Data Modelling & AI Data Structure & Algorithm

Gill’s 4th Order Method to solve Differential Equations

By Calisto Chipfumbu

23 June 2025

0

2

Given the following inputs:

An ordinary differential equation that defines the value of dy/dx in the form x and y.
$\LARGE \frac{dy}{dx} = f(x, y)$
Initial value of y, i.e., y(0).
$\LARGE y(0)= y_o$

The task is to find the value of the unknown function y at a given point x, i.e. y(x).
Examples:

Input: x0 = 0, y = 3.0, x = 5.0, h = 0.2
Output: y(x) = 3.410426
Input: x0 = 0, y = 1, x = 3, h = 0.3
Output: y(x) = 1.669395

Approach:
Gill’s method is used to find an approximate value of y for a given x. Below is the formula used to compute the next value y_n+1 from the previous value y_n.
Therefore:

y_n+1 = value of y at (x = n + 1)
y_n = value of y at (x = n)
where
  0 ? n ? (x - x₀)/h
  h is step height
  x_n+1 = x₀ + h

The essential formula to compute the value of y(n+1):
$\LARGE K_{1} = h*f(x_{n}, y_{n})$
$\LARGE K_{2} = h*f(x_n + \frac{1}{2}h, y_n + \frac{1}{2}k_1)$
$\LARGE K_{3} = h*f[x_n + \frac{1}{2}h, y_n + \frac{1}{2}(-1 + \sqrt{2})k_1 + (1 - \frac{1}{2}\sqrt{2})k_2]$
$\LARGE K_{4} = h*f[x_n + h, y_n - \frac{1}{2}\sqrt{2}k_2 + (1 + \frac{1}{2}\sqrt{2})k_3]$
$\LARGE y_{n+1} = y_{n} + \frac{1}/{6}[K_1 + (2 - \sqrt{2})K_{2} + (2 + \sqrt{2})K_3 + K_4 ] + Error Terms$
The formula basically computes the next value y_n+1 using current y_n:

K₁ is the increment based on the slope at the beginning of the interval, using y.
K₂ is the increment based on the slope, using
$\LARGE (y + \frac{1}{2}(K_1*h))$
K₃ is the increment based on the slope, using
$\LARGE (y + \frac{1}{2}(-1 + \sqrt{2})K_1 + (1 - \frac{1}{2}\sqrt{2})K_2)$
K₄ is the increment based on the slope, using
$(y - \frac{1}{2}\sqrt{2}K_2 + (1 + \frac{1}{2}\sqrt{2})K_3)$

The method is a fourth-order method, meaning that the local truncation error is of the order of O(h⁵).
Below is the implementation of the above approach:

C++

// C++ program to implement Gill's method
 
#include <bits/stdc++.h>
using namespace std;
 
// 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 Gill(float x0, float y0,
        float x, float h)
{
    // Count number of iterations
    // using step size or height h
    int n = (int)((x - x0) / h);
 
    // Value of K_i
    float k1, k2, k3, k4;
 
    // Initial value of y(0)
    float y = y0;
 
    // Iterate for number of iteration
    for (int i = 1; i <= n; i++) {
 
        // Apply Gill's Formulas to
        // find next value of y
 
        // Value of K1
        k1 = h * dydx(x0, y);
 
        // Value of K2
        k2 = h * dydx(x0 + 0.5 * h,
                    y + 0.5 * k1);
 
        // Value of K3
        k3 = h * dydx(x0 + 0.5 * h,
                    y + 0.5 * (-1 + sqrt(2)) * k1
                        + k2 * (1 - 0.5 * sqrt(2)));
 
        // Value of K4
        k4 = h * dydx(x0 + h,
                    y - (0.5 * sqrt(2)) * k2
                        + k3 * (1 + 0.5 * sqrt(2)));
 
        // Find the next value of y(n+1)
        // using y(n) and values of K in
        // the above steps
        y = y + (1.0 / 6) * (k1 + (2 - sqrt(2)) * k2 
                            + (2 + sqrt(2)) * k3 + k4);
 
