# c programming for Euler’s Method

## Euler’s Method Algorithm:

1. Start
2. Define function
3. Get the values of x0, y0, h and xn
*Here x0 and y0 are the initial conditions
h is the interval
xn is the required value
4. n = (xn – x0)/h + 1
5. Start loop from i=1 to n
6. y = y0 + h*f(x0,y0)
x = x + h
7. Print values of y0 and x0
8. Check if x < xn
If yes, assign x0 = x and y0 = y
If no, goto 9.
9. End loop i
10. Stop

source code:

#include<stdio.h>
#include<math.h>
float f(float x, float y)
{
float f;
f=(y-x)/(y+x);
return(f);
}
int main()
{
int i,n;
float x0,xn,y0,h,x,y;
printf(“n Enter the initial value of x and y i.e x0 and y0n”);
scanf(“%f%f”,&x0,&y0);
printf(“nEnter the value of x for which we need to find the value of yn”);
scanf(“%f”,&xn);
printf(“nEnter the value of hn”);
scanf(“%f”,&h);
n=((xn-x0)/h)+1;
for(i=1;i<=n;i++)
{
y =y0 + h*f(x0,y0);
x= x + h;
printf(” n the value of xo=%f”,x0);
printf(” n the value of yo=%fn”,y0);

if(x<xn)
{
x0=x;
y0=y;
}
}

printf(“n the value of y=%f”,y0);
return 0;
}

output:
Enter the initial value of x and y i.e x0 and y0
0
1

Enter the value of x for which we need to find the value of y
.1

Enter the value of h
0.02
the value of xo=0.000000
the value of yo=1.000000
the value of xo=0.020000
the value of yo=1.020000
the value of xo=0.040000
the value of yo=1.039231
the value of xo=0.060000
the value of yo=1.057748
the value of xo=0.080000
the value of yo=1.075601
the value of xo=0.100000
the value of yo=1.092832

the value of y=1.092832

## Hire a server Expert to resolve the issue Now.

Resolve this issue in just 5\$ from https://serverexpert.io