Euler method in c++ - c++

Question

The following code uses Euler's Method to approximate a value of y(x). My code currently accepts the endpoints a and b as user input and values for values for alpha which is the initial condition and the step size value which is h. Given my code I can now approximate a value of y, say y(8) given the initial condition y(0)=6.
However I have a small mistake in my code and I am not quite sure how to fix it and am looking for help. Right now my code does not check to ensure that the right endpoint b is an integer multiple of the stepsize h. Due to this the final approximation may not be for f(b) but for f(c) where c is the closest integer multiple of h to b. I am looking for some help on how to fix this, Thanks!
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
double dydx (double x, double y)
{
double f = y*(2-y)/(x+3);
return f;
}
int main()
{
double a,b,alpha,h,z;
cout<<"\nEnter the value of endpoint a: \n";
cin>>a;
cout<<"\nEnter the value of endpoint b: \n";
cin>>b;
cout<<"\nEnter the y value of the initial condition: \n";
cin>>alpha;
cout<<"\nEnter the stepsize, h: \n";
cin>>h;
cout<<"\n";
while((b-a)>0.0000001)
{
z=alpha+(h*dydx(a,alpha));
cout<<"z("<<a<<")="<<z<<endl;
alpha=z;
a=a+h;
}
cout<<"\nThe approximate solution of y("<<b<<") is "<<z<<"."<<endl;
return 0;
}

Answer 1

You can calculate step size h from number of steps n:
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
double dydx (double x, double y)
{
double f = y*(2-y)/(x+3);
return f;
}
int main()
{
double a,b,alpha,h,z;
cout<<"\nEnter the value of endpoint a: \n";
cin>>a;
cout<<"\nEnter the value of endpoint b: \n";
cin>>b;
cout<<"\nEnter the y value of the initial condition: \n";
cin>>alpha;
/*
* Obtains step size from number on steps
* h = 0.1 for [a; b] = [0; 8] can be given by n = 80
*/
int n = 0;
cout<<"\nEnter the number of steps, n: \n";
cin>>n;
h = (b - a) / n;
//------
cout<<"\n";
//-- Replaced 0.0000001 by h / 2.0 --
while((b-a)> h / 2.0)
{
z=alpha+(h*dydx(a,alpha));
alpha=z;
a=a+h;
/*
* z - function value in next point,
* so to output correct point a need to be incremented before this.
*/
cout<<"z("<<a<<")="<<z<<endl;
}
cout<<"\nThe approximate solution of y("<<b<<") is "<<z<<"."<<endl;
return 0;
}
Insert this instead of h input.