-5) (9,451.562459276841,4.0723166534917254e-5) (10,675.9999490167125,5.098330132113915e-5)</lang>

(See Euler method#Haskell for implementation of simple general ODE-solver)


Or, disaggregated a little, and expressed in terms of a single scanl: <lang haskell>rk4 :: Double -> Double -> Double -> Double rk4 y x dx = 

 let f x y = x * sqrt y
     k1 = dx * f x y
     k2 = dx * f (x + dx / 2.0) (y + k1 / 2.0)
     k3 = dx * f (x + dx / 2.0) (y + k2 / 2.0)
     k4 = dx * f (x + dx) (y + k3)
 in y + (k1 + 2.0 * k2 + 2.0 * k3 + k4) / 6.0

actual :: Double -> Double actual x = (1 / 16) * (x * x + 4) * (x * x + 4)

step :: Double step = 0.1

ixs :: [Int] ixs = [0 .. 100]

xys :: [(Double, Double)] xys = 

 scanl
   (\(x, y) _ -> (((x * 10) + (step * 10)) / 10, rk4 y x step))
   (0.0, 1.0)
   ixs

samples :: [(Double, Double, Double)] samples = 

 zip ixs xys >>=
 (\(i, (x, y)) ->
     [ (x, y, actual x - y)
     | 0 == mod i 10 ])

main :: IO () main = 

 (putStrLn . unlines) $
 (\(x, y, v) ->
     unwords
       [ "y" ++ justifyRight 3 ' ' ('(' : show (round x)) ++ ") = "
       , justifyLeft 19 ' ' (show y)
       , '±' : show v
       ]) <$>
 samples
 where
   justifyLeft n c s = take n (s ++ replicate n c)
   justifyRight n c s = drop (length s) (replicate n c ++ s)</lang>
Output:
y (0) =  1.0                 ±0.0
y (1) =  1.562499854278108   ±1.4572189210859676e-7
y (2) =  3.999999080520799   ±9.194792007782837e-7
y (3) =  10.562497090437551  ±2.9095624487496252e-6
y (4) =  24.999993765090636  ±6.234909363911356e-6
y (5) =  52.562489180302585  ±1.0819697415342944e-5
y (6) =  99.99998340540358   ±1.659459641700778e-5
y (7) =  175.56247648227125  ±2.3517728749311573e-5
y (8) =  288.9999684347986   ±3.156520142510999e-5
y (9) =  451.56245927683966  ±4.07231603389846e-5
y(10) =  675.9999490167097   ±5.098329029351589e-5

J

Solution: <lang j>NB.*rk4 a Solve function using Runge-Kutta method NB. y is: y(ta) , ta , tb , tstep NB. u is: function to solve NB. eg: fyp rk4 1 0 10 0.1 rk4=: adverb define 

'Y0 a b h'=. 4{. y
T=. a + i.@>:&.(%&h) b - a
Y=. Yt=. Y0
for_t. }: T do.
  ty=. t,Yt
  k1=. h * u ty
  k2=. h * u ty + -: h,k1 
  k3=. h * u ty + -: h,k2 
  k4=. h * u ty + h,k3
  Y=. Y, Yt=. Yt + (%6) * 1 2 2 1 +/@:* k1, k2, k3, k4  
end.

T ,. Y )</lang> Example: <lang j> fy=: (%16) * [: *: 4 + *: NB. f(t,y) 

  fyp=: (* %:)/                         NB. f'(t,y)
  report_whole=: (10 * i. >:10)&{       NB. report at whole-numbered t values
  report_err=: (, {: - [: fy {.)"1      NB. report errors

  report_err report_whole fyp rk4 1 0 10 0.1
0       1           0
1  1.5625 _1.45722e_7
2       4 _9.19479e_7
3 10.5625 _2.90956e_6
4      25 _6.23491e_6
5 52.5625 _1.08197e_5
6     100 _1.65946e_5
7 175.562 _2.35177e_5
8     289 _3.15652e_5
9 451.562 _4.07232e_5

10 676 _5.09833e_5</lang>

Alternative solution:

The following solution replaces the for loop as well as the calculation of the increments (ks) with an accumulating suffix. <lang j>rk4=: adverb define 

'Y0 a b h'=. 4{. y
T=. a + i.@>:&.(%&h) b-a
(,. [: h&(u nextY)@,/\. Y0 ,~ }.)&.|. T

)

NB. nextY a Calculate Yn+1 of a function using Runge-Kutta method NB. y is: 2-item numeric list of time t and y(t) NB. u is: function to use NB. x is: step size NB. eg: 0.001 fyp nextY 0 1 nextY=: adverb define 

tableau=. 1 0.5 0.5, x * u y
ks=. (x * [: u y + (* x&,))/\. tableau
({:y) + 6 %~ +/ 1 2 2 1 * ks

)</lang>

Use: 

report_err report_whole fyp rk4 1 0 10 0.1

Java

Translation of Ada via D

Works with: Java version 8

<lang java>import static java.lang.Math.*; import java.util.function.BiFunction;

public class RungeKutta { 

   static void runge(BiFunction<Double, Double, Double> yp_func, double[] t,
           double[] y, double dt) {

       for (int n = 0; n < t.length - 1; n++) {
           double dy1 = dt * yp_func.apply(t[n], y[n]);
           double dy2 = dt * yp_func.apply(t[n] + dt / 2.0, y[n] + dy1 / 2.0);
           double dy3 = dt * yp_func.apply(t[n] + dt / 2.0, y[n] + dy2 / 2.0);
           double dy4 = dt * yp_func.apply(t[n] + dt, y[n] + dy3);
           t[n + 1] = t[n] + dt;
           y[n + 1] = y[n] + (dy1 + 2.0 * (dy2 + dy3) + dy4) / 6.0;
       }
   }

   static double calc_err(double t, double calc) {
       double actual = pow(pow(t, 2.0) + 4.0, 2) / 16.0;
       return abs(actual - calc);
   }

   public static void main(String[] args) {
       double dt = 0.10;
       double[] t_arr = new double[101];
       double[] y_arr = new double[101];
       y_arr[0] = 1.0;

       runge((t, y) -> t * sqrt(y), t_arr, y_arr, dt);

       for (int i = 0; i < t_arr.length; i++)
           if (i % 10 == 0)
               System.out.printf("y(%.1f) = %.8f Error: %.6f%n",
                       t_arr[i], y_arr[i],
                       calc_err(t_arr[i], y_arr[i]));
   }

}</lang> 

y(0,0) = 1,00000000 Error: 0,000000
y(1,0) = 1,56249985 Error: 0,000000
y(2,0) = 3,99999908 Error: 0,000001
y(3,0) = 10,56249709 Error: 0,000003
y(4,0) = 24,99999377 Error: 0,000006
y(5,0) = 52,56248918 Error: 0,000011
y(6,0) = 99,99998341 Error: 0,000017
y(7,0) = 175,56247648 Error: 0,000024
y(8,0) = 288,99996843 Error: 0,000032
y(9,0) = 451,56245928 Error: 0,000041
y(10,0) = 675,99994902 Error: 0,000051

JavaScript

ES5

<lang JavaScript> function rk4(y, x, dx, f) { 

   var k1 = dx * f(x, y),
       k2 = dx * f(x + dx / 2.0,   +y + k1 / 2.0),
       k3 = dx * f(x + dx / 2.0,   +y + k2 / 2.0),
       k4 = dx * f(x + dx,         +y + k3);

   return y + (k1 + 2.0 * k2 + 2.0 * k3 + k4) / 6.0;

}

function f(x, y) { 

   return x * Math.sqrt(y);

}

function actual(x) { 

   return (1/16) * (x*x+4)*(x*x+4);

}

var y = 1.0, 

   x = 0.0,
   step = 0.1,
   steps = 0,
   maxSteps = 101,
   sampleEveryN = 10;

while (steps < maxSteps) { 

   if (steps%sampleEveryN === 0) {
       console.log("y(" + x + ") =  \t" + y + "\t ± " + (actual(x) - y).toExponential());
   }

   y = rk4(y, x, step, f);

   // using integer math for the step addition
   // to prevent floating point errors as 0.2 + 0.1 != 0.3
   x = ((x * 10) + (step * 10)) / 10;
   steps += 1;

} </lang>

Output:
y(0) =  	1	                 ± 0e+0
y(1) =  	1.562499854278108	 ± 1.4572189210859676e-7
y(2) =  	3.999999080520799	 ± 9.194792007782837e-7
y(3) =  	10.562497090437551	 ± 2.9095624487496252e-6
y(4) =  	24.999993765090636	 ± 6.234909363911356e-6
y(5) =  	52.562489180302585	 ± 1.0819697415342944e-5
y(6) =  	99.99998340540358	 ± 1.659459641700778e-5
y(7) =  	175.56247648227125	 ± 2.3517728749311573e-5
y(8) =  	288.9999684347986	 ± 3.156520142510999e-5
y(9) =  	451.56245927683966	 ± 4.07231603389846e-5
y(10) =  	675.9999490167097	 ± 5.098329029351589e-5

ES6

<lang javascript>(() => { 

 'use strict';

 // rk4 :: (Double -> Double -> Double) ->
 //          Double -> Double -> Double -> Double
 const rk4 = f => (y, x, dx) => {
   const
     k1 = dx * f(x, y),
     k2 = dx * f(x + dx / 2.0, y + k1 / 2.0),
     k3 = dx * f(x + dx / 2.0, y + k2 / 2.0),
     k4 = dx * f(x + dx, y + k3);
   return y + (k1 + 2.0 * k2 + 2.0 * k3 + k4) / 6.0;
 };

 // rk :: Double -> Double -> Double -> Double
 const rk = rk4((x, y) => x * Math.sqrt(y));

 // actual :: Double -> Double
 const actual = x => (1 / 16) * ((x * x) + 4) * ((x * x) + 4);



 // TEST -------------------------------------------------

 // main :: IO ()
 const main = () => {
   const
     step = 0.1,
     ixs = enumFromTo(0, 100),
     xys = scanl(
       xy => Tuple(
         ((xy[0] * 10) + (step * 10)) / 10, rk(xy[1], xy[0], step)
       ),
       Tuple(0.0, 1.0),
       ixs
     );

   // samples :: [(Double, Double, Double)]
   const samples = concatMap(
     tpl => 0 === tpl[0] % 10 ? (() => {
       const [x, y] = Array.from(tpl[1]);
       return [TupleN(x, y, actual(x) - y)];
     })() : [],
     zip(ixs, xys)
   );

   console.log(
     unlines(map(
       tpl => {
         const [x, y, v] = Array.from(tpl),
           [sn, sm] = splitOn('.', y.toString());
         return unwords([
           'y' + justifyRight(3, ' ', '(' + Math.round(x).toString()) +
           ') =',
           justifyRight(3, ' ', sn) + '.' + justifyLeft(15, ' ', sm || '0'),
           '± ' + v.toExponential()
         ]);
       },
       samples
     ))
   );
 };



