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Another repeated roots problem

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the solution of the initial value problem.

    x'=[3 9; -1 -3]x, x(0)=(2 4)T

    2. Relevant equations

    A good guess should be x=tert$ + ert#

    3. The attempt at a solution

    But ..........

    the only root is r=0, which gives me a solution x(1)(t)= C1(-3 1)T. Using the guess above, stuff goes wrong because I end up with tA$ + A# -$ = 0, which just gives me #=(-3 1)T as well.
    Last edited: Apr 24, 2010
  2. jcsd
  3. Apr 25, 2010 #2


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    I don't see how you are getting this. If your repeated root is r=0, then your solution is x=te0t$ + e0t#=t$+#...substituting your initial condition gives you (2,4)T=0$+#=#
  4. Apr 25, 2010 #3
    Then I'd have a solution of the form x=C1(-3 1)T + C2(2 4)T, which is different from the solution in the back of the book.
  5. Apr 25, 2010 #4


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    You need to give us a bit more detail. What you've written doesn't really make sense, and it's hard to figure out what you're doing since you've omitted all the steps.
    What do A, $, and # stand for? Is A a matrix? Similarly, what exactly are C1 and C2?
    I'm still not sure how you came up with the vector (-3 1)T, why the constant vectors are multiplied by C1 and C2, and where the t and the exponential went.
  6. Apr 25, 2010 #5
    $,# are vectors with constant entries
    A is a matrix.
    Boldness means a non-scalar, in my personal math language.

    The problem I'm having comes from the fact that 0 is an eigenvalue.

    Say there was one eigenvalue a≠0. One solution would be eat$, where $ is the eigenvector associated with a. The I'd guess a second solution x(2)(t)=teat$ + eat#.

    -----> x'= (ateat+eat)$ + aeat#.

    And plugging this back into x'=Ax,

    (ateat+eat)$ + aeat# = teatA$ + eatA#,

    which is true if


    eat is never equal to zero, so to simplify,


    Notice that I can solve for $, since there's the first equation is an eigenvalue problem. In the particular problem this thread is about, a=0 and $=(-3 1)T. Plug that into the second equation and you can solve for #. But since a=0, we just get A#=$ ------> #1 = -3#2 - 1; #2=#2 ------> # = (-3 1)T + k(-1 0)T

    ------> x= C1teat(-3 1)T + C2eat [ (-3 1)T + k(-1 0)T ]
    Last edited: Apr 25, 2010
  7. Apr 25, 2010 #6


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    Thanks for the explanation. Sorry, my brain is a bit slow this morning/afternoon. I think you're just getting the constants messed up a bit. First, the k is in the wrong place. You should have #=k(-3 1)T+(-1 0)T, so your solution should be

    x = C1[(-3 1)Tteat+[k(-3 1)T+(-1 0)T]eat] + C2(-3 1)Teat

    so when you rearrange everything, you get

    x = C1[(-3 1)Tteat+(-1 0)Teat] + (kC1+C2)(-3 1)Teat

    Because the C's are arbitrary, you can just set k=0, which you could have done right from the start to simplify the subsequent algebra. So finally, you get

    x = C1[(-3 1)Tteat+(-1 0)Teat] + C2(-3 1)Teat

    to which you now apply the initial conditions.
  8. Apr 25, 2010 #7
    But since the initial condition is at t0=0, the constant that goes with the t term will disappear, and there won't be a way to solve for it.

    The answer in the back of the book is x=2(1 2)T + 14(-3 1)Tt, so there is a t term---somehow.

    By the way, thank you for your cooperation. I have a midterm Wednesday, so I'm just making sure I do the homework questions correctly.
  9. Apr 25, 2010 #8


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    The other term multiplying C1 won't disappear, though, so you'll still have two equations and two unknowns.
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