Euler

Algorithm : Euler Method

Euler(h,y0,a,b)
{
//This algorithm takes the following inputs where h is the width of each division
y0 is the value of y when x=0 and a & b are the limits

while(a<=b)
{
sum=y0+h*f(a,y0)
a=a+h
print(i,a,sum)
y0=sum
sum=0				// Initializes sum to 0
i=i+1
}
}


C Program For Euler Method

Source Code:

#include<stdio.h>
#include<conio.h>
#include<math.h>

void euler(float,float,float,int);
float f(float,float);
void main()
{
 int n;
 float a,b,y0;
 clrscr();
 printf("\nSOLVE :\n");
 printf("\n (dy/dx)=xy using the given limits.\n");
 printf("\nEnter the upper limit:");
 scanf("%f",&b);
 printf("\nEnter the lower:");
 scanf("%f",&a);
 printf("\nEnter the value of y for the first interval:");
 scanf("%f",&y0);
 printf("\nENTER THE NUMBE3R OF SUB-DIVISIONS:");
 scanf("%d",&n);
 euler(a,b,y0,n);
 getch();
}

void euler(float a,float b,float y0,int n)
{
 float h,x,y1,y2;
 int i=0;
 h=(b-a)/n;
 x=a;
 printf("\nSOLUTION :");
 printf("\n\tXr\t\tYr\t");
 printf("\n---------------------------------");
 printf("\nr=%d\t%g\t\t%g",i,x,y0);
 y1=y0;
 for(i=1;i<=n;i++)
 {
  y2=y1+h*f(x,y1);
   x=a+i*h;
  printf("\nr=%d\t%g\t\t%g",i,x,y2);
  y1=y2;
 }
}

float f(float x,float y)
{
 return x*y;
}


  OUTPUT :


SOLVE :

 (dy/dx)=xy     using the given limits.

Enter the upper limit:1

Enter the lower:0

Enter the value of y for the first interval:1

ENTER THE NUMBE3R OF SUB-DIVISIONS:5

SOLUTION :

	Xr              Yr
---------------------------------
r=0     0               1
r=1     0.2             1
r=2     0.4             1.04
r=3     0.6             1.1232
r=4     0.8             1.25798
r=5     1               1.45926