 // GENERIC FUNCTIONS ----------------------------

 // Tuple (,) :: a -> b -> (a, b)
 const Tuple = (a, b) => ({
   type: 'Tuple',
   '0': a,
   '1': b,
   length: 2
 });

 // TupleN :: a -> b ...  -> (a, b ... )
 function TupleN() {
   const
     args = Array.from(arguments),
     lng = args.length;
   return lng > 1 ? Object.assign(
     args.reduce((a, x, i) => Object.assign(a, {
       [i]: x
     }), {
       type: 'Tuple' + (2 < lng ? lng.toString() : ),
       length: lng
     })
   ) : args[0];
 };

 // concatMap :: (a -> [b]) -> [a] -> [b]
 const concatMap = (f, xs) =>
   xs.reduce((a, x) => a.concat(f(x)), []);

 // enumFromTo :: Int -> Int -> [Int]
 const enumFromTo = (m, n) =>
   Array.from({
     length: 1 + n - m
   }, (_, i) => m + i)

 // justifyLeft :: Int -> Char -> String -> String
 const justifyLeft = (n, cFiller, s) =>
   n > s.length ? (
     s.padEnd(n, cFiller)
   ) : s;

 // justifyRight :: Int -> Char -> String -> String
 const justifyRight = (n, cFiller, s) =>
   n > s.length ? (
     s.padStart(n, cFiller)
   ) : s;

 // Returns Infinity over objects without finite length
 // this enables zip and zipWith to choose the shorter
 // argument when one is non-finite, like cycle, repeat etc

 // length :: [a] -> Int
 const length = xs => xs.length || Infinity;

 // map :: (a -> b) -> [a] -> [b]
 const map = (f, xs) => xs.map(f);

 // scanl :: (b -> a -> b) -> b -> [a] -> [b]
 const scanl = (f, startValue, xs) =>
   xs.reduce((a, x) => {
     const v = f(a[0], x);
     return Tuple(v, a[1].concat(v));
   }, Tuple(startValue, [startValue]))[1];

 // splitOn :: String -> String -> [String]
 const splitOn = (pat, src) => src.split(pat);

 // take :: Int -> [a] -> [a]
 // take :: Int -> String -> String
 const take = (n, xs) =>
   xs.constructor.constructor.name !== 'GeneratorFunction' ? (
     xs.slice(0, n)
   ) : [].concat.apply([], Array.from({
     length: n
   }, () => {
     const x = xs.next();
     return x.done ? [] : [x.value];
   }));

 // unlines :: [String] -> String
 const unlines = xs => xs.join('\n');

 // unwords :: [String] -> String
 const unwords = xs => xs.join(' ');

 // Use of `take` and `length` here allows for zipping with non-finite
 // lists - i.e. generators like cycle, repeat, iterate.

 // zip :: [a] -> [b] -> [(a, b)]
 const zip = (xs, ys) => {
   const lng = Math.min(length(xs), length(ys));
   return Infinity !== lng ? (() => {
     const bs = take(lng, ys);
     return take(lng, xs).map((x, i) => Tuple(x, bs[i]));
   })() : zipGen(xs, ys);
 };

 // MAIN ---
 return main();

})();</lang>

Output:
y (0) =   1.0               ± 0e+0
y (1) =   1.562499854278108 ± 1.4572189210859676e-7
y (2) =   3.999999080520799 ± 9.194792007782837e-7
y (3) =  10.562497090437551 ± 2.9095624487496252e-6
y (4) =  24.999993765090636 ± 6.234909363911356e-6
y (5) =  52.562489180302585 ± 1.0819697415342944e-5
y (6) =  99.99998340540358  ± 1.659459641700778e-5
y (7) = 175.56247648227125  ± 2.3517728749311573e-5
y (8) = 288.9999684347986   ± 3.156520142510999e-5
y (9) = 451.56245927683966  ± 4.07231603389846e-5
y(10) = 675.9999490167097   ± 5.098329029351589e-5

jq

In this section, two solutions are presented. They use "while" and/or "until" as defined in recent versions of jq (after version 1.4). To use either of the two programs with jq 1.4, simply include the lines in the following block: <lang jq>def until(cond; next): 

 def _until: if cond then . else (next|_until) end;
 _until;

def while(cond; update): 

 def _while:  if cond then ., (update | _while) else empty end;
 _while;</lang>

The Example Differential Equation and its Exact Solution

<lang jq># yprime maps [t,y] to a number, i.e. t * sqrt(y) def yprime: .[0] * (.[1] | sqrt);

  1. The exact solution of yprime:

def actual: 

 . as $t
 | (( $t*$t) + 4 )
 | . * . / 16;</lang>

dy/dt

The first solution presented here uses the terminology and style of the Perl 6 version.

Generic filters: <lang jq># n is the number of decimal places of precision def round(n): 

 (if . < 0 then -1 else 1 end) as $s
 | $s*10*.*n | if (floor % 10) > 4 then (.+5) else . end | ./10 | floor/n | .*$s;

def abs: if . < 0 then -. else . end;

  1. Is the input an integer?

def integerq: ((. - ((.+.01) | floor)) | abs) < 0.01;</lang>

dy(f) <lang jq>def dt: 0.1;

  1. Input: [t, y]; yp is a filter that accepts [t,y] as input

def runge_kutta(yp): 

 .[0] as $t | .[1] as $y
 | (dt * yp) as $a
 | (dt * ([ ($t + (dt/2)), $y + ($a/2) ] | yp)) as $b
 | (dt * ([ ($t + (dt/2)), $y + ($b/2) ] | yp)) as $c
 | (dt * ([ ($t + dt)    , $y + $c     ] | yp)) as $d
 | ($a + (2*($b + $c)) + $d) / 6
  1. Input: [t,y]

def dy(f): runge_kutta(f);</lang> Example: <lang jq># state: [t,y] [0,1] | while( .[0] <= 10; 

        .[0] as $t | .[1] as $y
        | [$t + dt, $y + dy(yprime) ] )

| .[0] as $t | .[1] as $y | if $t | integerq then 

    "y(\($t|round(1))) = \($y|round(10000)) ± \( ($t|actual) - $y | abs)" 
 else empty
 end</lang>
Output:

<lang sh>$ time jq -r -n -f rk4.pl.jq y(0) = 1 ± 0 y(1) = 1.5625 ± 1.4572189210859676e-07 y(2) = 4 ± 9.194792029987298e-07 y(3) = 10.5625 ± 2.9095624576314094e-06 y(4) = 25 ± 6.234909392333066e-06 y(5) = 52.5625 ± 1.081969734428867e-05 y(6) = 100 ± 1.659459609015812e-05 y(7) = 175.5625 ± 2.3517728038768837e-05 y(8) = 289 ± 3.156520000402452e-05 y(9) = 451.5625 ± 4.072315812209126e-05 y(10) = 675.9999 ± 5.0983286655537086e-05

real 0m0.048s user 0m0.013s sys 0m0.006s</lang>

newRK4Step

The second solution follows the nomenclature and style of the Go solution on this page.

In the following notes:

  • ypFunc denotes the type of a jq filter that maps [t, y] to a number;
  • ypStepFunc denotes the type of a jq filter that maps [t, y, dt] to a number.

The heart of the program is the filter newRK4Step(yp), which is of type ypStepFunc and performs a single step of the fourth-order Runge-Kutta method, provided yp is of type ypFunc. <lang jq># Input: [t, y, dt] def newRK4Step(yp): 

 .[0] as $t | .[1] as $y | .[2] as $dt
 | ($dt * ([$t, $y]|yp))              as $dy1
 | ($dt * ([$t+$dt/2, $y+$dy1/2]|yp)) as $dy2
 | ($dt * ([$t+$dt/2, $y+$dy2/2]|yp)) as $dy3
 | ($dt * ([$t+$dt, $y+$dy3]    |yp)) as $dy4
 | $y + ($dy1+2*($dy2+$dy3)+$dy4)/6


def printErr: # input: [t, y] 

 def abs: if . < 0 then -. else . end;
 .[0] as $t | .[1] as $y
 | "y(\($t)) = \($y) with error: \( (($t|actual) - $y) | abs )"

def main(t0; y0; tFinal; dtPrint): 

 def ypStep: newRK4Step(yprime) ;

 0.1 as $dtStep     # step value
 # [ t, y] is the state vector
 | [ t0, y0 ]
 | while( .[0] <= tFinal;
          .[0] as $t | .[1] as $y

| ($t + dtPrint) as $t1 | (((dtPrint/$dtStep) + 0.5) | floor) as $steps | [$steps, $t, $y] # state vector 

          | until( .[0] <= 1;

.[0] as $steps | .[1] as $t | .[2] as $y | [ ($steps - 1), ($t + $dtStep), ([$t, $y, $dtStep]|ypStep) ] 

                 )

| .[1] as $t | .[2] as $y | [$t1, ([ $t, $y, ($t1-$t)] | ypStep)] # adjust step to integer time 

        )
  | printErr # print results
  1. main(t0; y0; tFinal; dtPrint)

main(0; 1; 10; 1)</lang>

Output:

<lang sh>$ time jq -n -r -f runge-kutta.jq y(0) = 1 with error: 0 y(1) = 1.562499854278108 with error: 1.4572189210859676e-07 y(2) = 3.9999990805207974 with error: 9.194792025546406e-07 y(3) = 10.562497090437544 with error: 2.9095624558550526e-06 y(4) = 24.999993765090615 with error: 6.234909385227638e-06 y(5) = 52.562489180302656 with error: 1.081969734428867e-05 y(6) = 99.99998340540387 with error: 1.6594596132790684e-05 y(7) = 175.56247648227188 with error: 2.3517728124033965e-05 y(8) = 288.9999684347997 with error: 3.156520028824161e-05 y(9) = 451.56245927684154 with error: 4.0723158463151776e-05 y(10) = 675.9999490167129 with error: 5.0983287110284436e-05

real 0m0.023s user 0m0.014s sys 0m0.006s</lang>

Julia

Works with: Julia version 0.6

Using lambda expressions

Translation of: Python

<lang julia>f(x, y) = x * sqrt(y) theoric(t) = (t ^ 2 + 4.0) ^ 2 / 16.0

rk4(f) = (t, y, δt) -> # 1st (result) lambda 

        ((δy1) ->      # 2nd lambda
        ((δy2) ->      # 3rd lambda
        ((δy3) ->      # 4th lambda
        ((δy4) -> ( δy1 + 2δy2 + 2δy3 + δy4 ) / 6 # 5th and deepest lambda: calc y_{n+1}
        )(δt * f(t + δt, y + δy3))         # calc δy₄
        )(δt * f(t + δt / 2, y + δy2 / 2)) # calc δy₃
        )(δt * f(t + δt / 2, y + δy1 / 2)) # calc δy₂
        )(δt * f(t, y))                    # calc δy₁

