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Solving Differential - Equation of Motion

by aerowenn
Tags: differential, equation, motion, solving
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aerowenn
#1
Jan30-13, 11:48 PM
P: 19
Let's say I have the equation:

##\ddot {\theta}(t)(J + y(t)^2) + 2 \dot {\theta}(t) y(t) \dot y(t) + \ddot y(t)Jn##

It's the general form of an equation I'm working with to describe the motion of a beam. As you can see both ##{\theta}(t)## and y(t) are equations of t. J and Jn are just constants.

I'm wanting to solve for ##{\theta}(t)##, ##\dot {\theta}(t)##, and ##\ddot {\theta}(t)## as a function of time. These will correspond to position, velocity, and acceleration around an axis.

I'm not sure how to go about this (differential) generally, I'm wanting the solutions mentioned about to come out something like:

##{\theta}(t)## = (equation of t)

Any help would be greatly appreciated!
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jedishrfu
#2
Jan31-13, 12:30 AM
P: 2,769
have you tried using e^iat style functions for theta(t)?

I ask because its sometimes used when you periodic motion which in your case is rotating about an axis.
HallsofIvy
#3
Jan31-13, 10:03 AM
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That's not a differential equation. Is one of those "+" signs supposed to be an "=" sign? Or is the whole formula equal to something, say "0"?

aerowenn
#4
Jan31-13, 10:07 AM
P: 19
Solving Differential - Equation of Motion

Quote Quote by HallsofIvy View Post
That's not a differential equation. Is one of those "+" signs supposed to be an "=" sign? Or is the whole formula equal to something, say "0"?
Terribly sorry about that, you are correct. All of that is equal to 0.

As for the other response, I thought of that, but I'm not sure the general solution to second order differential equations applies here. Both functions are dependent on "t".


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