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A Homogeneous Linear System w/ Constant Coefficients

  1. Apr 10, 2010 #1
    I'll make this post short.

    The problem just asks me to something in the form x'=Ax (A is a 2x2 of constants) and then describe the behavior of the solution as t approaches infinity.

    My solution is x=C1e-2t(2/3 1)T + C2e-t(1 1)T.

    Since both vectors are multiplied by 1/et, my solution just goes to (0 0)T as t->∞, right?
     
  2. jcsd
  3. Apr 10, 2010 #2

    tiny-tim

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    Hi Jamin2112! :smile:
    Yes …

    but from which direction? :wink:

    (and if your solution is for x, not xT, why is it a transpose? :confused:)
     
  4. Apr 10, 2010 #3
    Should indeed be xT

    From the sketch I drew, it approaches the origin along y=x in quadrant 1 and goes away from the origin along y=x in quadrant 3. Correct?
     
  5. Apr 10, 2010 #4

    tiny-tim

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    Yes, it approaches the origin along y=x. :smile:

    But what do you mean by "goes away from the origin" …

    isn't there only one way of getting to t = ∞ ? :confused:
    (and what if C2 = 0 ? :wink:)
     
  6. Apr 10, 2010 #5
    The equation was x'=[1 -2; 3 -4]x (wrote the matrix MATLAB-style)

    So at x=(1 1)T, x'=(-1 -1)T, and so on ...

    And at x=(-1 -1)T, x'=(1 1)T, and so on ...

    See what I mean?
     
  7. Apr 10, 2010 #6
    In general, I'm confused about these questions about what x(t) approaches as t approaches infinity.

    For a different problem on my homework, I got a solution x= 2e3t(1 5)T -e-t(1 1)T. Obviously, the first part, 2e3t(1 5)T, will dominate as t --> ∞. Does that mean the x approaches (∞ 5∞)T?
     
  8. Apr 10, 2010 #7

    tiny-tim

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    What happens as t approaches ∞ is the same for matrices as it is for ordinary numbers …

    so yes in your example, 2e3t(1 5)T, will dominate, and that approaches infinity along the (1 5)T direction. :smile:
     
  9. Apr 10, 2010 #8
    I see. Can I ask one more question?
     
  10. Apr 10, 2010 #9

    tiny-tim

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    uhhh? :redface: depends what it is!! :biggrin:
     
  11. Apr 10, 2010 #10
    Consider the vectors x(1)(t)=(t 1)T and x(2)(t)=(t2 2t)T.

    (a) Compute the Wronskian x(1) and x(2).
    (b) In what intervals are x(1) and x(2) linearly independent?
    (c) What conclusion can be drawn about the coefficients in the system of homogeneous differential equations satisfied by x(1) and x(2)?
    (d) Find this system of equations and verify the conclusions of part (c).

    ------------------------*snip*----------------------

    Wronskian ≠ 0 <==> system has a solution

    W|(t 1)T, (t2 2t)T|= t2, which only equals zero when t=0.

    Making x(1), x(2) into a system of homogenous differential equations would look like

    C1(t 1)T + C2(t2 2t)T=(0 0)T.

    I don't understand what the point of (c) is. Where are they going with this? Please explain in as much or as little detail as possible parts (c) and (d). I'm ready to learn.
     
  12. Apr 11, 2010 #11

    tiny-tim

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    Hi Jamin2112! :smile:

    (just got up :zzz: …)

    Sorry, Wronskians are not my field :redface:

    you'd better start another thread on this one. :smile:
     
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