δy = rk4(f) t₀, δt, tmax = 0.0, 0.1, 10.0 y₀ = 1.0

t, y = t₀, y₀ while t ≤ tmax 

   if t ≈ round(t) @printf("y(%4.1f) = %10.6f\terror: %12.6e\n", t, y, abs(y - theoric(t))) end
   y += δy(t, y, δt)
   t += δt

end</lang>

Output:
y( 0.0) =   1.000000	error: 0.000000e+00
y( 1.0) =   1.562500	error: 1.457219e-07
y( 2.0) =   3.999999	error: 9.194792e-07
y( 3.0) =  10.562497	error: 2.909562e-06
y( 4.0) =  24.999994	error: 6.234909e-06
y( 5.0) =  52.562489	error: 1.081970e-05
y( 6.0) =  99.999983	error: 1.659460e-05
y( 7.0) = 175.562476	error: 2.351773e-05
y( 8.0) = 288.999968	error: 3.156520e-05
y( 9.0) = 451.562459	error: 4.072316e-05
y(10.0) = 675.999949	error: 5.098329e-05

Alternative version

Translation of: Python

<lang julia>function rk4(f::Function, x₀::Float64, y₀::Float64, x₁::Float64, n) 

   vx = Vector{Float64}(n + 1)
   vy = Vector{Float64}(n + 1)
   vx[1] = x = x₀
   vy[1] = y = y₀
   h = (x₁ - x₀) / n
   for i in 1:n
       k₁ = h * f(x, y)
       k₂ = h * f(x + 0.5h, y + 0.5k₁)
       k₃ = h * f(x + 0.5h, y + 0.5k₂)
       k₄ = h * f(x + h, y + k₃)
       vx[i + 1] = x = x₀ + i * h
       vy[i + 1] = y = y + (k₁ + 2k₂ + 2k₃ + k₄) / 6
   end
   return vx, vy

end

vx, vy = rk4(f, 0.0, 1.0, 10.0, 100) for (x, y) in Iterators.take(zip(vx, vy), 10) 

   @printf("%4.1f %10.5f %+12.4e\n", x, y, y - theoric(x))

end</lang>

Kotlin

<lang scala>// version 1.1.2

typealias Y = (Double) -> Double typealias Yd = (Double, Double) -> Double

fun rungeKutta4(t0: Double, tz: Double, dt: Double, y: Y, yd: Yd) { 

   var tn = t0
   var yn = y(tn)
   val z = ((tz  - t0) / dt).toInt()
   for (i in 0..z) {
       if (i % 10 == 0) {
           val exact = y(tn)
           val error = yn - exact
           println("%4.1f  %10f  %10f  %9f".format(tn, yn, exact, error))
       }
       if (i == z) break
       val dy1 = dt * yd(tn, yn)
       val dy2 = dt * yd(tn + 0.5 * dt, yn + 0.5 * dy1)
       val dy3 = dt * yd(tn + 0.5 * dt, yn + 0.5 * dy2)
       val dy4 = dt * yd(tn + dt, yn + dy3)
       yn += (dy1 + 2.0 * dy2 + 2.0 * dy3 + dy4) / 6.0
       tn += dt
   }

}

fun main(args: Array<String>) { 

   println("  T        RK4        Exact      Error")
   println("----  ----------  ----------  ---------")
   val y = fun(t: Double): Double {
       val x = t * t + 4.0
       return x * x / 16.0
   }
   val yd = fun(t: Double, yt: Double) = t * Math.sqrt(yt)
   rungeKutta4(0.0, 10.0, 0.1, y, yd)

}</lang>

Output:
  T        RK4        Exact      Error
----  ----------  ----------  ---------
 0.0    1.000000    1.000000   0.000000
 1.0    1.562500    1.562500  -0.000000
 2.0    3.999999    4.000000  -0.000001
 3.0   10.562497   10.562500  -0.000003
 4.0   24.999994   25.000000  -0.000006
 5.0   52.562489   52.562500  -0.000011
 6.0   99.999983  100.000000  -0.000017
 7.0  175.562476  175.562500  -0.000024
 8.0  288.999968  289.000000  -0.000032
 9.0  451.562459  451.562500  -0.000041
10.0  675.999949  676.000000  -0.000051

Liberty BASIC

<lang lb> '[RC] Runge-Kutta method 'initial conditions x0 = 0 y0 = 1 'step h = 0.1 'number of points N=101

y=y0 FOR i = 0 TO N-1 

   x = x0+ i*h
   IF x = INT(x) THEN
       actual = exactY(x)
       PRINT "y("; x ;") = "; y; TAB(20); "Error = ";  actual - y
   END IF

   k1 = h*dydx(x,y)
   k2 = h*dydx(x+h/2,y+k1/2)
   k3 = h*dydx(x+h/2,y+k2/2)
   k4 = h*dydx(x+h,y+k3)
   y = y + 1/6 * (k1 + 2*k2 + 2*k3 + k4)

NEXT i

function dydx(x,y) 

   dydx=x*sqr(y)

end function

function exactY(x) 

   exactY=(x^2 + 4)^2 / 16

end function </lang>

Output:
y(0) = 1           Error = 0
y(1) = 1.56249985  Error = 0.14572189e-6
y(2) = 3.99999908  Error = 0.9194792e-6
y(3) = 10.5624971  Error = 0.29095624e-5
y(4) = 24.9999938  Error = 0.62349094e-5
y(5) = 52.5624892  Error = 0.10819697e-4
y(6) = 99.9999834  Error = 0.16594596e-4
y(7) = 175.562476  Error = 0.23517729e-4
y(8) = 288.999968  Error = 0.31565201e-4
y(9) = 451.562459  Error = 0.4072316e-4
y(10) = 675.999949 Error = 0.5098329e-4

Mathematica

<lang Mathematica>(* Symbolic solution *) DSolve[{y'[t] == t*Sqrt[y[t]], y[0] == 1}, y, t] Table[{t, 1/16 (4 + t^2)^2}, {t, 0, 10}]

(* Numerical solution I (not RK4) *) Table[{t, y[t], Abs[y[t] - 1/16*(4 + t^2)^2]}, {t, 0, 10}] /. 

First@NDSolve[{y'[t] == t*Sqrt[y[t]], y[0] == 1}, y, {t, 0, 10}]

(* Numerical solution II (RK4) *) f[{t_, y_}] := {1, t Sqrt[y]} h = 0.1; phi[y_] := Module[{k1, k2, k3, k4}, 

 k1 = h*f[y];
 k2 = h*f[y + 1/2 k1];
 k3 = h*f[y + 1/2 k2];
 k4 = h*f[y + k3];
 y + k1/6 + k2/3 + k3/3 + k4/6]

solution = NestList[phi, {0, 1}, 101]; Table[{y1, y2, Abs[y2 - 1/16 (y1^2 + 4)^2]}, 

 {y,  solution1 ;; 101 ;; 10}]

</lang>

MATLAB

The normally-used built-in solver is the ode45 function, which uses a non-fixed-step solver with 4th/5th order Runge-Kutta methods. The MathWorks Support Team released a package of fixed-step RK method ODE solvers on MATLABCentral. The ode4 function contained within uses a 4th-order Runge-Kutta method. Here is code that tests both ode4 and my own function, shows that they are the same, and compares them to the exact solution. <lang MATLAB>function testRK4Programs 

   figure
   hold on
   t = 0:0.1:10;
   y = 0.0625.*(t.^2+4).^2;
   plot(t, y, '-k')
   [tode4, yode4] = testODE4(t);
   plot(tode4, yode4, '--b')
   [trk4, yrk4] = testRK4(t);
   plot(trk4, yrk4, ':r')
   legend('Exact', 'ODE4', 'RK4')
   hold off
   fprintf('Time\tExactVal\tODE4Val\tODE4Error\tRK4Val\tRK4Error\n')
   for k = 1:10:length(t)
       fprintf('%.f\t\t%7.3f\t\t%7.3f\t%7.3g\t%7.3f\t%7.3g\n', t(k), y(k), ...
           yode4(k), abs(y(k)-yode4(k)), yrk4(k), abs(y(k)-yrk4(k)))
   end

end

function [t, y] = testODE4(t) 

   y0 = 1;
   y = ode4(@(tVal,yVal)tVal*sqrt(yVal), t, y0);

end

function [t, y] = testRK4(t) 

   dydt = @(tVal,yVal)tVal*sqrt(yVal);
   y = zeros(size(t));
   y(1) = 1;
   for k = 1:length(t)-1
       dt = t(k+1)-t(k);
       dy1 = dt*dydt(t(k), y(k));
       dy2 = dt*dydt(t(k)+0.5*dt, y(k)+0.5*dy1);
       dy3 = dt*dydt(t(k)+0.5*dt, y(k)+0.5*dy2);
       dy4 = dt*dydt(t(k)+dt, y(k)+dy3);
       y(k+1) = y(k)+(dy1+2*dy2+2*dy3+dy4)/6;
   end

end</lang>

Output:
Time	ExactVal	ODE4Val		ODE4Error	RK4Val		RK4Error
0	  1.000		  1.000		      0		  1.000		      0
1	  1.563		  1.562		1.46e-007	  1.562		1.46e-007
2	  4.000		  4.000		9.19e-007	  4.000		9.19e-007
3	 10.563		 10.562		2.91e-006	 10.562		2.91e-006
4	 25.000		 25.000		6.23e-006	 25.000		6.23e-006
5	 52.563		 52.562		1.08e-005	 52.562		1.08e-005
6	100.000		100.000		1.66e-005	100.000		1.66e-005
7	175.563		175.562		2.35e-005	175.562		2.35e-005
8	289.000		289.000		3.16e-005	289.000		3.16e-005
9	451.563		451.562		4.07e-005	451.562		4.07e-005
10	676.000		676.000		5.10e-005	676.000		5.10e-005

Maxima

<lang maxima>/* Here is how to solve a differential equation */ 'diff(y, x) = x * sqrt(y); ode2(%, y, x); ic1(%, x = 0, y = 1); factor(solve(%, y)); /* [y = (x^2 + 4)^2 / 16] */

/* The Runge-Kutta solver is builtin */

load(dynamics)$ sol: rk(t * sqrt(y), y, 1, [t, 0, 10, 1.0])$ plot2d([discrete, sol])$

/* An implementation of RK4 for one equation */

rk4(f, x0, y0, x1, n) := block([h, x, y, vx, vy, k1, k2, k3, k4], 

  h: bfloat((x1 - x0) / (n - 1)),
  x: x0,
  y: y0,
  vx: makelist(0, n + 1),
  vy: makelist(0, n + 1),
  vx[1]: x0,
  vy[1]: y0,
  for i from 1 thru n do (
     k1: bfloat(h * f(x, y)),
     k2: bfloat(h * f(x + h / 2, y + k1 / 2)),
     k3: bfloat(h * f(x + h / 2, y + k2 / 2)),
     k4: bfloat(h * f(x + h, y + k3)),
     vy[i + 1]: y: y + (k1 + 2 * k2 + 2 * k3 + k4) / 6,
     vx[i + 1]: x: x + h
  ),
  [vx, vy]

)$

[x, y]: rk4(lambda([x, y], x * sqrt(y)), 0, 1, 10, 101)$

plot2d([discrete, x, y])$

s: map(lambda([x], (x^2 + 4)^2 / 16), x)$

for i from 1 step 10 thru 101 do print(x[i], " ", y[i], " ", y[i] - s[i]);</lang>

МК-61/52

ПП	38	П1	ПП	30	П2	ПП	35	П3	2
*	ПП	30	ИП2	ИП3	+	2	*	+	ИП1
+	3	/	ИП7	+	П7	П8	С/П	БП	00
ИП6	ИП5	+	П6	<->	ИП7	+	П8

ИП8	КвКор	ИП6	*

ИП5	*	В/О

Input: 1/2 (h/2) - Р5, 1 (y0) - Р8 and Р7, 0 (t0) - Р6.


