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Complex Eigenvectors

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data

    The higher order equation y"+y=0 can be written as a unknown d/dt[y y']=[y' y"]=[y' -y]

    If this is du/dt=Au, what is the 2x2 matrix A? Find its eigenvectors and eigenvalues, and compute the solution THAT STARTS FROM y(0)=2, y'(0)=0.

    2. Relevant equations

    y'=Ay
    y(0)=y0

    3. The attempt at a solution

    I found matrix A
    [0 1
    -1 0].
    The eigenvalues are i and -i, and the eigenvectors
    [1 -i]^T
    [1 i]^T

    I found the geneal solution to be:
    y(t) = c1eit[1 i]^T+c2e-it[1 -i]^T

    Which is equivalent,
    y(t)=c1[cos(t) -sin(t)]^T + c2[sin(t) cos(t)]^T

    I just don't know how to incorporate the initial conditions that y(0)=2 and y'(0)=0???

    Any ideas???
     
  2. jcsd
  3. Nov 6, 2011 #2
    You know the function y(t), and you know what y(0) equals (and you know y'(t) from differentiating your original y(t) equation), so then try plugging in for t and see what you get.
     
  4. Nov 6, 2011 #3
    That's what I tried doing but I'm getting a funky solution:

    y(0)=2=c1*[1 0]^T + c2*[0 1]^T, so this is saying that [c1 c2]^T=2 ?????

    When I differentiate y(t) as in previous post,
    y'(t)=[c2*cos(t)-c1*sin(t) -c1*cos(t)-c2*sin(t)]^T
    y'(0)=0=[c2 -c1]^T=0

    Totally lost!
     
  5. Nov 7, 2011 #4

    I like Serena

    User Avatar
    Homework Helper

    Hi tatianaiistb! :smile:

    Your solution is not y(t), but [y(t) y'(t)] which is a vector as it should be.
     
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