// D_URR_Szymanek_Piotr.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include <math.h>

using namespace std;

double f1(double t, double x, double y);
double f2(double t, double x, double y);
void k1(double t, double x, double y);
void k2(double t, double x, double y);
void k3(double t, double x, double y);
void k4(double t, double x, double y);
void x1();
int r = 2;
double h = pow(10, -4);
double K1[2], K2[2], K3[2], K4[2], X[2] = { 1,1 };

int main()
{
	int i = 0;
	double t = 0, t0 = 1;
	while (t != 3)
	{
		t = t0 + i*h;
		k1(t, X[0], X[1]);
		k2(t, X[0], X[1]);
		k3(t, X[0], X[1]);
		k4(t, X[0], X[1]);
		x1();
		i++;
	}
	cout <<"Tk= "<< t << endl;
	cout << "X0= " << X[0] << endl;
	cout << "X1= " << X[1] << endl;
	cout << "h=" << h << endl;
	system("PAUSE");
	return 0;
}

double f1(double t, double x, double y)
{
	return (-x / t)+ y;
}

double f2(double t, double x, double y)
{
	return (-2.0 *x) / (t*t) + y / t;
}

void k1(double t, double x, double y)
{
	double pom = 0;
	for (int i = 0; i<r; i++)
	{
		switch (i)
		{
		case 0: pom = h*f1(t, x, y); break;
		case 1: pom = h*f2(t, x, y); break;
		}
		K1[i] = pom;
	}
}

void k2(double t, double x, double y)
{
	double pom = 0;
	for (int i = 0; i<r; i++)
	{
		switch (i)
		{
		case 0: pom = h*f1(t + 0.5*h, x + 0.5*K1[0], y + 0.5*K1[1]); break;
		case 1: pom = h*f2(t + 0.5*h, x + 0.5*K1[0], y + 0.5*K1[1]); break;
		}
		K2[i] = pom;
	}
}

void k3(double t, double x, double y)
{
	double pom = 0;
	for (int i = 0; i<r; i++)
	{
		switch (i)
		{
		case 0: pom = h*f1(t + 0.5*h, x + 0.5*K2[0], y + 0.5*K2[1]); break;
		case 1: pom = h*f2(t + 0.5*h, x + 0.5*K2[0], y + 0.5*K2[1]); break;
		}
		K3[i] = pom;
	}
}

void k4(double t, double x, double y)
{
	double pom = 0;
	for (int i = 0; i<r; i++)
	{
		switch (i)
		{
		case 0: pom = h*f1(t + h, x + K3[0], y + K3[1]); break;
		case 1: pom = h*f2(t + h, x + K3[0], y + K3[1]); break;
		}
		K4[i] = pom;
	}
}

void x1()
{
	for (int i = 0; i<r; i++)
		X[i] = X[i] + 1.0 / 6 * (K1[i] + 2 * K2[i] + 2 * K3[i] + K4[i]);
}