# Just a differential equation.

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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That's right, but it's easier to just integrate the whole equation twice with respect to $t$. – Hans Lundmark Sep 23 '12 at 17:47
@HansLundmark Thanks.. looks like I took the long route but at least I didn't do anything silly/wrong .. – Ronald Sep 23 '12 at 18:41

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 $C=0$.