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Dif.equation system

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data
    System1:
    dx/dt=x+y
    dy/dt=8x-y


    2. Relevant equations



    3. The attempt at a solution
    detreminant=(1-λ)(-1-λ)=(λ-3)(λ+3);λ[itex]_{1}[/itex]=-3 and λ[itex]_{2}[/itex]=3

    So system 2:
    (1-λ)[itex]\alpha[/itex]+[itex]\beta[/itex]=0
    8[itex]\alpha[/itex]+(-1-λ)[itex]\beta[/itex]=0

    When i put λ[itex]_{1}[/itex]=-3 in system 2 -> [itex]\alpha[/itex] and [itex]\beta[/itex]=0.
    the same goes for λ[itex]_{2}[/itex]

    That menas that solution in form of y=C_1*[itex]\beta[/itex]_1*exp(λ[itex]_{1}[/itex]*t)+C_2*[itex]\beta[/itex]_2*exp(λ[itex]_{2}[/itex]*t) is equal to 0. Thats wrong.

    Where is my mistake?
     
    Last edited by a moderator: Oct 21, 2013
  2. jcsd
  3. Oct 21, 2013 #2

    tiny-tim

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    hi prehisto! :smile:

    so far so good! …
    now solve either line to get β = 2α, so your eigenvector is any multiple of x + 2y :wink:
     
  4. Oct 21, 2013 #3
    ok,that means that i can chose α1=1 β1=2 and
    α2=1 β2=-4

    y=C11*exp(λ1*t)+C22*exp(λ2*t)
    x=C11*exp(λ1*t)+C22*exp(λ2*t)
    Is this form of solution correct or I have to use something else?
     
  5. Oct 21, 2013 #4

    tiny-tim

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    i think it would be better if you checked by starting with the eigenvector equations

    x + 2y = Ae3t
    x - 4y = Ae-3t
     
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