Initial values lead to periodic solution

mmnoname
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Homework Statement





Homework Equations



How do you know when a given set of initial values for an ODE produce a periodic solution?

The Attempt at a Solution



I tried playing around with it for a bit and then searching the net but came up with nothing
 
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mmnoname said:

Homework Statement





Homework Equations



How do you know when a given set of initial values for an ODE produce a periodic solution?

The Attempt at a Solution



I tried playing around with it for a bit and then searching the net but came up with nothing


What do you mean with periodic solution? Probbably i am not used to this term. DO you mean like for example that sinx has a period of 2pi, and so its values after 2pi start to repeat, sth like this, right? or?-
 
yup that's it. Also by the wording of the question it sounds as if it's asking for all cases not just the particular ODE given. But if it makes any difference, it's a second order ODE that's pretty long so I'd rather not type it up unless you really need it
 
Are you saying that even if you have a genereal solution of a diff eq. for example

y(x)=e^{x}+c_1e^{-x}+c_2xe^{x} then for y(a)=b , y'(a)=k. you are saying that for what values of a,b,k the solution y(x) will be periodic, right?

I just took that example at the top of my head, but we run into those kind of problems all the time.

If you are saying this, then i have no clue either, sorry! Unles there are at least one function on the general solution that is by itself periodic, otherwise i would also be interested to know.
 
I don't think we are even suppose to know the general solution but we have to figure it out from the question
 
mmnoname said:
I don't think we are even suppose to know the general solution but we have to figure it out from the question

AH, sorry, i don't think i am getting you right. Is that question all you were given?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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