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{{task}} Given the example Differential equation: :$y\text{'}\left(t\right) = t \times \sqrt \left\{y\left(t\right)\right\}$ With initial condition: :$t_0 = 0$ and $y_0 = y\left(t_0\right) = y\left(0\right) = 1$ This equation has an exact solution: :$y\left(t\right) = \tfrac\left\{1\right\}\left\{16\right\}\left(t^2 +4\right)^2$ ;Task Demonstrate the commonly used explicit [[wp:Runge–Kutta_methods#Common_fourth-order_Runge.E2.80.93Kutta_method|fourth-order Runge–Kutta method]] to solve the above differential equation.

• Solve the given differential equation over the range $t = 0 \ldots 10$ with a step value of $\delta t=0.1$ (101 total points, the first being given)
• Print the calculated values of $y$ at whole numbered $t$'s ($0.0, 1.0, \ldots 10.0$) along with error as compared to the exact solution. ;Method summary Starting with a given $y_n$ and $t_n$ calculate: :$\delta y_1 = \delta t\times y\text{'}\left(t_n, y_n\right)\quad$ :$\delta y_2 = \delta t\times y\text{'}\left(t_n + \tfrac\left\{1\right\}\left\{2\right\}\delta t , y_n + \tfrac\left\{1\right\}\left\{2\right\}\delta y_1\right)$ :$\delta y_3 = \delta t\times y\text{'}\left(t_n + \tfrac\left\{1\right\}\left\{2\right\}\delta t , y_n + \tfrac\left\{1\right\}\left\{2\right\}\delta y_2\right)$ :$\delta y_4 = \delta t\times y\text{'}\left(t_n + \delta t , y_n + \delta y_3\right)\quad$ then: :$y_\left\{n+1\right\} = y_n + \tfrac\left\{1\right\}\left\{6\right\} \left(\delta y_1 + 2\delta y_2 + 2\delta y_3 + \delta y_4\right)$ :$t_\left\{n+1\right\} = t_n + \delta t\quad$

with Ada.Text_IO; use Ada.Text_IO;
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;


{{out}}

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


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



{{out}}


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


∇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]
∇



{{out}}


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


# syntax: GAWK -f RUNGE-KUTTA_METHOD.AWK
# 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)
}



{{out}}


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



=

## BBC BASIC

=

      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%


{{out}}

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



==={{header|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=TSQR(Y) 150 LET K2=(T+.05)SQR(Y+.05K1) 160 LET K3=(T+.05)SQR(Y+.05K2) 170 LET K4=(T+.1)SQR(Y+.1K3) 180 LET Y=Y+.1(K1+2*(K2+K3)+K4)/6 190 NEXT  ## C 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; }  {{out}} (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#  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; } } }  ## C++ Using Lambdas /* * compiled with gcc 5.4: * g++-mp-5 -std=c++14 -o rk4 rk4.cc * */ # include <iostream> # 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; }  ## Common 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))  ## Crystal {{trans|Run Basic and Ruby output}} y, t = 1, 0 while t <= 10 k1 = t * Math.sqrt(y) k2 = (t + 0.05) * Math.sqrt(y + 0.05 * k1) k3 = (t + 0.05) * Math.sqrt(y + 0.05 * k2) k4 = (t + 0.1) * Math.sqrt(y + 0.1 * k3) printf("y(%4.1f)\t= %12.6f \t error: %12.6e\n", t, y, (((t**2 + 4)**2 / 16) - y )) if (t.round - t).abs < 1.0e-5 y += 0.1 * (k1 + 2 * (k2 + k3) + k4) / 6 t += 0.1 end  {{out}}  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  ## D {{trans|Ada}} 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])); }  {{out}} 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 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;
}
}


{{out}}


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


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


{{out}}


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 Sharp

{{works with|F# interactive (fsi.exe)}}


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) )


{{out}}


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

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



{{out}}


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

{{trans|BBC BASIC}}

' 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


{{out}}

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


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  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}} 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)) }  {{out}}  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  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;} }  {{out}}  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. 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)


Example executed in GHCi:

*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)


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

Or, disaggregated a little, and expressed in terms of a single scanl:

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)


{{Out}}

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:'''

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
'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
)


'''Example:'''

   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 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


'''Alternative solution:'''

The following solution replaces the for loop as well as the calculation of the increments (ks) with an accumulating suffix.

