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Translating a linear system to have critical points at the origin?

  1. Sep 18, 2012 #1
    So suppose you have a nice simple linear system
    dx/dt = a1x + b1y + c1
    dy/dt = a2x + b2y + c2

    and a critical point (x0, y0) ≠ (0,0). If you want to analyze the stability in the usual linear method, then I assume there should be some transformation to have the critical point at the origin. So is this system equivalent to the system that results from simply removing c1, c2? I see that in terms of taking the Jacobian, the constants don't matter at all. Furthermore, I also see that if there is no b's in either equation of a line, then of course they can only intersect at (0,0) -> that is a critical point as well. So I guess what I am asking is if the linear system for a given a1,a2,b1,b2 but and c1,c2 are all equivalent? Thanks!

    Okay, I thought about it more and realized that substituting x' = x - x_0 and y' = y - y_0 where x_0, y_0 are the critical points. So can one just ignore the constants or is this intermediate step needed for rigor?
    Last edited: Sep 18, 2012
  2. jcsd
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