Runge-Kutta 4th Order Method to Solve Differential Equation

Given following inputs,

  • An ordinary differential equation that defines value of dy/dx in the form x and y.
  • Initial value of y, i.e., y(0)

Thus we are given below.


 \frac{\mathrm{dx} }{\mathrm{d} y} = f(x, y),y(0)= y_o

The task is to find value of unknown function y at a given point x.



The Runge-Kutta method finds approximate value of y for a given x. Only first order ordinary differential equations can be solved by using the Runge Kutta 4th order method.

Below is the formula used to compute next value yn+1 from previous value yn. The value of n are 0, 1, 2, 3, ….(x – x0)/h. Here h is step height and xn+1 = x0 + h

. Lower step size means more accuracy.

  K_1 = hf(x_n, y_n)  K_2 = hf(x_n+\frac{h}{2}, y_n+\frac{k_1}{2})  K_3 = hf(x_n+\frac{h}{2}, y_n+\frac{k_2}{2})  K_4 = hf(x_n+h, y_n+k_3)  y_n_+_1 = y_n + k_1/6 + k_2/3 + k_3/3 + k_4/6 + O(h^{5})
The formula basically computes next value yn+1 using current yn plus weighted average of four increments.

  • k1 is the increment based on the slope at the beginning of the interval, using y
  • k2 is the increment based on the slope at the midpoint of the interval, using y + hk1/2.
  • k3 is again the increment based on the slope at the midpoint, using using y + hk2/2.
  • k4 is the increment based on the slope at the end of the interval, using y + hk3.

The method is a fourth-order method, meaning that the local truncation error is on the order of O(h5), while the total accumulated error is order O(h4).

Source: https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods

Below is implementation for the above formula.

C/C++

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// C program to implement Runge Kutta method
#include<stdio.h>
  
// A sample differential equation "dy/dx = (x - y)/2"
float dydx(float x, float y)
{
    return((x - y)/2);
}
  
// Finds value of y for a given x using step size h
// and initial value y0 at x0.
float rungeKutta(float x0, float y0, float x, float h)
{
    // Count number of iterations using step size or
    // step height h
    int n = (int)((x - x0) / h);
  
    float k1, k2, k3, k4, k5;
  
    // Iterate for number of iterations
    float y = y0;
    for (int i=1; i<=n; i++)
    {
        // Apply Runge Kutta Formulas to find
        // next value of y
        k1 = h*dydx(x0, y);
        k2 = h*dydx(x0 + 0.5*h, y + 0.5*k1);
        k3 = h*dydx(x0 + 0.5*h, y + 0.5*k2);
        k4 = h*dydx(x0 + h, y + k3);
  
        // Update next value of y
        y = y + (1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4);;
  
        // Update next value of x
        x0 = x0 + h;
    }
  
    return y;
}
  
// Driver method
int main()
{
    float x0 = 0, y = 1, x = 2, h = 0.2;
    printf("\nThe value of y at x is : %f",
            rungeKutta(x0, y, x, h));
    return 0;
}

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Java

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// Java program to implement Runge Kutta method
import java.io.*;
class differential
{
    double dydx(double x, double y)
    {
        return ((x - y) / 2);
    }
      
    // Finds value of y for a given x using step size h
    // and initial value y0 at x0.
    double rungeKutta(double x0, double y0, double x, double h)
    {
        differential d1 = new differential();
        // Count number of iterations using step size or
        // step height h
        int n = (int)((x - x0) / h);
  
        double k1, k2, k3, k4, k5;
  
        // Iterate for number of iterations
        double y = y0;
        for (int i = 1; i <= n; i++) 
        {
            // Apply Runge Kutta Formulas to find
            // next value of y
            k1 = h * (d1.dydx(x0, y));
            k2 = h * (d1.dydx(x0 + 0.5 * h, y + 0.5 * k1));
            k3 = h * (d1.dydx(x0 + 0.5 * h, y + 0.5 * k2));
            k4 = h * (d1.dydx(x0 + h, y + k3));
  
            // Update next value of y
            y = y + (1.0 / 6.0) * (k1 + 2 * k2 + 2 * k3 + k4);
              
            // Update next value of x
            x0 = x0 + h;
        }
        return y;
    }
      
    public static void main(String args[])
    {
        differential d2 = new differential();
        double x0 = 0, y = 1, x = 2, h = 0.2;
          
        System.out.println("\nThe value of y at x is : " 
                    + d2.rungeKutta(x0, y, x, h));
    }
}
  