        // Update next value of x
        x0 = x0 + h;
    }
 
    // Return the final value of dy/dx
    return y;
}
 
// Driver Code
int main()
{
    float x0 = 0, y = 3.0,
        x = 5.0, h = 0.2;
 
    printf("y(x) = %.6f",
        Gill(x0, y, x, h));
    return 0;
}

Java

// Java program to implement Gill's method
class GFG{
 
// A sample differential equation
// "dy/dx = (x - y)/2"
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 Gill(double x0, double y0,
                   double x, double h)
{
     
    // Count number of iterations
    // using step size or height h
    int n = (int)((x - x0) / h);
 
    // Value of K_i
    double k1, k2, k3, k4;
 
    // Initial value of y(0)
    double y = y0;
 
    // Iterate for number of iteration
    for(int i = 1; i <= n; i++)
    {
         
       // Apply Gill's Formulas to
       // find next value of y
        
       // Value of K1
       k1 = h * dydx(x0, y);
        
       // Value of K2
       k2 = h * dydx(x0 + 0.5 * h,
                      y + 0.5 * k1);
        
       // Value of K3
       k3 = h * dydx(x0 + 0.5 * h,
                      y + 0.5 * (-1 + Math.sqrt(2)) *
                     k1 + k2 * (1 - 0.5 * Math.sqrt(2)));
        
       // Value of K4
       k4 = h * dydx(x0 + h,
                      y - (0.5 * Math.sqrt(2)) *
                     k2 + k3 * (1 + 0.5 * Math.sqrt(2)));
        
       // Find the next value of y(n+1)
       // using y(n) and values of K in
       // the above steps
       y = y + (1.0 / 6) * (k1 + (2 - Math.sqrt(2)) * 
                            k2 + (2 + Math.sqrt(2)) *
                            k3 + k4);
        
       // Update next value of x
       x0 = x0 + h;
    }
 
    // Return the final value of dy/dx
    return y;
}
 
// Driver Code
public static void main(String[] args)
{
    double x0 = 0, y = 3.0,
            x = 5.0, h = 0.2;
 
    System.out.printf("y(x) = %.6f", Gill(x0, y, x, h));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement Gill's method
from math import sqrt
 
# 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 Gill(x0, y0, x, h):
     
    # Count number of iterations
    # using step size or height h
    n = ((x - x0) / h)
 
    # Initial value of y(0)
    y = y0
 
    # Iterate for number of iteration
    for i in range(1, int(n + 1), 1):
         
        # Apply Gill's Formulas to
        # find next value of y
 
        # Value of K1
        k1 = h * dydx(x0, y)
 
        # Value of K2
        k2 = h * dydx(x0 + 0.5 * h, 
                       y + 0.5 * k1)
 
        # Value of K3
        k3 = h * dydx(x0 + 0.5 * h,
                       y + 0.5 * (-1 + sqrt(2)) *
                      k1 + k2 * (1 - 0.5 * sqrt(2)))
 
        # Value of K4
        k4 = h * dydx(x0 + h, y - (0.5 * sqrt(2)) *
                    k2 + k3 * (1 + 0.5 * sqrt(2)))
 
        # Find the next value of y(n+1)
        # using y(n) and values of K in
        # the above steps
        y = y + (1 / 6) * (k1 + (2 - sqrt(2)) *
                           k2 + (2 + sqrt(2)) *
                           k3 + k4)
 
        # Update next value of x
        x0 = x0 + h
 
    # Return the final value of dy/dx
    return y
 
# Driver Code
if __name__ == '__main__':
     
    x0 = 0
    y = 3.0
    x = 5.0
    h = 0.2
 
    print("y(x) =", round(Gill(x0, y, x, h), 6))
 
# This code is contributed by Surendra_Gangwar

C#

// C# program to implement Gill's method
using System;
 
class GFG{
  
// A sample differential equation
// "dy/dx = (x - y)/2"
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 Gill(double x0, double y0,
                   double x, double h)
{
      
    // Count number of iterations
    // using step size or height h
    int n = (int)((x - x0) / h);
  
    // Value of K_i
    double k1, k2, k3, k4;
  
    // Initial value of y(0)
    double y = y0;
  
    // Iterate for number of iteration
    for(int i = 1; i <= n; i++)
    {
          
       // Apply Gill's Formulas to
       // find next value of y
         
       // Value of K1
       k1 = h * dydx(x0, y);
         
