Write a C program to numerical solution of ODE using Runge-Kutta method by MistarAV

 #include<stdio.h>
#include<conio.h>
#define f(x,y) (y*y-x*x)/(y*y+x*x)
int main()
{
float x0, y0, xn, h, yn, k1, k2, k3, k4, k;
int i, n;
clrscr();
printf("Enter Initial Condition\n");
printf("x0 = ");
scanf("%f", &x0);
printf("y0 = ");
scanf("%f", &y0);
printf("Enter calculation point xn = ");
scanf("%f", &xn);
printf("Enter number of steps: ");
scanf("%d", &n);
/* Calculating step size (h) */
h = (xn-x0)/n;
/* Runge Kutta Method */
printf("\nx0\ty0\tyn\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;
printf("%0.4f\t%0.4f\t%0.4f\n",x0,y0,yn);
x0 = x0+h;
y0 = yn;
}
/* Displaying result */
printf("\nValue of y at x = %0.2f is %0.3f",xn, yn);
getch();
return 0;
}

Output

Enter Initial Condition
x0 = 0
y0 = 1
Enter calculation point xn = 0.4
Enter number of steps: 2
x0 y0 yn
0.0000 1.0000 1.1960
0.2000 1.1960 1.3753
Value of y at x = 0.40 is 1.375