Industrial Mathematics - Alexiades
                      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. The functions 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. Observe how the error decreases.

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