# Runge-Kutta 2nd order method to solve Differential equations //

Given the following inputs:

1. An ordinary differential equation that defines the value of dy/dx in the form x and y. 2. Initial value of y, i.e., y(0). The task is to find the value of unknown function y at a given point x, i.e. y(x).

Example:

Input: x0 = 0, y0 = 1, x = 2, h = 0.2
Output: y(x) = 0.645590
Input: x0 = 2, y0 = 1, x = 4, h = 0.4;
Output: y(x) = 4.122991   

Approach:

The Runge-Kutta method finds an approximate value of y for a given x. Only first-order ordinary differential equations can be solved by using the Runge-Kutta 2nd-order method.

Below is the formula used to compute the next value yn+1 from the previous value yn. Therefore:

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

The essential formula to compute the value of y(n+1):

K1 = h * f(x,y)
K2 = h * f(x/2, y/2) or K1/2
yn+1 = yn + K2 + (h3)

The formula basically computes the next value yn+1 using current yn plus the weighted average of two 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 + h*K1/2).

The method is a second-order method, meaning that the local truncation error is in the order of O(h3), while the total accumulated error is order of O(h4).

Below is the implementation of the above approach:

## C++

 // C++ program to implement Runge // Kutta method   #include    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 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;       // 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);           // Update next value of y         y = y + (1.0 / 6.0) * (k1 + 2 * k2);           // Update next value of x         x0 = x0 + h;     }       return y; }   // Driver Code int main() {     float x0 = 0, y = 1, x = 2, h = 0.2;       cout << fixed << setprecision(6)          << "y(x) = " << rungeKutta(x0, y, x, h);     return 0; }   // This code is contributed by shivani

## Java

 // Java program to implement Runge // Kutta 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 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;           // 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);               // Update next value of y             y = y + (1.0 / 6.0) * (k1 + 2 * k2);               // Update next value of x             x0 = x0 + h;         }           return y;     }       // Driver Code     public static void main(String[] args)     {         double x0 = 0, y = 1, x = 2, h = 0.2;           System.out.println(rungeKutta(x0, y, x, h));     } }   // This code is contributed by Yash_R

## Python3

 # Python3 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 = round((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)           # Update next value of y         y = y + (1.0 / 6.0) * (k1 + 2 * k2)           # Update next value of x         x0 = x0 + h       return y     # Driver Code if __name__ == "__main__":       x0 = 0     y = 1     x = 2     h = 0.2       print("y(x) =", rungeKutta(x0, y, x, h))   # This code is contributed by Yash_R

## C

 // C program to implement Runge // Kutta method   #include    // 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;       // 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);           // Update next value of y         y = y + (1.0 / 6.0) * (k1 + 2 * k2);           // Update next value of x         x0 = x0 + h;     }       return y; }   // Driver Code int main() {     float x0 = 0, y = 1, x = 2, h = 0.2;       printf("y(x) = %f", rungeKutta(x0, y, x, h));     return 0; }

## C#

 // C# program to implement Runge // Kutta 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 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;           // 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);               // Update next value of y             y = y + (1.0 / 6.0) * (k1 + 2 * k2);               // Update next value of x             x0 = x0 + h;         }           return y;     }       // Driver Code     public static void Main(string[] args)     {         double x0 = 0, y = 1, x = 2, h = 0.2;           Console.WriteLine(rungeKutta(x0, y, x, h));     } }   // This code is contributed by Yash_R

## Javascript

 

Output:

y(x) = 0.645590

Time Complexity: O(n)
Auxiliary Space: O(1)

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