Nim

<lang nim>import math

proc fn(t, y: float): float = 

   result = t * math.sqrt(y)

proc solution(t: float): float = 

   result = (t^2 + 4)^2 / 16

proc rk(start, stop, step: float) = 

   let nsteps = int(round((stop - start) / step)) + 1
   let delta = (stop - start) / float(nsteps - 1)
   var cur_y = 1.0
   for i in 0..(nsteps - 1):
       let cur_t = start + delta * float(i)

       if abs(cur_t - math.round(cur_t)) < 1e-5:
           echo "y(", cur_t, ") = ", cur_y, ", error = ", solution(cur_t) - cur_y
       
       let dy1 = step * fn(cur_t, cur_y)
       let dy2 = step * fn(cur_t + 0.5 * step, cur_y + 0.5 * dy1)
       let dy3 = step * fn(cur_t + 0.5 * step, cur_y + 0.5 * dy2)
       let dy4 = step * fn(cur_t + step, cur_y + dy3)

       cur_y += (dy1 + 2.0 * (dy2 + dy3) + dy4) / 6.0

rk(start=0.0, stop=10.0, step=0.1)</lang>

Output:
y(0.0) = 1.0, error = 0.0
y(1.0) = 1.562499854278108, error = 1.457218921085968e-007
y(2.0) = 3.9999990805208, error = 9.194792003341945e-007
y(3.0) = 10.56249709043755, error = 2.909562448749625e-006
y(4.0) = 24.99999376509064, error = 6.234909363911356e-006
y(5.0) = 52.56248918030259, error = 1.081969741534294e-005
y(6.0) = 99.99998340540358, error = 1.659459641700778e-005
y(7.0) = 175.5624764822713, error = 2.351772874931157e-005
y(8.0) = 288.9999684347986, error = 3.156520142510999e-005
y(9.0) = 451.5624592768397, error = 4.07231603389846e-005
y(10.0) = 675.9999490167097, error = 5.098329029351589e-005

Objeck

<lang objeck>class RungeKuttaMethod { 

 function : Main(args : String[]) ~ Nil {
   x0 := 0.0; x1 := 10.0; dx := .1;
   
   n := 1 + (x1 - x0)/dx;
   y := Float->New[n->As(Int)];
   
   y[0] := 1;
   for(i := 1; i < n; i++;) {
     y[i] := Rk4(Rate(Float, Float) ~ Float, dx, x0 + dx * (i - 1), y[i-1]);
   };
   
   for(i := 0; i < n; i += 10;) {
     x := x0 + dx * i;
     y2 := (x * x / 4 + 1)->Power(2.0);
     
     x_value := x->As(Int); 
     y_value := y[i]; 
     rel_value := y_value/y2 - 1.0;
     "y({$x_value})={$y_value}; error: {$rel_value}"->PrintLine();
   };
 }

 function : native : Rk4(f : (Float, Float) ~ Float, dx : Float, x : Float, y : Float) ~ Float {
   k1 := dx * f(x, y);
   k2 := dx * f(x + dx / 2, y + k1 / 2);
   k3 := dx * f(x + dx / 2, y + k2 / 2);
   k4 := dx * f(x + dx, y + k3);
   
   return y + (k1 + 2 * k2 + 2 * k3 + k4) / 6;
 }
 
 function : native : Rate(x : Float, y : Float) ~ Float {
   return x * y->SquareRoot();
 }

}</lang>

Output: 

y(0)=1.0; error: 0.0
y(1)=1.563; error: -0.0000000933
y(2)=3.1000; error: -0.000000230
y(3)=10.563; error: -0.000000275
y(4)=24.1000; error: -0.000000249
y(5)=52.563; error: -0.000000206
y(6)=99.1000; error: -0.000000166
y(7)=175.563; error: -0.000000134
y(8)=288.1000; error: -0.000000109
y(9)=451.563; error: -0.0000000902
y(10)=675.1000; error: -0.0000000754

OCaml

<lang ocaml>let y' t y = t *. sqrt y let exact t = let u = 0.25*.t*.t +. 1.0 in u*.u

let rk4_step (y,t) h = 

 let k1 = h *. y' t y in
 let k2 = h *. y' (t +. 0.5*.h) (y +. 0.5*.k1) in
 let k3 = h *. y' (t +. 0.5*.h) (y +. 0.5*.k2) in
 let k4 = h *. y' (t +. h) (y +. k3) in
 (y +. (k1+.k4)/.6.0 +. (k2+.k3)/.3.0, t +. h)

let rec loop h n (y,t) = 

 if n mod 10 = 1 then
   Printf.printf "t = %f,\ty = %f,\terr = %g\n" t y (abs_float (y -. exact t));
 if n < 102 then loop h (n+1) (rk4_step (y,t) h)

let _ = loop 0.1 1 (1.0, 0.0)</lang>

Output:
t = 0.000000,	y = 1.000000,	err = 0
t = 1.000000,	y = 1.562500,	err = 1.45722e-07
t = 2.000000,	y = 3.999999,	err = 9.19479e-07
t = 3.000000,	y = 10.562497,	err = 2.90956e-06
t = 4.000000,	y = 24.999994,	err = 6.23491e-06
t = 5.000000,	y = 52.562489,	err = 1.08197e-05
t = 6.000000,	y = 99.999983,	err = 1.65946e-05
t = 7.000000,	y = 175.562476,	err = 2.35177e-05
t = 8.000000,	y = 288.999968,	err = 3.15652e-05
t = 9.000000,	y = 451.562459,	err = 4.07232e-05
t = 10.000000,	y = 675.999949,	err = 5.09833e-05

Octave

<lang octave>

  1. Applying the Runge-Kutta method (This code must be implement on a different file than the main one).

function temp = rk4(func,x,pvi,h) 

   K1 = h*func(x,pvi);
   K2 = h*func(x+0.5*h,pvi+0.5*K1);
   K3 = h*func(x+0.5*h,pvi+0.5*K2);
   K4 = h*func(x+h,pvi+K3);
   temp = pvi + (K1 + 2*K2 + 2*K3 + K4)/6;

endfunction

  1. Main Program.

f = @(t) (1/16)*((t.^2 + 4).^2); df = @(t,y) t*sqrt(y);

pvi = 1.0; h = 0.1; Yn = pvi;

for x = 0:h:10-h 

   pvi = rk4(df,x,pvi,h);
   Yn = [Yn pvi];

endfor

fprintf('Time \t Exact Value \t ODE4 Value \t Num. Error\n');

for i=0:10 

   fprintf('%d \t %.5f \t %.5f \t %.4g \n',i,f(i),Yn(1+i*10),f(i)-Yn(1+i*10));

endfor </lang>

Output:
Time     Exact Value     ODE4 Value      Num. Error
0        1.00000         1.00000         0
1        1.56250         1.56250         1.457e-007
2        4.00000         4.00000         9.195e-007
3        10.56250        10.56250        2.91e-006
4        25.00000        24.99999        6.235e-006
5        52.56250        52.56249        1.082e-005
6        100.00000       99.99998        1.659e-005
7        175.56250       175.56248       2.352e-005
8        289.00000       288.99997       3.157e-005
9        451.56250       451.56246       4.072e-005
10       676.00000       675.99995       5.098e-005

PARI/GP

Translation of: C

<lang parigp>rk4(f,dx,x,y)={ 

 my(k1=dx*f(x,y), k2=dx*f(x+dx/2,y+k1/2), k3=dx*f(x+dx/2,y+k2/2), k4=dx*f(x+dx,y+k3));
 y + (k1 + 2*k2 + 2*k3 + k4) / 6

}; rate(x,y)=x*sqrt(y); go()={ 

 my(x0=0,x1=10,dx=.1,n=1+(x1-x0)\dx,y=vector(n));
 y[1]=1;
 for(i=2,n,y[i]=rk4(rate, dx, x0 + dx * (i - 1), y[i-1]));
 print("x\ty\trel. err.\n------------");
 forstep(i=1,n,10,
   my(x=x0+dx*i,y2=(x^2/4+1)^2);
   print(x "\t" y[i] "\t" y[i]/y2 - 1)
 )

}; go()</lang>

Output:
x       y       rel. err.
------------
0.100000000     1       -0.00498131231
1.10000000      1.68999982      -0.00383519474
2.10000000      4.40999894      -0.00237694942
3.10000000      11.5599968      -0.00146924588
4.10000000      27.0399933      -0.000961094862
5.10000000      56.2499884      -0.000666538719
6.10000000      106.089982      -0.000485427212
7.10000000      184.959975      -0.000367681962
8.10000000      302.759966      -0.000287408941
9.10000000      470.889955      -0.000230470905

Pascal

Translation of: Ada

This code has been compiled using Free Pascal 2.6.2.

