Matrices

ok here is my code for initialising a matrix to be solved.

Code:

//All Header Files Used
#include <iostream>
#include <string>
#include <cmath>


//initialising  matrix to the size of 100
double A[100][100];
//declaring functions for use in the program
void Finite_Difference(int n,double a,double b,double c,double dx,double A[ ][100]);


using namespace std;

//Function Definition to Calculate the Finite Difference Matrix
void Finite_Difference(int n,double a,double b,double c,double dx, double A[ ][100])
{
	
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<n;j++)
		{
			if (i == j)
			{
				A[i][j]=((2*a) - c*(pow(dx,2)));
			}

			else if( j == (i+1))
			{
				A[i][j] = (-a + (b/2)*dx);
			}

			else if (j == (i-1))
			{
				A[i][j] = (-a - (b/2)*dx);
			}

			else 
				A[i][j] = 0;
		}
	}
}


int main()
{
		cout<<" Welcome to the Boundary Value Problem Solver"<<endl;
		cout<<" This Solver Will be Using the Finite Difference Method"<<endl;
		cout<<"	The Differential Equation to be Solved is in the Form of:"<<endl;
		cout<<endl;
		cout<<"			ay'' + b' + cy + x = 0"<<endl;
		cout<<" With Boundary Conditions y(0) = d & y(20) = e"<<endl;
		double const1,const2,const3,bound1,bound2,dx;
		bool v1 = false;
		while(!v1)
		{
			cout<<"Please Enter the Constant 'a':\n";
			cin>>const1;

			//Determined whether the input is valid or invalid
			if(cin.fail())
			{
				cin.clear();
				cin.ignore(INT_MAX, '\n');
				cout << "The input was invalid. Please try again" << endl;
			}
			
			else
			{
				v1 = true;
			}
		}

		bool v2 = false;
		while(!v2)
		{
			cout<<"Please Enter the Constant 'b':\n";
			cin>>const2;

			//Determined whether the input is valid or invalid
			if(cin.fail())
			{
				cin.clear();
				cin.ignore(INT_MAX, '\n');
				cout << "The input was invalid. Please try again" << endl;
			}
			
			else
			{
				v2 = true;
			}
		}

		bool v3 = false;
		while(!v3)
		{
			cout<<"Please Enter the Constant 'c':\n";
			cin>>const3;

			//Determined whether the input is valid or invalid
			if(cin.fail())
			{
				cin.clear();
				cin.ignore(INT_MAX, '\n');
				cout << "The input was invalid. Please try again" << endl;
			}
			
			else
			{
				v3 = true;
			}
		}

		bool v4 = false;
		while(!v4)
		{
			cout<<"Please Enter the Boundary Condtion at y(0):\n";
			cin>>bound1;

			//Determined whether the input is valid or invalid
			if(cin.fail())
			{
				cin.clear();
				cin.ignore(INT_MAX, '\n');
				cout << "The input was invalid. Please try again" << endl;
			}
			
			else
			{
				v4 = true;
			}
		}

		bool v5 = false;
		while(!v5)
		{
			cout<<"Please Enter the Boundary Condition at y(20):\n";
			cin>>bound2;

			//Determined whether the input is valid or invalid
			if(cin.fail())
			{
				cin.clear();
				cin.ignore(INT_MAX, '\n');
				cout << "The input was invalid. Please try again" << endl;
			}
			
			else
			{
				v5 = true;
			}
		}

		bool v6 = false;
		while(!v6)
		{
			cout<<"Please Enter the Interval Size:\n";
			cin>>dx;

			//Determined whether the input is valid or invalid
			if(cin.fail())
			{
				cin.clear();
				cin.ignore(INT_MAX, '\n');
				cout << "The input was invalid. Please try again" << endl;
			}
			else if(dx<=0)
			{
				cout<< "The Interval Must be Greater than 0"<<endl;
				cout<< "Please Try Again."<<endl;
			}
			else
			{
				v6 = true;
			}
		}

		double dim = ((bound2-bound1)/dx) - 1;
		int size = (int)dim;

		//Initialises the A matrix to 0 everywhere
		for(int i=0;i<size-1;i++)
		{
			for(int j=0;j<size-1;j++)
			{
				A[i][j]=0;
			}
		}

		//creates the Finite_Difference matrix
		Finite_Difference(size,const1,const2,const3,dx,A);

return 0;

Anyways, with values of: a= 8, b=-2, c=-1, y(0) = 5, y(20)= 8 & dx =2


the matrix should then be:

assuming x = 2a -c(dx^2)
y = a + (b/2)dx
z = -a - (b/2)*dx

then the matrix should look like this:

x z 0 0 0 0 0 0 0
y x z 0 0 0 0 0 0
0 y x z 0 0 0 0 0
0 0 y x z 0 0 0 0
0 0 0 y x z 0 0 0
0 0 0 0 y x z 0 0
0 0 0 0 0 y x z 0
0 0 0 0 0 0 y x z
0 0 0 0 0 0 0 y x

But this is not what i'm getting for my output. Can anyone see what i'm not seeing.

Thanks for your time.