# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 next value yn+1 from 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):   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 on the order of O(h3), while the total accumulated error is order O(h4).

Below is the implementation of the above approach:

## 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;  }

## 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  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

Output:

y(x) = 0.645590


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