<lang pascal>program RungeKuttaExample;

uses sysutils;

type 

   TDerivative = function (t, y : Real) : Real;

procedure RungeKutta(yDer : TDerivative; 

                    var t, y : array of Real;
                    dt   : Real);

var 

   dy1, dy2, dy3, dy4 : Real;
   idx                : Cardinal;

begin 

   for idx := Low(t) to High(t) - 1 do
   begin
       dy1 := dt * yDer(t[idx],            y[idx]);
       dy2 := dt * yDer(t[idx] + dt / 2.0, y[idx] + dy1 / 2.0);
       dy3 := dt * yDer(t[idx] + dt / 2.0, y[idx] + dy2 / 2.0);
       dy4 := dt * yDer(t[idx] + dt,       y[idx] + dy3);
       
       t[idx + 1] := t[idx] + dt;
       y[idx + 1] := y[idx] + (dy1 + 2.0 * (dy2 + dy3) + dy4) / 6.0;
   end;

end;

function CalcError(t, y : Real) : Real; var 

   trueVal : Real;

begin 

   trueVal := sqr(sqr(t) + 4.0) / 16.0;
   CalcError := abs(trueVal - y);

end;

procedure Print(t, y : array of Real; 

               modnum : Integer);

var 

   idx : Cardinal;

begin 

   for idx := Low(t) to High(t) do
   begin
       if idx mod modnum = 0 then
       begin
           WriteLn(Format('y(%4.1f) = %12.8f  Error: %12.6e', 
               [t[idx], y[idx], CalcError(t[idx], y[idx])]));
       end;
   end;

end;

function YPrime(t, y : Real) : Real; begin 

   YPrime := t * sqrt(y);

end;

const 

   dt = 0.10;
   N = 100;

var 

   tArr, yArr : array [0..N] of Real;

begin 

   tArr[0] := 0.0;
   yArr[0] := 1.0;
   
   RungeKutta(@YPrime, tArr, yArr, dt);
   Print(tArr, yArr, 10);

end.</lang>

Output:
y( 0.0) =   1.00000000  Error: 0.00000E+000
y( 1.0) =   1.56249985  Error: 1.45722E-007
y( 2.0) =   3.99999908  Error: 9.19479E-007
y( 3.0) =  10.56249709  Error: 2.90956E-006
y( 4.0) =  24.99999377  Error: 6.23491E-006
y( 5.0) =  52.56248918  Error: 1.08197E-005
y( 6.0) =  99.99998341  Error: 1.65946E-005
y( 7.0) = 175.56247648  Error: 2.35177E-005
y( 8.0) = 288.99996843  Error: 3.15652E-005
y( 9.0) = 451.56245928  Error: 4.07232E-005
y(10.0) = 675.99994902  Error: 5.09833E-005

Perl

There are many ways of doing this. Here we define the runge_kutta function as a function of and , returning a closure which itself takes as argument and returns the next .

Notice how we have to use sprintf to deal with floating point rounding. See perlfaq4. <lang perl>sub runge_kutta { 

   my ($yp, $dt) = @_;
   sub {

my ($t, $y) = @_; my @dy = $dt * $yp->( $t , $y ); push @dy, $dt * $yp->( $t + $dt/2, $y + $dy[0]/2 ); push @dy, $dt * $yp->( $t + $dt/2, $y + $dy[1]/2 ); push @dy, $dt * $yp->( $t + $dt , $y + $dy[2] ); return $t + $dt, $y + ($dy[0] + 2*$dy[1] + 2*$dy[2] + $dy[3]) / 6; 

   }

}

my $RK = runge_kutta sub { $_[0] * sqrt $_[1] }, .1;

for( 

   my ($t, $y) = (0, 1);
   sprintf("%.0f", $t) <= 10;
   ($t, $y) = $RK->($t, $y)

) { 

   printf "y(%2.0f) = %12f ± %e\n", $t, $y, abs($y - ($t**2 + 4)**2 / 16)
   if sprintf("%.4f", $t) =~ /0000$/;

}</lang>

Output:
y( 0) =     1.000000 ± 0.000000e+00
y( 1) =     1.562500 ± 1.457219e-07
y( 2) =     3.999999 ± 9.194792e-07
y( 3) =    10.562497 ± 2.909562e-06
y( 4) =    24.999994 ± 6.234909e-06
y( 5) =    52.562489 ± 1.081970e-05
y( 6) =    99.999983 ± 1.659460e-05
y( 7) =   175.562476 ± 2.351773e-05
y( 8) =   288.999968 ± 3.156520e-05
y( 9) =   451.562459 ± 4.072316e-05
y(10) =   675.999949 ± 5.098329e-05

Perl 6

Works with: rakudo version 2016.03

<lang perl6>sub runge-kutta(&yp) { 

   return -> \t, \y, \δt {
       my $a = δt * yp( t, y );
       my $b = δt * yp( t + δt/2, y + $a/2 );
       my $c = δt * yp( t + δt/2, y + $b/2 );
       my $d = δt * yp( t + δt, y + $c );
       ($a + 2*($b + $c) + $d) / 6;
   }

}

constant δt = .1; my &δy = runge-kutta { $^t * sqrt($^y) };

loop ( 

   my ($t, $y) = (0, 1);
   $t <= 10;
   ($t, $y) »+=« (δt, δy($t, $y, δt))

) { 

   printf "y(%2d) = %12f ± %e\n", $t, $y, abs($y - ($t**2 + 4)**2 / 16)
   if $t %% 1;

}</lang>

Output:
y( 0) =     1.000000 ± 0.000000e+00
y( 1) =     1.562500 ± 1.457219e-07
y( 2) =     3.999999 ± 9.194792e-07
y( 3) =    10.562497 ± 2.909562e-06
y( 4) =    24.999994 ± 6.234909e-06
y( 5) =    52.562489 ± 1.081970e-05
y( 6) =    99.999983 ± 1.659460e-05
y( 7) =   175.562476 ± 2.351773e-05
y( 8) =   288.999968 ± 3.156520e-05
y( 9) =   451.562459 ± 4.072316e-05
y(10) =   675.999949 ± 5.098329e-05

Phix

Translation of: ERRE

<lang Phix>constant dt = 0.1 atom y = 1.0 printf(1," x true/actual y calculated y relative error\n") printf(1," --- ------------- ------------- --------------\n") for i=0 to 100 do 

   atom t = i*dt
   if integer(t) then
       atom act = power(t*t+4,2)/16
       printf(1,"%4.1f  %14.9f  %14.9f   %.9e\n",{t,act,y,abs(y-act)})
   end if
   atom k1 = t*sqrt(y),
        k2 = (t+dt/2)*sqrt(y+dt/2*k1),
        k3 = (t+dt/2)*sqrt(y+dt/2*k2),
        k4 = (t+dt)*sqrt(y+dt*k3)
   y += dt*(k1+2*(k2+k3)+k4)/6

end for</lang>

Output:
  x    true/actual y   calculated y    relative error
 ---   -------------   -------------   --------------
 0.0     1.000000000     1.000000000   0.000000000e+0
 1.0     1.562500000     1.562499854   1.457218921e-7
 2.0     4.000000000     3.999999081   9.194791999e-7
 3.0    10.562500000    10.562497090   2.909562447e-6
 4.0    25.000000000    24.999993765   6.234909363e-6
 5.0    52.562500000    52.562489180   1.081969741e-5
 6.0   100.000000000    99.999983405   1.659459641e-5
 7.0   175.562500000   175.562476482   2.351772874e-5
 8.0   289.000000000   288.999968435   3.156520142e-5
 9.0   451.562500000   451.562459277   4.072316033e-5
10.0   676.000000000   675.999949017   5.098329030e-5

PL/I

<lang PL/I> Runge_Kutta: procedure options (main); /* 10 March 2014 */ 

  declare (y, dy1, dy2, dy3, dy4) float (18);
  declare t fixed decimal (10,1);
  declare dt float (18) static initial (0.1);

  y = 1;
  do t = 0 to 10 by 0.1;
     dy1 = dt * ydash(t, y);
     dy2 = dt * ydash(t + dt/2, y + dy1/2);
     dy3 = dt * ydash(t + dt/2, y + dy2/2);
     dy4 = dt * ydash(t + dt,   y + dy3);

     if mod(t, 1.0) = 0 then
        put skip edit('y(', trim(t), ')=', y, ', error = ', abs(y - (t**2 + 4)**2 / 16 ))
                     (3 a, column(9), f(16,10), a, f(13,10));      
     y = y + (dy1 + 2*dy2 + 2*dy3 + dy4)/6;
  end;


ydash: procedure (t, y) returns (float(18)); 

  declare (t, y) float (18) nonassignable;
  return ( t*sqrt(y) );

end ydash;

end Runge_kutta; </lang>

Output:
y(0.0)=     1.0000000000, error =  0.0000000000
y(1.0)=     1.5624998543, error =  0.0000001457
y(2.0)=     3.9999990805, error =  0.0000009195
y(3.0)=    10.5624970904, error =  0.0000029096
y(4.0)=    24.9999937651, error =  0.0000062349
y(5.0)=    52.5624891803, error =  0.0000108197
y(6.0)=    99.9999834054, error =  0.0000165946
y(7.0)=   175.5624764823, error =  0.0000235177
y(8.0)=   288.9999684348, error =  0.0000315652
y(9.0)=   451.5624592768, error =  0.0000407232
y(10.0)=  675.9999490167, error =  0.0000509833

PowerShell

Works with: PowerShell version 4.0

<lang PowerShell> function Runge-Kutta (${function:F}, ${function:y}, $y0, $t0, $dt, $tEnd) { 

   function RK ($tn,$yn)  {
       $y1 = $dt*(F -t $tn -y $yn)
       $y2 = $dt*(F -t ($tn + (1/2)*$dt) -y ($yn + (1/2)*$y1))
       $y3 = $dt*(F -t ($tn + (1/2)*$dt) -y ($yn + (1/2)*$y2))
       $y4 = $dt*(F -t ($tn + $dt) -y ($yn + $y3))
       $yn + (1/6)*($y1 + 2*$y2 + 2*$y3 + $y4)
   }
   function time ($t0, $dt, $tEnd)  {
       $end = [MATH]::Floor(($tEnd - $t0)/$dt)
       foreach ($_ in 0..$end) { $_*$dt + $t0 }
   }
   $time, $yn, $t = (time $t0 $dt $tEnd), $y0, 0
   foreach ($tn in $time) {
       if($t -eq $tn) {
           [pscustomobject]@{
               t = "$tn"
               y = "$yn"
               error = "$([MATH]::abs($yn - (y $tn)))"
           }
           $t += 1
       }
       $yn = RK $tn $yn
   }

} function F ($t,$y) { 

   $t * [MATH]::Sqrt($y)

} function y ($t) { 

   (1/16) * [MATH]::Pow($t*$t + 4,2)

} $y0 = 1 $t0 = 0 $dt = 0.1 $tEnd = 10 Runge-Kutta F y $y0 $t0 $dt $tEnd </lang> Output: 

t                                    y                                    error                              
-                                    -                                    -----                              
0                                    1                                    0                                  
1                                    1.56249985427811                     1.45721892108597E-07               
2                                    3.9999990805208                      9.19479200778284E-07               
3                                    10.5624970904376                     2.90956244874963E-06               
4                                    24.9999937650906                     6.23490936391136E-06               
5                                    52.5624891803026                     1.08196974153429E-05               
6                                    99.9999834054036                     1.65945964170078E-05               
7                                    175.562476482271                     2.35177287493116E-05               
8                                    288.999968434799                     3.156520142511E-05                 
9                                    451.56245927684                      4.07231603389846E-05               
10                                   675.99994901671                      5.09832902935159E-05

PureBasic

Translation of: BBC Basic

<lang PureBasic>EnableExplicit Define.i i Define.d y=1.0, k1=0.0, k2=0.0, k3=0.0, k4=0.0, t=0.0

If OpenConsole() 

 For i=0 To 100
   t=i/10
   If Not i%10
     PrintN("y("+RSet(StrF(t,0),2," ")+") ="+RSet(StrF(y,4),9," ")+#TAB$+"Error ="+RSet(StrF(Pow(Pow(t,2)+4,2)/16-y,10),14," "))            
   EndIf    
   k1=t*Sqr(y)
   k2=(t+0.05)*Sqr(y+0.05*k1)
   k3=(t+0.05)*Sqr(y+0.05*k2)
   k4=(t+0.10)*Sqr(y+0.10*k3)    
   y+0.1*(k1+2*(k2+k3)+k4)/6    
 Next  
 Print("Press return to exit...") : Input()

EndIf End</lang>

Output:
y( 0) =   1.0000        Error =  0.0000000000
y( 1) =   1.5625        Error =  0.0000001457
y( 2) =   4.0000        Error =  0.0000009195
y( 3) =  10.5625        Error =  0.0000029096
y( 4) =  25.0000        Error =  0.0000062349
y( 5) =  52.5625        Error =  0.0000108197
y( 6) = 100.0000        Error =  0.0000165946
y( 7) = 175.5625        Error =  0.0000235177
y( 8) = 289.0000        Error =  0.0000315652
y( 9) = 451.5625        Error =  0.0000407232
y(10) = 675.9999        Error =  0.0000509833
Press return to exit...