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


Use: report_err report_whole fyp rk4 1 0 10 0.1

## 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]));
}
}

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


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;
}



{{out}}


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

(() => {
'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;

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


{{Out}}

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:

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;


### The Example Differential Equation and its Exact Solution

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

# The exact solution of yprime:
def actual:
. as $t | (($t*$t) + 4 ) | . * . / 16;  ### dy/dt The first solution presented here uses the terminology and style of the Perl 6 version. '''Generic filters:''' # n is the number of decimal places of precision def round(n): (if . < 0 then -1 else 1 end) as$s
| $s*10*.*n | if (floor % 10) > 4 then (.+5) else . end | ./10 | floor/n | .*$s;

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

# Is the input an integer?
def integerq: ((. - ((.+.01) | floor)) | abs) < 0.01;


'''dy(f)'''

def dt: 0.1;

# 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)) + $d) / 6 ; # Input: [t,y] def dy(f): runge_kutta(f);  ''' Example''': # state: [t,y] [0,1] | while( .[0] <= 10; .[0] as$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


{{out}}

$time jq -r -n -f rk4.pl.jq y(0) = 1 ± 0 y(1) = 1.5625 ± 1.4572189210859676e-07 y(2) = 4 ± 9.194792029987298e-07 y(3) = 10.5625 ± 2.9095624576314094e-06 y(4) = 25 ± 6.234909392333066e-06 y(5) = 52.5625 ± 1.081969734428867e-05 y(6) = 100 ± 1.659459609015812e-05 y(7) = 175.5625 ± 2.3517728038768837e-05 y(8) = 289 ± 3.156520000402452e-05 y(9) = 451.5625 ± 4.072315812209126e-05 y(10) = 675.9999 ± 5.0983286655537086e-05 real 0m0.048s user 0m0.013s sys 0m0.006s  ### newRK4Step The second solution follows the nomenclature and style of the Go solution on this page. In the following notes: • ypFunc denotes the type of a jq filter that maps [t, y] to a number; • ypStepFunc denotes the type of a jq filter that maps [t, y, dt] to a number. The heart of the program is the filter newRK4Step(yp), which is of type ypStepFunc and performs a single step of the fourth-order Runge-Kutta method, provided yp is of type ypFunc. # Input: [t, y, dt] def newRK4Step(yp): .[0] as$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 ; # main(t0; y0; tFinal; dtPrint) main(0; 1; 10; 1)  {{out}} $ 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


## Julia

{{works with|Julia|0.6}}

== Using lambda expressions == {{trans|Python}}

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


{{out}}

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 == {{trans|Python}}

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


## Kotlin

// 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)
}


{{out}}


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


'[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



{{Out}}


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

(* 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[{y[[1]], y[[2]], Abs[y[[2]] - 1/16 (y[[1]]^2 + 4)^2]},
{y,  solution[[1 ;; 101 ;; 10]]}]



## 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 [http://www.mathworks.com/matlabcentral/answers/98293-is-there-a-fixed-step-ordinary-differential-equation-ode-solver-in-matlab-8-0-r2012b#answer_107643 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.

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


{{out}}


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

/* 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]);  =={{header|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 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)  {{out}} 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 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();
}
}


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

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)


{{out}}

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


#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

#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



{{out}}


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

{{trans|C}}

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()


{{out}}

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

This code has been compiled using Free Pascal 2.6.2.

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.


{{out}}

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 $y\text{'}$ and $\delta t$, returning a closure which itself takes $\left(t, y\right)$ as argument and returns the next $\left(t, y\right)$.

Notice how we have to use sprintf to deal with floating point rounding. See perlfaq4.

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", $t) =~ /0000$/;
}


{{out}}

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|2016.03}}

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; }  {{out}} 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 {{trans|ERRE}} 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  {{out}}  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  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;  {{out}}  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|4.0}}  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  $tEnd  Output:  t y error - - ----- 0 1 0 1 1.56249985427811 1.45721892108597E-07 2 3.9999990805208 9.19479200778284E-07 3 10.5624970904376 2.90956244874963E-06 4 24.9999937650906 6.23490936391136E-06 5 52.5624891803026 1.08196974153429E-05 6 99.9999834054036 1.65945964170078E-05 7 175.562476482271 2.35177287493116E-05 8 288.999968434799 3.156520142511E-05 9 451.56245927684 4.07231603389846E-05 10 675.99994901671 5.09832902935159E-05  ## PureBasic {{trans|BBC Basic}} EnableExplicit Define.i i Define.d y=1.0, k1=0.0, k2=0.0, k3=0.0, k4=0.0, t=0.0 If OpenConsole() For i=0 To 100 t=i/10 If Not i%10 PrintN("y("+RSet(StrF(t,0),2," ")+") ="+RSet(StrF(y,4),9," ")+#TAB$+"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
EndIf
End


{{out}}

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


## Python

### using lambda

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 )



{{Out}}

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

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


## 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


## Racket

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

The Runge-Kutta method


(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))))))



The method modifier which divides each time-step into ''n'' sub-steps:


(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))))



Usage:


(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))))



{{out}}


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:


> (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)")



[[File:runge-kutta.png]]

## 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)      ╚═══════════════════════════════════════════╝


/*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  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. {{out|output|text= 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  {{out|output|text= 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  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  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 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  {{Out}}  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 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  {{out}} 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. 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; } }   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 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) } } }   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 {{trans|Perl 6}} 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; }  {{out}}  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 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'  {{out}} 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 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 | +----------------------------------------------+  ## Swift {{trans|C}} 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)) }  {{out}}  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 package require Tcl 8.5 # 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
}
}


{{out}}

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

{{trans|OCaml}}

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
}


{{out}}


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

`