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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 $:))$

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    @H$a$nsLundmark Thanks.. looks like I took the long route $b$ut at least I didn't do anything silly/wrong ..2012-09-23

1 Answers 1

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It looks correct what you have written. So you have $ \ddot{Z}(t)=A+B\cos(\omega t) $ So $ \dot{Z}(t)=At+B\frac{1}{\omega}\sin(\omega t) + C $ So $ Z(t)=\frac{A}{2}t^2-B\frac{1}{\omega^2}\cos(\omega t) + Ct + D $ You get each step by finding the anti-derivative (i.e. integrating). If $\dot{Z}(0) = 0$, then indeed $C=0$.

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    @Ronald: no problem. Glad to help.2012-09-23