Improved Euler's Method and Euler's Method with their C Program

Improved Euler’s Method

The Improved Euler’s method, also known as the Heun formula or the average slope method, gives a more accurate approximation than the Euler rule and gives an explicit formula for computing y_n+1. The basic idea is to correct some errors of the original Euler's method. The syntax of the Improved Euler’s method is similar to that of the trapezoid rule, but the y value of the function in terms of y_n+1 consists of the sum of the y value and the product of h and the function in terms of x_n and y_n.


Improved Euler formula or the average slope method is commonly referred to as the Heun method:

yn+1=yn

+h/2[f(xn,yn)+f(xn+1,yn+hf(xn,yn))],n=0,1,2,….

Since it is actually the simplest version of the predictor-corrector method, the recurrence can be written as

pn+1

yn+1=yn+hf(xn,yn),

yn+1

=yn+h/2[f(xn,yn)+f(xn+1,pn+1)],n=0,1,2,….

C program to find the value using Heun's method

#include<stdio.h>
#include<math.h>
#include<conio.h>
#define f(x,y) 2*(x)+(y)

int main(){
 float x,xp,x0,y0,y,h,m1,m2;
 printf("\n Heuns Method\n");
 printf("\nEnter initial values of x and y\n");
 scanf("%f%f",&x0,&y0);
 printf("\nEnter x at which function to be evaluated\n");
 scanf("%f",&xp);
 printf("\nEnter the step size \n");
 scanf("%f",&h);
 y=y0;
 x=x0;
 for(x=x0;x<xp;x=x+h){
  m1=f(x,y);
  printf("\nm1=%f",m1);
  m2=f(x+h,y+(h*m1));
  printf("\nm2=%f",m2);
  y=y+h/2*(m1+m2);
  printf("\ny=%f",y);
 }
 
 printf("Function value at x=%f is %f\n",xp,y);
 getch();
 return 0; 
 
}

Euler's Method

Euler's Method assumes our solution is written in the form of a Taylor’s Series.

That is, we'll have a function of the form:

$This gives us a reasonably good approximation if we take plenty of terms, and if the value of$h is reasonably small.

For Euler's Method, we just take the first 2 terms only.

$The last term is just$$h times our dx/dy$ ​ expression, so we can write Euler's Method as follows:

$$How do we use this formula?

We start with some known value for y, which we could call y0​. It has this value when X$=x$0​. (We make use of the initial value $($x0​,Y0​)).

The result of using this formula is the value for y, one h step to the right of the current value. Let's call it $Y$1So we have:

$$where
$Y$1​ is the next estimated solution value;
Y0​ is the current value;
h is the interval between steps; and
$f$(x0​,y0​) is the value of the derivative at the starting point,(x0​,y0​).

Next value: 

To get the next value $Y$2​, we would use the value we just found for $Y$1​ as follows:

$$Where
Y2​ is the next estimated solution value;
Y1​ is the current value;
$h$ is the interval between steps;
$X$1​=X0​+h; and
$f$(x1​,y1​) is the value of the derivative at the current (x1​,y1​) point.

We continue this process for as many steps as required.

C Program for Euler's Method

#include<stdio.h>
#include<math.h>
#include<conio.h>
#define f(x,y) 2*(x)+(y)
//#define f(x,y) 3*(x)*(x)+1

int main(){
 float x,xp,x0,y0,y,h,m1,m2,m3,m4;
 printf("\n Euler's method'\n");
 printf("\nEnter initial values of x and y\n");
 scanf("%f%f",&x0,&y0);
 printf("\nEnter x at which function to be evaluated\n");
 scanf("%f",&xp);
 printf("\nEnter the step size \n");
 scanf("%f",&h);
 y=y0;
 x=x0;
 for(x=x0;x<xp;x=x+h){
  m1=f(x,y);
  printf("\nm1=%f",m1);
  y=y+h*(m1);
  printf("\ny=%f",y);
 }
 
 printf("Function value at x=%f is %f\n",xp,y);
 getch();
 return 0; 
 
}