//simple demo of runga-kutta method to solve diff-eq


//this example solves the equation y'= x^2 + y where f(0)=-1

public class rungakutta
{
    
    public static void main ( String args[] )
    {
		
	//for this example we are going to get the answer
	//at 0.10  since f(0) is our starting point the 
	//difference between 0 and .1 is .1
	double h = .1;
	
	//now we do 4 iterations to get our answer
	double x0 = 0;
	double y0 = -1;
	
	double k1 = (solve( x0, y0 ) )*h;
	double k2 = (solve( x0+(h/2), y0+(k1/2)) )*h;
	double k3 = (solve( x0+(h/2), y0+(k2/2)) )*h;
	double k4 = (solve( x0+h, y0+k3) )*h;
	
	double y1 = y0 + k4;
	
	
	System.out.println ( "k1 " + k1 + "  k2 " + k2 + "  k3 " + k3 + "  k4 " + k4 );
	System.out.println ( "result " + y1 );
	
	    
    }
    
    
    static double solve ( double x, double y )
    {
	System.out.println ( "x " + x + "   y " + y );
	
	// x^2 + y this is the diff-eq we are solving
	return ( x*x + y );
    }
    
}