System of coupled second order differential equations.

  • Thread starter Deadstar
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  • #1
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Hey folks I'm looking for a way to find the characteristic equation for a second order coupled system of differential equations such as...

[tex]\ddot{x} + A\dot{y} + Bx = 0[/tex]

[tex]\ddot{y} + C\dot{x} + Dy = 0[/tex]

Where x and y are functions of time.

I know I can solve it by setting x and y to standard results (trig, exponential) but I'd like to know a method to solving this rather than plug and solve for coefficients.

Specifically I'd like to know how to find the characteristic equation for this. I've tried setting it to a first order system but I can't see it leading anywhere (or perhaps I just did it wrong...).

I don't want a full answer, just the name of a method or something like that.
 

Answers and Replies

  • #2
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Actually I think I've got it...

Setting [tex]x_1 = x[/tex], [tex]y_1 = y[/tex], then converting into four first order differential equations. Find eigenvalues of 4x4 matrix etc... Think that's it.
 
  • #3
vela
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It's not entirely clear what you did other than add a subscript to x and y, but I suspect you have the right approach.
 
  • #4
lurflurf
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rewrite the system (I will use E for your D and D for differentiation)
(D^2+B)x+(AD)y=0
(CD)x+(D^2+E)y=0
then the characteristic polynomial is
(D^2+B)(D^2+E)-ACD^2=
D^4+(B+E-AC)D^2+BE
as expected.
 

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