- #1
kidsmoker
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Homework Statement
http://img51.imageshack.us/img51/853/39983853.jpg
2. The attempt at a solution
Q3.1
I get the general solution as
[tex]x(t) = Ae^{3t}+Be^{-t} + cost - 2sint[/tex] .
Q3.2
Letting
[tex]y=\dot{x}[/tex]
and using the general solution, we get
[tex]y=3Ae^{3t}-Be^{-t}-sint-2cost[/tex] .
Therefore the solution in the form they ask for is
[tex](Ae^{3t}+Be^{-t} + cost - 2sint, 3Ae^{3t}-Be^{-t}-sint-2cost, t)[/tex] .
Or am I misunderstanding?
Q3.3
[tex]x_{0}=x(0)=A+b-1 , y_{0}=y(0)=3A-B-2[/tex] .
Solving these simultaneously gives
[tex]A=0.25(x_{0}+y_{0}+1) , B=0.25(3x_{0}-y_{0}-5)[/tex]
so the Poincaré mapping is
[tex](0.25(x_{0}+y_{0}+1)e^{3t}+0.25(3x_{0}-y_{0}-5)e^{-t}+cost-2sint, 0.75(x_{0}+y_{0}+1)e^{3t} - 0.25(3x_{0}-y_{0}-5)e^{-t}-sint - 2cost, t)[/tex]
Is that correct so far?
To find the fixed points, do I let x(2pi)=x(0), y(2pi)=y(0) and solve for x(0) and y(0)? I tried this but I get something ridiculously complicated, so I'm worried I'm not understanding the question correctly at all...
Please help! :-( Thanks.
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