Unit 5.1 Euler's Method

// program
#include<stdio.h>
#define f(x,y) (y-x)/(y+x)
int main(){
    float x0,y0,h,xn,x,y;
    int n;
    printf("Enter values x0, y0, h and xn \n");
    scanf("%f %f %f %f",&x0,&y0,&h,&xn);
    n = (xn-x0)/h;
    for(int i=1 ; i<=n ; i++){
        printf("\nvalues of x0 = %f & y0 = %f",x0,y0);
        y = y0 + (h*f(x0,y0));
        x = x0 + h;

        if(x<xn){
            x0 = x;
            y0 = y;
        }
        else
            break;
    }
    printf("\nOutput x = %f and y = %f",x,y);
    return 0;
}
/*
Ouput
Enter values x0, y0, h and xn
0 1 0.02 0.1

values of x0 = 0.000000 & y0 = 1.000000
values of x0 = 0.020000 & y0 = 1.020000
values of x0 = 0.040000 & y0 = 1.039231
values of x0 = 0.060000 & y0 = 1.057748
values of x0 = 0.080000 & y0 = 1.075601

Output x = 0.100000 and y = 1.092832
*/


// Algorithm


start

define function // define f(x,y) x*typede

Get the values of x0,y0,h and xn

*Here x0 and y0 are the initial conditions
h is the interval
xn is the requred value
// x0= 0 y0=1 , h=0.1 , xn=0.4

n=(xn-x0)/h // n=(0.4-0)/0.1 = 4

start loop from i =1 to n

Print values of y0 and x0 // 1,0 // 1,0.1 // 1,0.1,0.2 // 1.0302,-.0302

y=y0  + h*f(x0,y0)//1.061106
x=x0+h//0.4

check if x<xn // 0.4<0.4.4
if yes
then
    x0 = x
    y0 = y
if no goto 10

End loop i

Print x and y

Stop