This eqn came toward the end of a much bigger problem, and I'm a bit rusty with these differential equations.. But maybe I got it right (probably not .. )
Anyway..
$$\ddot{Z}(t)=A+Bcos(\omega t)$$
to the best of my knowledge this is a Second Order inhomogenous non-linear ordinary differential equation (quite a mouthful) and can be solved as follows
Soln to homogenous part: $$\ddot{Z}=0 \ \ \ => \ \ \ Z=Ct+D$$
Then the soln to the particular case I wasn't quite as sure but this is what I tried:
let $Z = pt^2 + qcos(\omega t)$ where p, q are arbitrary
then $$\dot Z = 2pt - q\omega sin(\omega t)$$
$$\ddot Z = 2p - q(\omega)^2 cos(\omega t)$$
and thus:
$$2p - q(\omega)^2 cos(\omega t) = A+Bcos(\omega t)$$
$=>$
$$p=A/2 ; q = \frac{-B}{(\omega)^2}$$
which would give us our soln:
$$Z =Ct + D + (A/2)t^2 - \frac{B}{(\omega)^2}cos(\omega t)$$
Is this right ?! and if so, is this the most efficient method of solving this ODE?
.....
If this is right, I have an intial condition that : $$ t=0, => \dot Z = 0 $$
which solves to $C=0$ and
$$Z = D + (A/2)t^2 - \frac{B}{(\omega)^2}cos(\omega t)$$ Which is obviously not unique, and I was wondering about the significance of the undetermined parameter D. Does it just mean that the set of functions that satisfy $\ddot{Z}=A+Bcos(\omega t)$ and $ t=0, => \dot Z = 0 $ are all equivalent with only a translation up the Z axis.
Thanks a lot $:))$