Runge Kutta (RK) Fourth Order Using C++ with Output

Implementation of Runge Kutta (RK) Fourth Order method for solving ordinary differential equation using C++ programming language with output is given below.

Output of this is program is solution for dy/dx = (y2 - x2)/(y2+x2) with initial condition y = 1 for x = 0 i.e. y(0) = 1 and we are trying to evaluate this differential equation at y = 0.6 in three steps i.e. n = 3. ( Here y = 0.6 i.e. y(0.6) = ? is our calculation point)

C++ Program for RK-4 Method


#include<iostream>

/* Defining ordinary differential equation to be solved */
#define f(x,y) (y*y-x*x)/(y*y+x*x)

using namespace std;

#include

/* defining ordinary differential equation to be solved */
#define f(x,y) (y*y-x*x)/(y*y+x*x)

using namespace std;
int main()
{
 float x0, y0, xn, h, yn, k1, k2, k3, k4, k;
 int i, n;

 cout<<"Enter Initial Condition"<< endl;
 cout<<"x0 = ";
 cin>> x0;
 cout<<"y0 = ";
 cin >> y0;
 cout<<"Enter calculation point xn = ";
 cin>>xn;
 cout<<"Enter number of steps: ";
 cin>> n;

 /* Calculating step size (h) */
 h = (xn-x0)/n;

 /* Runge Kutta Method */
 cout<<"\nx0\ty0\tyn\n";
 cout<<"------------------\n";
 for(i=0; i < n; i++)
 {
  k1 = h * (f(x0, y0));
  k2 = h * (f((x0+h/2), (y0+k1/2)));
  k3 = h * (f((x0+h/2), (y0+k2/2)));
  k4 = h * (f((x0+h), (y0+k3)));
  k = (k1+2*k2+2*k3+k4)/6;
  yn = y0 + k;
  cout<< x0<<"\t"<< y0<<"\t"<< yn<< endl;
  x0 = x0+h;
  y0 = yn;
 }

 /* Displaying result */
 cout<<"\nValue of y at x = "<< xn<< " is " << yn;

 return 0;
}

RK-4 C++ Program Output

	
Enter Initial Condition
x0 = 0
y0 = 1
Enter calculation point xn = 0.6
Enter number of steps: 3

x0      y0      yn
------------------
0       1       1.196
0.2     1.196   1.37527
0.4     1.37527 1.53311

Value of y at x = 0.6 is 1.53311

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