       // Value of K2
       k2 = h * dydx(x0 + 0.5 * h,
                      y + 0.5 * k1);
         
       // Value of K3
       k3 = h * dydx(x0 + 0.5 * h,
                      y + 0.5 * (-1 + Math.Sqrt(2)) *
                     k1 + k2 * (1 - 0.5 * Math.Sqrt(2)));
         
       // Value of K4
       k4 = h * dydx(x0 + h,
                      y - (0.5 * Math.Sqrt(2)) *
                     k2 + k3 * (1 + 0.5 * Math.Sqrt(2)));
         
       // Find the next value of y(n+1)
       // using y(n) and values of K in
       // the above steps
       y = y + (1.0 / 6) * (k1 + (2 - Math.Sqrt(2)) * 
                            k2 + (2 + Math.Sqrt(2)) *
                            k3 + k4);
         
       // Update next value of x
       x0 = x0 + h;
    }
  
    // Return the final value of dy/dx
    return y;
}
  
// Driver Code
public static void Main(String[] args)
{
    double x0 = 0, y = 3.0,
            x = 5.0, h = 0.2;
  
    Console.Write("y(x) = {0:F6}", Gill(x0, y, x, h));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

<script>
 
// Javascript program to implement Gill's 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 Gill(x0, y0, x, h)
{
       
    // Count number of iterations
    // using step size or height h
    let n = ((x - x0) / h);
   
    // Value of K_i
    let k1, k2, k3, k4;
   
    // Initial value of y(0)
    let y = y0;
   
    // Iterate for number of iteration
    for(let i = 1; i <= n; i++)
    {
           
       // Apply Gill's Formulas to
       // find next value of y
          
       // Value of K1
       k1 = h * dydx(x0, y);
          
       // Value of K2
       k2 = h * dydx(x0 + 0.5 * h,
                      y + 0.5 * k1);
          
       // Value of K3
       k3 = h * dydx(x0 + 0.5 * h,
                      y + 0.5 * (-1 + Math.sqrt(2)) *
                     k1 + k2 * (1 - 0.5 * Math.sqrt(2)));
          
       // Value of K4
       k4 = h * dydx(x0 + h,
                      y - (0.5 * Math.sqrt(2)) *
                     k2 + k3 * (1 + 0.5 * Math.sqrt(2)));
          
       // Find the next value of y(n+1)
       // using y(n) and values of K in
       // the above steps
       y = y + (1.0 / 6) * (k1 + (2 - Math.sqrt(2)) * 
                            k2 + (2 + Math.sqrt(2)) *
                            k3 + k4);
          
       // Update next value of x
       x0 = x0 + h;
    }
   
    // Return the final value of dy/dx
    return y;
}
 
// Driver Code
     
    let x0 = 0, y = 3.0,
            x = 5.0, h = 0.2;
   
    document.write("y(x) = ", Gill(x0, y, x, h).toFixed(6));
         
</script>

Output:

y(x) = 3.410426

Time Complexity: O(n^3/2)

Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Gill’s 4th Order Method to solve Differential Equations

C++

Java

Python3

C#

Javascript

How to Debug Slow Search Requests in Milvus

When Context Engineering Is Done Right, Hallucinations Can Be the Spark of AI Creativity

Getting Started with langgraph-up-react: A Practical LangGraph Template

LEAVE A REPLY Cancel reply

Most Popular

The chip that will comfortably dethrone Apple. [Video]

I have used dozens of Android phones, but these Pixel features still absolutely amaze me

Don’t waste money — the Samsung Galaxy S24 Ultra still gets the job done

I still reach for the Microsoft Surface Duo over today’s foldables; here’s why

EDITOR PICKS

The chip that will comfortably dethrone Apple. [Video]

I have used dozens of Android phones, but these Pixel features still absolutely amaze me

Don’t waste money — the Samsung Galaxy S24 Ultra still gets the job done

POPULAR POSTS

The chip that will comfortably dethrone Apple. [Video]

I have used dozens of Android phones, but these Pixel features still absolutely amaze me

Don’t waste money — the Samsung Galaxy S24 Ultra still gets the job done

POPULAR CATEGORY

ABOUT US

FOLLOW US