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 $:))$