Consider the solution to x''= -x with the initial conditions x(0) = 0, x'(0) = 1. Find the alpha and omega limit sets for this solution.
(Then rewrite as a 2x2 system)
If you solve the ODE you get
The definition of an omega limit point is if there exists a sequence that goes to infinity such that
Since there exists such a sequence for any
This is the omega set.
The alpha set is the same except the limit must go to negative infinity.
for the next part let
Now use the ODE to write this as a first order system.
Ty for your fast response.
I also got the x(t) = sin(t) after staring at the problem for a little.
But, I think that the way you wrote the limit set and what I got are different.
-limit set = {sin(t) : t [-2pi, 0]}
-limit set = {sin(t) : t [0, 2pi]}
Unless, what you wrote and what I wrote mean the same thing.
Also, for the last part, I seemed to have forgotten a word. "Rewrite the equation as a 2x2 system". I am not sure which equation it is referring to.
But would it be:
(cos(t) sint(t)
-sin(t) cos(t))
sin(t) and cos(t) take on all values between -1 and 1.
There is only one equation given: x''= -x. As TheEmptySet suggested, let y= x' so your equation becomes y'= -x and you have the system of equations x'= y, and y'= -x.Also, for the last part, I seemed to have forgotten a word. "Rewrite the equation as a 2x2 system". I am not sure which equation it is referring to.
But would it be:
(cos(t) sint(t)
-sin(t) cos(t))