If the differential equations
$$ y'' + 0.5yy'z = \sin(x) $$
$$ y' - z'' + zy = \cos(y), \quad 0 \leq x \leq 6 $$
are rewritten as a system of (n) first-order differential equations, then (n).
I answered 6 because I thought we need to know have for y, y', y'', z, z', z''. But the actual answer is 4. I would appreciate it if someone could explain this for me!
Explicitly, those equations are equivalent to the following system of first order ODEs: $$ \begin{cases} y'=p, \\ p'+0.5ypz=\sin(x), \\ z'=q, \\ p-q'+zy = \cos(y). \end{cases} $$