Python

using lambda

<lang Python>def RK4(f): 

   return lambda t, y, dt: (
           lambda dy1: (
           lambda dy2: (
           lambda dy3: (
           lambda dy4: (dy1 + 2*dy2 + 2*dy3 + dy4)/6
           )( dt * f( t + dt  , y + dy3   ) )

)( dt * f( t + dt/2, y + dy2/2 ) ) )( dt * f( t + dt/2, y + dy1/2 ) ) )( dt * f( t , y ) )

def theory(t): return (t**2 + 4)**2 /16

from math import sqrt dy = RK4(lambda t, y: t*sqrt(y))

t, y, dt = 0., 1., .1 while t <= 10: 

   if abs(round(t) - t) < 1e-5:

print("y(%2.1f)\t= %4.6f \t error: %4.6g" % ( t, y, abs(y - theory(t)))) 

   t, y = t + dt, y + dy( t, y, dt )

</lang>

Output:
y(0.0)	= 1.000000 	 error:    0
y(1.0)	= 1.562500 	 error: 1.45722e-07
y(2.0)	= 3.999999 	 error: 9.19479e-07
y(3.0)	= 10.562497 	 error: 2.90956e-06
y(4.0)	= 24.999994 	 error: 6.23491e-06
y(5.0)	= 52.562489 	 error: 1.08197e-05
y(6.0)	= 99.999983 	 error: 1.65946e-05
y(7.0)	= 175.562476 	 error: 2.35177e-05
y(8.0)	= 288.999968 	 error: 3.15652e-05
y(9.0)	= 451.562459 	 error: 4.07232e-05
y(10.0)	= 675.999949 	 error: 5.09833e-05

Alternate solution

<lang python>from math import sqrt

def rk4(f, x0, y0, x1, n): 

   vx = [0] * (n + 1)
   vy = [0] * (n + 1)
   h = (x1 - x0) / float(n)
   vx[0] = x = x0
   vy[0] = y = y0
   for i in range(1, n + 1):
       k1 = h * f(x, y)
       k2 = h * f(x + 0.5 * h, y + 0.5 * k1)
       k3 = h * f(x + 0.5 * h, y + 0.5 * k2)
       k4 = h * f(x + h, y + k3)
       vx[i] = x = x0 + i * h
       vy[i] = y = y + (k1 + k2 + k2 + k3 + k3 + k4) / 6
   return vx, vy

def f(x, y): 

   return x * sqrt(y)

vx, vy = rk4(f, 0, 1, 10, 100) for x, y in list(zip(vx, vy))[::10]: 

   print("%4.1f %10.5f %+12.4e" % (x, y, y - (4 + x * x)**2 / 16))

0.0    1.00000  +0.0000e+00
1.0    1.56250  -1.4572e-07
2.0    4.00000  -9.1948e-07
3.0   10.56250  -2.9096e-06
4.0   24.99999  -6.2349e-06
5.0   52.56249  -1.0820e-05
6.0   99.99998  -1.6595e-05
7.0  175.56248  -2.3518e-05
8.0  288.99997  -3.1565e-05
9.0  451.56246  -4.0723e-05

10.0 675.99995 -5.0983e-05</lang>

R

<lang r>rk4 <- function(f, x0, y0, x1, n) { 

   vx <- double(n + 1)
   vy <- double(n + 1)
   vx[1] <- x <- x0
   vy[1] <- y <- y0
   h <- (x1 - x0)/n
   for(i in 1:n) {
       k1 <- h*f(x, y)
       k2 <- h*f(x + 0.5*h, y + 0.5*k1)
       k3 <- h*f(x + 0.5*h, y + 0.5*k2)
       k4 <- h*f(x + h, y + k3)
       vx[i + 1] <- x <- x0 + i*h
       vy[i + 1] <- y <- y + (k1 + k2 + k2 + k3 + k3 + k4)/6
   }
   cbind(vx, vy)

}

sol <- rk4(function(x, y) x*sqrt(y), 0, 1, 10, 100) cbind(sol, sol[, 2] - (4 + sol[, 1]^2)^2/16)[seq(1, 101, 10), ] 

     vx         vy              
[1,]  0   1.000000  0.000000e+00
[2,]  1   1.562500 -1.457219e-07
[3,]  2   3.999999 -9.194792e-07
[4,]  3  10.562497 -2.909562e-06
[5,]  4  24.999994 -6.234909e-06
[6,]  5  52.562489 -1.081970e-05
[7,]  6  99.999983 -1.659460e-05
[8,]  7 175.562476 -2.351773e-05
[9,]  8 288.999968 -3.156520e-05

[10,] 9 451.562459 -4.072316e-05 [11,] 10 675.999949 -5.098329e-05</lang>

Racket

See Euler method#Racket for implementation of simple general ODE-solver.

The Runge-Kutta method <lang racket> (define (RK4 F δt) 

 (λ (t y) 
   (define δy1 (* δt (F t y)))
   (define δy2 (* δt (F (+ t (* 1/2 δt)) (+ y (* 1/2 δy1)))))
   (define δy3 (* δt (F (+ t (* 1/2 δt)) (+ y (* 1/2 δy2)))))
   (define δy4 (* δt (F (+ t δt) (+ y δy1))))
   (list (+ t δt) 
         (+ y (* 1/6 (+ δy1 (* 2 δy2) (* 2 δy3) δy4))))))

</lang>

The method modifier which divides each time-step into n sub-steps: <lang racket> (define ((step-subdivision n method) F h) 

 (λ (x . y) (last (ODE-solve F (cons x y) 
                             #:x-max (+ x h) 
                             #:step (/ h n)
                             #:method method))))

</lang>

Usage: <lang racket> (define (F t y) (* t (sqrt y)))

(define (exact-solution t) (* 1/16 (sqr (+ 4 (sqr t)))))

(define numeric-solution 

   (ODE-solve F '(0 1) #:x-max 10 #:step 1 #:method (step-subdivision 10 RK4)))

(for ([s numeric-solution]) 

 (match-define (list t y) s)
 (printf "t=~a\ty=~a\terror=~a\n" t y (- y (exact-solution t))))

</lang>

Output:
t=0	y=1	                error=0
t=1	y=1.562499854278108	error=-1.4572189210859676e-07
t=2	y=3.999999080520799	error=-9.194792007782837e-07
t=3	y=10.562497090437551	error=-2.9095624487496252e-06
t=4	y=24.999993765090636	error=-6.234909363911356e-06
t=5	y=52.562489180302585	error=-1.0819697415342944e-05
t=6	y=99.99998340540358	error=-1.659459641700778e-05
t=7	y=175.56247648227125	error=-2.3517728749311573e-05
t=8	y=288.9999684347986	error=-3.156520142510999e-05
t=9	y=451.56245927683966	error=-4.07231603389846e-05
t=10	y=675.9999490167097	error=-5.098329029351589e-05

Graphical representation:

<lang racket> > (require plot) > (plot (list (function exact-solution 0 10 #:label "Exact solution") 

             (points numeric-solution #:label "Runge-Kutta method"))
  #:x-label "t" #:y-label "y(t)")

</lang>

REXX

       The Runge─Kutta method is used to solve the following differential equation:

  ╔═══════════════╗             ______     ╔══ the exact solution:  y(t)= (t²+4)²/16 ══╗
  ╚═══════════════╝   y'(t)=t² √ y(t)      ╚═══════════════════════════════════════════╝

<lang rexx>/*REXX program uses the Runge─Kutta method to solve the equation: y'(t)=t² √[y(t)] */ numeric digits 40; f=digits() % 4 /*use 40 decimal digs, but only show 10*/ x0=0; x1=10; dx= .1 /*define variables: X0 X1 DX */ n=1 + (x1-x0) / dx y.=1; do m=1 for n-1; p=m-1; y.m=RK4(dx, x0 + dx*p, y.p) 

                        end   /*m*/             /*   [↑]  use 4th order Runge─Kutta.   */

w=digits() % 2 /*W: width used for displaying numbers.*/ say center('X', f, "═") center('Y', w+2, "═") center("relative error", w+8, '═') /*hdr*/ 

               do i=0  to n-1  by 10;  x=(x0 + dx*i) / 1;           $=y.i/(x*x/4+1)**2 -1
               say  center(x, f)     fmt(y.i)     left(, 2 + ($>=0) )        fmt($)
               end   /*i*/                      /*└┴┴┴───◄─────── aligns positive #'s. */

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ fmt: z=right( format( arg(1), w, f), w); hasE=pos('E', z)\==0; has.=pos(., z)\==0 

     jus=has. & \hasE;        if jus  then z=left( strip( strip(z, 'T', 0),  "T", .),  w)
     return translate(right(z, (z>=0) +  w  +  5*hasE  +  2*(jus & (z<0) ) ),  'e',  "E")

/*──────────────────────────────────────────────────────────────────────────────────────*/ RK4: procedure; parse arg dx,x,y; dxH=dx/2; k1= dx * (x ) * sqrt(y ) 

                                                 k2= dx  *  (x + dxH)  *  sqrt(y + k1/2)
                                                 k3= dx  *  (x + dxH)  *  sqrt(y + k2/2)
                                                 k4= dx  *  (x + dx )  *  sqrt(y + k3  )
     return y + (k1 + k2*2 + k3*2 + k4) / 6

