Lab 6
	   Euler scheme for systems of ODEs  
	            (preparation for project 2)

It is easy to modify your Euler scheme code for a system of (two) ODEs:

      y1'(t) = f1(t, y1(t), y2(t)),    y1(t0) = y10
      y2'(t) = f2(t, y1(t), y2(t)),    y2(t0) = y20

Copy the file with your scalar Euler scheme code to a file with a new name 
(e.g. Euler2.m), and make the necessary changes.  
You can simply rename the single old unknown to y1n, and introduce 
the 2nd unknown y2n (and y20, and f2), and a 2nd Euler update for y2. 
f1 and f2 should be calculated in a FCN subprogram.

Debug your code on the easy problem, for 0 ≤ t ≤ 1:
            y1′ = y2 ,  y1(0) = 1,
            y2′ = y1 ,  y2(0) = 0.
This can be solved exactly: Setting z=y1, we have z'= y1' = y2, and z''= y2'= y1 = z,  
so this system is equivalent to the 2nd order linear ODE  z'' − z = 0,  with ICs:  z(0)=1, z'(0)=0.  
Solve it to show that the unique solution is: y1(t) ≡ z(t) = cosh(t) , y2(t) ≡ z'(t) = sinh(t)

For debugging, calculate Y1exact(tn) = cosh(tn)
and the error ERRn = ABS(Y1exact - y1n).  At each time step, output 
	tn        y1n       Y1exactn       ERRn
and calculate the worst overall error: ERRmax = MAX( ERRmax, ERRn).  
Output ERRmax at the end. After debugging, comment out the printing!
Make runs with N=1000 , N=5000 , N=10000  time-steps.

Submit ONLY the following:
	from run with N=1000:  input , y1(1) , ERRmax
	-----------------------------------------------
	from run with N=5000:  input , y1(1) , ERRmax
	-----------------------------------------------
	your code