Unit 5.4 Runga Kutta 4th order

algorithm

start
input x0,y0,h,last point n
m1 = f(xi,yi)
m2 = f(xi+h/2,yi+m1*h/2)
m3 = f(xi+h/2 , yi+m2*h/2);
m4 = f(xi+h,yi+m3*h)
y(i+1) = yi + ( (m1 + 2m2 + 2m3 + m4)/6)
display Output
stop

// program
#include <stdio.h>
#include<math.h>
#define f(x,y) (x*y)
int main(){
    int n,i;
    float x0,xn,y0,h,s,s1,s2,s3,s4;
    printf("Enter the x0,y0,h & xn\n");
    scanf("%f %f %f %f",&x0,&y0,&h,&xn);
    while(x0<xn){
        s1 = h*f(x0,y0);
        s2 = h*f((x0+(h/2)),(y0+(s1/2)));
        s3 = h*f((x0+(h/2)),(y0+(s2/2)));
        s4 = h*f((x0+h),(y0+s3));
        s = (s1 + (2*s2) + (2*s3) + s4)/6;
        y0 = y0+s;
        x0 = x0+h;
        printf("\ny(%f) =  %f",x0,y0);
    }
    printf("\n\nFinal Output y(%f) =  %f",x0,y0);
    return 0;
}

/*
Enter the x0,y0,h & xn
1 2 0.2 1.4

y(1.200000) =  2.492143
y(1.400000) =  3.232107

Final Output y(1.400000) =  3.232107
*/