/*──────────────────────────────────────────────────────────────────────────────────────*/ sqrt: procedure; parse arg x; if x=0 then return 0; d=digits(); m.=9; numeric form; h=d+6 

     numeric digits;  parse value format(x,2,1,,0) 'E0' with g 'E' _ .;  g=g * .5'e'_ % 2
       do j=0  while h>9;      m.j=h;               h=h%2+1;       end /*j*/
       do k=j+5  to 0  by -1;  numeric digits m.k;  g=(g+x/g)*.5;  end /*k*/;    return g</lang>

Programming note:   the   fmt   function is used to align the output with attention paid to the different ways some
REXXes format numbers that are in floating point representation.


output   when using Regina REXX:
════X═════ ══════════Y═══════════ ═══════relative error═══════
    0               1                         0
    1               1.5624998543             -9.3262010935e-8
    2               3.9999990805             -2.2986980019e-7
    3              10.5624970904             -2.7546153356e-7
    4              24.9999937651             -2.4939637459e-7
    5              52.5624891803             -2.0584442174e-7
    6              99.9999834054             -1.6594596403e-7
    7             175.5624764823             -1.3395644713e-7
    8             288.9999684348             -1.0922215040e-7
    9             451.5624592768             -9.0182777476e-8
    10            675.9999490167             -7.5419068846e-8
output   when using PC/REXX, Personal REXX, ROO, or R4 REXX:
════X═════ ══════════Y═══════════ ═══════relative error═══════
    0               1                         0
    1               1.5624998543             -0.0000000933
    2               3.9999990805             -0.0000002299
    3              10.5624970904             -0.0000002755
    4              24.9999937651             -0.0000002494
    5              52.5624891803             -0.0000002058
    6              99.9999834054             -0.0000001659
    7             175.5624764823             -0.000000134
    8             288.9999684348             -0.0000001092
    9             451.5624592768             -0.0000000902
    10            675.9999490167             -0.0000000754

Ring

<lang ring> decimals(8) y = 1.0 for i = 0 to 100 

   t = i  / 10
   if t = floor(t) 
      actual = (pow((pow(t,2) + 4),2)) / 16
      see "y(" + t + ") = " + y + "  error = " + (actual - y) + nl ok
   k1 =  t * sqrt(y)
   k2 = (t + 0.05) * sqrt(y + 0.05 * k1)
   k3 = (t + 0.05) * sqrt(y + 0.05 * k2)
   k4 = (t + 0.10) * sqrt(y + 0.10 * k3)
   y += 0.1 * (k1 + 2 * (k2 + k3) + k4) / 6

next </lang>

Output: 

y(0) = 1  error = 0
y(1) = 1.56249985  error = 0.00000015
y(2) = 3.99999908  error = 0.00000092
y(3) = 10.56249709  error = 0.00000291
y(4) = 24.99999377  error = 0.00000623
y(5) = 52.56248918  error = 0.00001082
y(6) = 99.99998341  error = 0.00001659
y(7) = 175.56247648  error = 0.00002352
y(8) = 288.99996843  error = 0.00003157
y(9) = 451.56245928  error = 0.00004072
y(10) = 675.99994902  error = 0.00005098

Ruby

<lang ruby>def calc_rk4(f) 

 return ->(t,y,dt){
        ->(dy1   ){
        ->(dy2   ){
        ->(dy3   ){
        ->(dy4   ){ ( dy1 + 2*dy2 + 2*dy3 + dy4 ) / 6 }.call(
          dt * f.call( t + dt  , y + dy3   ))}.call(
          dt * f.call( t + dt/2, y + dy2/2 ))}.call(
          dt * f.call( t + dt/2, y + dy1/2 ))}.call(
          dt * f.call( t       , y         ))}

end

TIME_MAXIMUM, WHOLE_TOLERANCE = 10.0, 1.0e-5 T_START, Y_START, DT = 0.0, 1.0, 0.10

def my_diff_eqn(t,y) ; t * Math.sqrt(y)  ; end def my_solution(t ) ; (t**2 + 4)**2 / 16  ; end def find_error(t,y) ; (y - my_solution(t)).abs  ; end def is_whole?(t ) ; (t.round - t).abs < WHOLE_TOLERANCE ; end

dy = calc_rk4( ->(t,y){my_diff_eqn(t,y)} )

t, y = T_START, Y_START while t <= TIME_MAXIMUM 

 printf("y(%4.1f)\t= %12.6f \t error: %12.6e\n",t,y,find_error(t,y)) if is_whole?(t)
 t, y = t + DT, y + dy.call(t,y,DT)

end</lang>

Output:
y( 0.0)	=     1.000000 	 error: 0.000000e+00
y( 1.0)	=     1.562500 	 error: 1.457219e-07
y( 2.0)	=     3.999999 	 error: 9.194792e-07
y( 3.0)	=    10.562497 	 error: 2.909562e-06
y( 4.0)	=    24.999994 	 error: 6.234909e-06
y( 5.0)	=    52.562489 	 error: 1.081970e-05
y( 6.0)	=    99.999983 	 error: 1.659460e-05
y( 7.0)	=   175.562476 	 error: 2.351773e-05
y( 8.0)	=   288.999968 	 error: 3.156520e-05
y( 9.0)	=   451.562459 	 error: 4.072316e-05
y(10.0)	=   675.999949 	 error: 5.098329e-05

Run BASIC

<lang Runbasic>y = 1 while t <= 10 

  k1	=  t        * sqr(y)
  k2	= (t + .05) * sqr(y + .05 * k1)
  k3	= (t + .05) * sqr(y + .05 * k2)
  k4	= (t + .1)  * sqr(y + .1  * k3)

if right$(using("##.#",t),1) = "0" then print "y(";using("##",t);") ="; using("####.#######", y);chr$(9);"Error ="; (((t^2 + 4)^2) /16) -y 

   y = y + .1 *(k1 + 2 * (k2 + k3) + k4) / 6
  t = t + .1

wend end</lang>

Output:
y( 0) =   1.0000000	Error =0
y( 1) =   1.5624999	Error =1.45721892e-7
y( 2) =   3.9999991	Error =9.19479203e-7
y( 3) =  10.5624971	Error =2.90956246e-6
y( 4) =  24.9999938	Error =6.23490939e-6
y( 5) =  52.5624892	Error =1.08196973e-5
y( 6) =  99.9999834	Error =1.65945961e-5
y( 7) = 175.5624765	Error =2.3517728e-5
y( 8) = 288.9999684	Error =3.15652e-5
y( 9) = 451.5624593	Error =4.07231581e-5
y(10) = 675.9999490	Error =5.09832864e-5

Rust

This is a translation of the javascript solution with some minor differences. <lang rust>fn runge_kutta4(fx: &Fn(f64, f64) -> f64, x: f64, y: f64, dx: f64) -> f64 { 

   let k1 = dx * fx(x, y);
   let k2 = dx * fx(x + dx / 2.0, y + k1 / 2.0);
   let k3 = dx * fx(x + dx / 2.0, y + k2 / 2.0);
   let k4 = dx * fx(x + dx, y + k3);

   y + (k1 + 2.0 * k2 + 2.0 * k3 + k4) / 6.0

}

fn f(x: f64, y: f64) -> f64 { 

   x * y.sqrt()

}

fn actual(x: f64) -> f64 { 

   (1.0 / 16.0) * (x * x + 4.0).powi(2)

}

fn main() { 

   let mut y = 1.0;
   let mut x = 0.0;
   let step = 0.1;
   let max_steps = 101;
   let sample_every_n = 10;

   for steps in 0..max_steps {
       if steps % sample_every_n == 0 {
           println!("y({}):\t{:.10}\t\t {:E}", x, y, actual(x) - y)
       }

       y = runge_kutta4(&f, x, y, step);

       x = ((x * 10.0) + (step * 10.0)) / 10.0;
   }

}</lang> 

y(0):	1.0000000000		 0E0
y(1):	1.5624998543		 1.4572189210859676E-7
y(2):	3.9999990805		 9.194792007782837E-7
y(3):	10.5624970904		 2.9095624487496252E-6
y(4):	24.9999937651		 6.234909363911356E-6
y(5):	52.5624891803		 1.0819697415342944E-5
y(6):	99.9999834054		 1.659459641700778E-5
y(7):	175.5624764823		 2.3517728749311573E-5
y(8):	288.9999684348		 3.156520142510999E-5
y(9):	451.5624592768		 4.07231603389846E-5
y(10):	675.9999490167		 5.098329029351589E-5

Scala

<lang scala>object Main extends App { 

  val f = (t: Double, y: Double) => t * Math.sqrt(y) // Runge-Kutta solution
  val g = (t: Double) => Math.pow(t * t + 4, 2) / 16 // Exact solution
  new Calculator(f, Some(g)).compute(100, 0, .1, 1)

}

class Calculator(f: (Double, Double) => Double, g: Option[Double => Double] = None) { 

  def compute(counter: Int, tn: Double, dt: Double, yn: Double): Unit = {
     if (counter % 10 == 0) {
        val c = (x: Double => Double) => (t: Double) => {
           val err = Math.abs(x(t) - yn)
           f" Error: $err%7.5e"
        }
        val s = g.map(c(_)).getOrElse((x: Double) => "") // If we don't have exact solution, just print nothing
        println(f"y($tn%4.1f) = $yn%12.8f${s(tn)}") // Else, print Error estimation here
     }
     if (counter > 0) {
        val dy1 = dt * f(tn, yn)
        val dy2 = dt * f(tn + dt / 2, yn + dy1 / 2)
        val dy3 = dt * f(tn + dt / 2, yn + dy2 / 2)
        val dy4 = dt * f(tn + dt, yn + dy3)
        val y = yn + (dy1 + 2 * dy2 + 2 * dy3 + dy4) / 6
        val t = tn + dt
        compute(counter - 1, t, dt, y)
     }
  }

}</lang> 

y( 0.0) =   1.00000000 Error: 0.00000e+00
y( 1.0) =   1.56249985 Error: 1.45722e-07
y( 2.0) =   3.99999908 Error: 9.19479e-07
y( 3.0) =  10.56249709 Error: 2.90956e-06
y( 4.0) =  24.99999377 Error: 6.23491e-06
y( 5.0) =  52.56248918 Error: 1.08197e-05
y( 6.0) =  99.99998341 Error: 1.65946e-05
y( 7.0) = 175.56247648 Error: 2.35177e-05
y( 8.0) = 288.99996843 Error: 3.15652e-05
y( 9.0) = 451.56245928 Error: 4.07232e-05
y(10.0) = 675.99994902 Error: 5.09833e-05

Sidef

Translation of: Perl 6

<lang ruby>func runge_kutta(yp) { 

   func (t, y, δt) {
       var a = (δt * yp(t, y));
       var b = (δt * yp(t + δt/2, y + a/2));
       var c = (δt * yp(t + δt/2, y + b/2));
       var d = (δt * yp(t + δt, y + c));
       (a + 2*(b + c) + d) / 6;
   }