// This code is contributed by Prateek Bhindwar

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Python

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# Python program to implement Runge Kutta method
# A sample differential equation "dy / dx = (x - y)/2"
def dydx(x, y):
    return ((x - y)/2)
  
# Finds value of y for a given x using step size h
# and initial value y0 at x0.
def rungeKutta(x0, y0, x, h):
    # Count number of iterations using step size or
    # step height h
    n = (int)((x - x0)/h) 
    # Iterate for number of iterations
    y = y0
    for i in range(1, n + 1):
        "Apply Runge Kutta Formulas to find next value of y"
        k1 = h * dydx(x0, y)
        k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1)
        k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * k2)
        k4 = h * dydx(x0 + h, y + k3)
  
        # Update next value of y
        y = y + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4)
  
        # Update next value of x
        x0 = x0 + h
    return y
  
# Driver method
x0 = 0
y = 1
x = 2
h = 0.2
print 'The value of y at x is:', rungeKutta(x0, y, x, h)
  
# This code is contributed by Prateek Bhindwar

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C#

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// C# program to implement Runge
// Kutta method
using System;
  
class GFG {
      
    static double dydx(double x, double y)
    {
        return ((x - y) / 2);
    }
      
    // Finds value of y for a given x
    // using step size h and initial
    // value y0 at x0.
    static double rungeKutta(double x0,
                double y0, double x, double h)
    {
      
        // Count number of iterations using
        // step size or step height h
        int n = (int)((x - x0) / h);
  
        double k1, k2, k3, k4;
  
        // Iterate for number of iterations
        double y = y0;
          
        for (int i = 1; i <= n; i++) 
        {
              
            // Apply Runge Kutta Formulas
            // to find next value of y
            k1 = h * (dydx(x0, y));
              
            k2 = h * (dydx(x0 + 0.5 * h,
                             y + 0.5 * k1));
                               
            k3 = h * (dydx(x0 + 0.5 * h,
                            y + 0.5 * k2));
                              
            k4 = h * (dydx(x0 + h, y + k3));
  
            // Update next value of y
            y = y + (1.0 / 6.0) * (k1 + 2 
                       * k2 + 2 * k3 + k4);
              
            // Update next value of x
            x0 = x0 + h;
        }
          
        return y;
    }
      
    // Driver code
    public static void Main()
    {
          
        double x0 = 0, y = 1, x = 2, h = 0.2;
          
        Console.WriteLine("\nThe value of y"
                             + " at x is : "
                 + rungeKutta(x0, y, x, h));
    }
}
  
// This code is contributed by Sam007.

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PHP

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<?php
// PHP program to implement 
// Runge Kutta method
  
// A sample differential equation
// "dy/dx = (x - y)/2"
function dydx($x, $y)
{
    return(($x - $y) / 2);
}
  
// Finds value of y for a
// given x using step size h
// and initial value y0 at x0.
function rungeKutta($x0, $y0, $x, $h)
{
      
    // Count number of iterations
    // using step size or step 
    // height h
    $n = (($x - $x0) / $h);
  
    $k1; $k2; $k3; $k4; $k5;
  
    // Iterate for number 
    // of iterations
    $y = $y0;
    for($i = 1; $i <= $n; $i++)
    {
          
        // Apply Runge Kutta
        // Formulas to find
        // next value of y
        $k1 = $h * dydx($x0, $y);
        $k2 = $h * dydx($x0 + 0.5 * $h,
                        $y + 0.5 * $k1);
        $k3 = $h * dydx($x0 + 0.5 * $h
                        $y + 0.5 * $k2);
        $k4 = $h * dydx($x0 + $h, $y + $k3);
  
        // Update next value of y
        $y = $y + (1.0 / 6.0) * ($k1 + 2 * 
                    $k2 + 2 * $k3 + $k4);;
  
        // Update next value of x
        $x0 = $x0 + $h;
    }
  
    return $y;
}
  
    // Driver method
    $x0 = 0;
    $y = 1; 
    $x = 2; 
    $h = 0.2;
    echo "The value of y at x is : ",
          rungeKutta($x0, $y, $x, $h);
  
// This code is contributed by anuj_67.
?>

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Output:

The value of y at x is : 1.103639

Time Complexity of above solution is O(n) where n is (x-x0)/h.

Some useful resources for detailed examples and more explanation.
http://w3.gazi.edu.tr/~balbasi/mws_gen_ode_txt_runge4th.pdf
https://www.youtube.com/watch?v=kUcc8vAgoQ0

This article is contributed by Arpit Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : Sam007, vt_m



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