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 y0\n");

scanf("%f%f",&x0,&y0);

printf("\nEnter the value of x for which we need to find the value of y\n");

scanf("%f",&xn);

printf("\nEnter the value of h\n");

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=%f\n",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