}

define δt = 0.1; var δy = runge_kutta(func(t, y) { t * y.sqrt });

var(t, y) = (0, 1); loop { 

   t.is_int &&
       printf("y(%2d) = %12f ± %e\n", t, y, abs(y - ((t**2 + 4)**2 / 16)));
   t <= 10 || break;
   y += δy(t, y, δt);
   t += δt;

}</lang>

Output:
y( 0) =     1.000000 ± 0.000000e+00
y( 1) =     1.562500 ± 1.457219e-07
y( 2) =     3.999999 ± 9.194792e-07
y( 3) =    10.562497 ± 2.909562e-06
y( 4) =    24.999994 ± 6.234909e-06
y( 5) =    52.562489 ± 1.081970e-05
y( 6) =    99.999983 ± 1.659460e-05
y( 7) =   175.562476 ± 2.351773e-05
y( 8) =   288.999968 ± 3.156520e-05
y( 9) =   451.562459 ± 4.072316e-05
y(10) =   675.999949 ± 5.098329e-05

Standard ML

<lang sml>fun step y' (tn,yn) dt = 

   let
       val dy1 = dt * y'(tn,yn)
       val dy2 = dt * y'(tn + 0.5 * dt, yn + 0.5 * dy1)
       val dy3 = dt * y'(tn + 0.5 * dt, yn + 0.5 * dy2)
       val dy4 = dt * y'(tn + dt, yn + dy3)
   in
       (tn + dt, yn + (1.0 / 6.0) * (dy1 + 2.0*dy2 + 2.0*dy3 + dy4))
   end

(* Suggested test case *) fun testy' (t,y) = 

   t * Math.sqrt y

fun testy t = 

   (1.0 / 16.0) * Math.pow(Math.pow(t,2.0) + 4.0, 2.0)

(* Test-runner that iterates the step function and prints the results. *) fun test t0 y0 dt steps print_freq y y' = 

   let
       fun loop i (tn,yn) =
           if i = steps then ()
           else
               let
                   val (t1,y1) = step y' (tn,yn) dt
                   val y1' = y tn
                   val () = if i mod print_freq = 0 then
                                (print ("Time: " ^ Real.toString tn ^ "\n");
                                 print ("Exact: " ^ Real.toString y1' ^ "\n");
                                 print ("Approx: " ^ Real.toString yn ^ "\n");
                                 print ("Error: " ^ Real.toString (y1' - yn) ^ "\n\n"))
                            else ()
                in
                    loop (i+1) (t1,y1)
               end
   in
       loop 0 (t0,y0)
   end

(* Run the suggested test case *) val () = test 0.0 1.0 0.1 101 10 testy testy'</lang>

Output:
Time: 0.0
Exact: 1.0
Approx: 1.0
Error: ~1.11022302463E~16

Time: 1.0
Exact: 1.5625
Approx: 1.56249985428
Error: 1.45722452549E~07

Time: 2.0
Exact: 4.0
Approx: 3.99999908052
Error: 9.19479203443E~07

Time: 3.0
Exact: 10.5625
Approx: 10.5624970904
Error: 2.90956245586E~06

Time: 4.0
Exact: 25.0
Approx: 24.9999937651
Error: 6.23490938878E~06

Time: 5.0
Exact: 52.5625
Approx: 52.5624891803
Error: 1.08196973727E~05

Time: 6.0
Exact: 100.0
Approx: 99.9999834054
Error: 1.65945961186E~05

Time: 7.0
Exact: 175.5625
Approx: 175.562476482
Error: 2.35177280956E~05

Time: 8.0
Exact: 289.0
Approx: 288.999968435
Error: 3.15651997767E~05

Time: 9.0
Exact: 451.5625
Approx: 451.562459277
Error: 4.07231581221E~05

Time: 10.0
Exact: 676.0
Approx: 675.999949017
Error: 5.09832866555E~05

Stata

<lang stata>function rk4(f, t0, y0, t1, n) { h = (t1-t0)/(n-1) a = J(n, 2, 0) a[1, 1] = t = t0 a[1, 2] = y = y0 for (i=2; i<=n; i++) { k1 = h*(*f)(t, y) k2 = h*(*f)(t+0.5*h, y+0.5*k1) k3 = h*(*f)(t+0.5*h, y+0.5*k2) k4 = h*(*f)(t+h, y+k3) t = t+h y = y+(k1+2*k2+2*k3+k4)/6 a[i, 1] = t a[i, 2] = y } return(a) }

function f(t, y) { return(t*sqrt(y)) }

a = rk4(&f(), 0, 1, 10, 101) t = a[., 1] a = a, a[., 2]:-(t:^2:+4):^2:/16 a[range(1,101,10), .] 

                  1              2              3
    +----------------------------------------------+
  1 |             0              1              0  |
  2 |             1    1.562499854   -1.45722e-07  |
  3 |             2    3.999999081   -9.19479e-07  |
  4 |             3    10.56249709   -2.90956e-06  |
  5 |             4    24.99999377   -6.23491e-06  |
  6 |             5    52.56248918   -.0000108197  |
  7 |             6    99.99998341   -.0000165946  |
  8 |             7    175.5624765   -.0000235177  |
  9 |             8    288.9999684   -.0000315652  |
 10 |             9    451.5624593   -.0000407232  |
 11 |            10     675.999949   -.0000509833  |
    +----------------------------------------------+</lang>

Swift

Translation of: C

<lang Swift>import Foundation

func rk4(dx: Double, x: Double, y: Double, f: (Double, Double) -> Double) -> Double { 

   let k1 = dx * f(x, y)
   let k2 = dx * f(x + dx / 2, y + k1 / 2)
   let k3 = dx * f(x + dx / 2, y + k2 / 2)
   let k4 = dx * f(x + dx, y + k3)

   return y + (k1 + 2 * k2 + 2 * k3 + k4) / 6

}

var y = [Double]() var x: Double = 0.0 var y2: Double = 0.0

var x0: Double = 0.0 var x1: Double = 10.0 var dx: Double = 0.1

var i = 0 var n = Int(1 + (x1 - x0) / dx)

y.append(1) for i in 1..<n { 

   y.append(rk4(dx, x: x0 + dx * (Double(i) - 1), y: y[i - 1]) { (x: Double, y: Double) -> Double in
       return x * sqrt(y)
   })

}

print(" x y rel. err.") print("------------------------------")

for (var i = 0; i < n; i += 10) { 

   x = x0 + dx * Double(i)
   y2 = pow(x * x / 4 + 1, 2)

   print(String(format: "%2g  %11.6g    %11.5g", x, y[i], y[i]/y2 - 1))

}</lang>

Output:
 x         y        rel. err.
------------------------------
 0            1              0
 1       1.5625    -9.3262e-08
 2            4    -2.2987e-07
 3      10.5625    -2.7546e-07
 4           25     -2.494e-07
 5      52.5625    -2.0584e-07
 6          100    -1.6595e-07
 7      175.562    -1.3396e-07
 8          289    -1.0922e-07
 9      451.562    -9.0183e-08
10          676    -7.5419e-08

Tcl

<lang tcl>package require Tcl 8.5

  1. Hack to bring argument function into expression

proc tcl::mathfunc::dy {t y} {upvar 1 dyFn dyFn; $dyFn $t $y}

proc rk4step {dyFn y* t* dt} { 

   upvar 1 ${y*} y ${t*} t
   set dy1 [expr {$dt * dy($t,       $y)}]
   set dy2 [expr {$dt * dy($t+$dt/2, $y+$dy1/2)}]
   set dy3 [expr {$dt * dy($t+$dt/2, $y+$dy2/2)}]
   set dy4 [expr {$dt * dy($t+$dt,   $y+$dy3)}]
   set y [expr {$y + ($dy1 + 2*$dy2 + 2*$dy3 + $dy4)/6.0}]
   set t [expr {$t + $dt}]

}

proc y {t} {expr {($t**2 + 4)**2 / 16}} proc δy {t y} {expr {$t * sqrt($y)}}

proc printvals {t y} { 

   set err [expr {abs($y - [y $t])}]
   puts [format "y(%.1f) = %.8f\tError: %.8e" $t $y $err]

}

set t 0.0 set y 1.0 set dt 0.1 printvals $t $y for {set i 1} {$i <= 101} {incr i} { 

   rk4step  δy  y t  $dt
   if {$i%10 == 0} {

printvals $t $y 

   }

}</lang>

Output:
y(0.0) = 1.00000000	Error: 0.00000000e+00
y(1.0) = 1.56249985	Error: 1.45721892e-07
y(2.0) = 3.99999908	Error: 9.19479203e-07
y(3.0) = 10.56249709	Error: 2.90956245e-06
y(4.0) = 24.99999377	Error: 6.23490939e-06
y(5.0) = 52.56248918	Error: 1.08196973e-05
y(6.0) = 99.99998341	Error: 1.65945961e-05
y(7.0) = 175.56247648	Error: 2.35177280e-05
y(8.0) = 288.99996843	Error: 3.15652000e-05
y(9.0) = 451.56245928	Error: 4.07231581e-05
y(10.0) = 675.99994902	Error: 5.09832864e-05

zkl

Translation of: OCaml

<lang zkl>fcn yp(t,y) { t * y.sqrt() } fcn exact(t){ u:=0.25*t*t + 1.0; u*u }

fcn rk4_step([(y,t)],h){ 

  k1:=h * yp(t,y);
  k2:=h * yp(t + 0.5*h, y + 0.5*k1);
  k3:=h * yp(t + 0.5*h, y + 0.5*k2);
  k4:=h * yp(t + h, y + k3);
  T(y + (k1+k4)/6.0 + (k2+k3)/3.0, t + h);

}

fcn loop(h,n,[(y,t)]){ 

  if(n % 10 == 1)
     print("t = %f,\ty = %f,\terr = %g\n".fmt(t,y,(y - exact(t)).abs()));
  if(n < 102) return(loop(h,(n+1),rk4_step(T(y,t),h))) //tail recursion

}</lang>

Output:
loop(0.1,1,T(1.0, 0.0))
t = 0.000000,	y = 1.000000,	err = 0
t = 1.000000,	y = 1.562500,	err = 1.45722e-07
t = 2.000000,	y = 3.999999,	err = 9.19479e-07
t = 3.000000,	y = 10.562497,	err = 2.90956e-06
t = 4.000000,	y = 24.999994,	err = 6.23491e-06
t = 5.000000,	y = 52.562489,	err = 1.08197e-05
t = 6.000000,	y = 99.999983,	err = 1.65946e-05
t = 7.000000,	y = 175.562476,	err = 2.35177e-05
t = 8.000000,	y = 288.999968,	err = 3.15652e-05
t = 9.000000,	y = 451.562459,	err = 4.07232e-05
t = 10.000000,	y = 675.999949,	err = 5.09833e-05
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