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Homework Help: Eigenvalue method for solving system of differential equations

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Here is the question along with my work. I attempted to solve for the actual solution using both eigenvectors. From what I have been taught it should yield the same answer... But as you can see (circled in red) the solutions are clearly different. Is this normal or maybe because I am making an algebra mistake?

    I know for sure the solution in attempt 1 is correct because that is what wolfram alpha gives me. But attempt 2 for some reason has a different answer.
  2. jcsd
  3. Dec 9, 2012 #2


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    I really don't understand what you mean by "solving with V1" and "solving with V2". The general solution, from which you get the solution with the given initial values, must be a linear combination of V1 and[ V2. You cannot get a solution from either one alone.
  4. Dec 9, 2012 #3


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    This is not true. You can retrieve a real valued solution from either [itex]x^{(1)}[/itex] or [itex]x^{(2)}[/itex] by taking the real parts of either one by using Euler's formula.
  5. Dec 9, 2012 #4
    Yup, take a look at the first example from Pauls notes. They said you only need one eigenvector:


    So am I doung something wrong or am I suppose to get a slightly different answer?
  6. Dec 9, 2012 #5


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    In your solution with [itex]v_2[/itex], you should have substituted [itex]\mu = -3\sqrt 3[/itex] into the your formula. Since [itex]\sin(-3\sqrt3 t) = - \sin(3 \sqrt 3 t)[/itex], the result is that every term multiplied by [itex]\sin (3 \sqrt3 t)[/itex] has the wrong sign.

    Make that correction and you should get the same result as with [itex]v_1[/itex].
    Last edited: Dec 9, 2012
  7. Dec 9, 2012 #6
    Wow! After doing these for such a long time I did not know that. I always thought you just took the magnitude of the eigenvalue... At least that's what it looked like in the generic formula I wrote and copied from the professor.

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