Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $:))$

share|improve this question
4  
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
add comment

1 Answer 1

up vote 2 down vote accepted

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 $C=0$.

share|improve this answer
    
thanks a lot.. I always seem to miss the obvious path but at least I know my intuition was right ! –  Ronald Sep 23 '12 at 18:46
    
@Ronald: no problem. Glad to help. –  Thomas Sep 23 '12 at 18:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.