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Math Help - Solving First order differential equations using matrix methods

  1. #1
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    Solving First order differential equations using matrix methods

    I have been trying to solve this equation using matrix methods to find the general solution for x1(t) and x2(t)
    (There should be a dot above the x1, x2 to show first order)

    x1 = 4x1+ 7x2

    x2 = -
    6x1-9x2

    Now i have been shown how to do a second order question so i have used a similar method with this which gets me to this point.

    det (a-lambda^2I) = (4-lambda^2)(-9-lambda^2)+42
    (lambda^2+2)(lambda^2+3)

    lambda^2 = -2 and lambda^2=-3

    This is where its different to the question i have been through

    As this would mean lambda is j1.414 and j1.732

    The second order question i have been through comes out with lambda +or- 2j and +or- j which is then obvious how to continue.

    Could somone point me in the right direction or to where ive gone wrong?

    I have an exam on wednesday and realy need to be able to work with any combination of question they give me, It could be first or second order.
    I didnt think it mattered which kind i was until differentiating the general solution part.


    Thank you!
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  2. #2
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    Check this out.
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  3. #3
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    Ive just been reading through that thread and i can't understand unfortunatly, How would i go through my question step by step? are my eigen values correct?
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  4. #4
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    You're setting up the wrong characteristic equation. Your system looks like this:

    \dot{\vec{x}}=A\vec{x}, where

    A=\begin{bmatrix}4 &7\\-6 &-9\end{bmatrix}.

    The solution to the system is \vec{x}(t)=e^{At}\vec{x}(0).

    We now have to make sense of the exponential. You use the series definition:

    e^{At}=I+tA+\frac{(tA)^{2}}{2}+\frac{(tA)^{3}}{6}+  \dots

    It's the usual series expansion of the exponential function. We can easily find the nth power of A if we can find an invertible P such that A=PDP^{-1}, where D is diagonal. For then

    A^{n}=(PDP^{-1})(PDP^{-1})\dots(PDP^{-1})=PDD\dots DP^{-1}=PD^{n}P^{-1}.

    Taking the nth power of a diagonal matrix is the same as taking the nth power of the diagonal elements, component-wise. Solving A=PDP^{-1} is the diagonalization problem. To diagonalize, assuming it's possible to do this, you find the eigenvalues and eigenvectors. The eigenvalues make up the diagonal of D, and the eigenvectors make up the columns of P. Hence the importance of the eigenvalues for a system of ODE's.

    Now, to find the eigenvalues, you have to set \det(A-\lambda I)=0, not \det(A-\lambda^{2}I)=0. That's where you first went wrong. So carry that correction through, and see what you come up with.
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  5. #5
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    Ok ive done it using dec (A-lambdaI)=0 which would give me Lambda= -2 and -3
    Then putting that back into the matrix i get.

    For lambda -2 6x1=-7x2
    Lambda -3 x1=-x2

    x1 = 6(Ae^-2jt + Be^-3jt)
    x2 = -7(Ae^-2jt -Be^-3jt)

    to be honest im lost, i do not understand the exponential part.
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  6. #6
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    So your eigenvectors turn out to be

    \vec{x}_{-2}=\begin{bmatrix}1\\ -6/7\end{bmatrix}, and

    \vec{x}_{-3}=\begin{bmatrix}1\\ -1\end{bmatrix}.

    Therefore, your invertible P can be written as

    P=\begin{bmatrix}1 &1\\-6/7 &-1\end{bmatrix}.

    To form your final solution, you write

    \vec{x}(t)=e^{At}\vec{x}(0)=Pe^{Dt}P^{-1}\vec{x}(0).

    But D=\begin{bmatrix}-2 &0\\0 &-3\end{bmatrix}. So its exponential

    e^{Dt}=\begin{bmatrix}e^{-2t} &0\\0 &e^{-3t}\end{bmatrix}.

    Can you finish from here?
    Last edited by Ackbeet; August 20th 2010 at 01:33 PM. Reason: Incomplete.
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  7. #7
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    I dont understand why there is a 1 in the first vector bracket? i only get either 6 and -7 or 7 and -6.
    Then i realy do not understand where the final solution comes from I can see that D and e^Dt are from the two eigen values so do you then work the determinant for that out?

    Please forgive my lack of understanding but i realy am trying to understand.
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  8. #8
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    I dont understand why there is a 1 in the first vector bracket? i only get either 6 and -7 or 7 and -6.
    Any nonzero scalar multiple of an eigenvector is an eigenvector. Why is that, might you ask? Well, recall that a vector \vec{x} is an eigenvector of the matrix A if and only if \vec{x}\not=0 and A\vec{x}=\lambda \vec{x} for some scalar \lambda, which is called the eigenvalue corresponding to \vec{x}. So, suppose that \vec{x} is an eigenvector of the matrix A with corresponding eigenvalue \lambda. Let y be a nonzero scalar. Then A(y\vec{x})=yA\vec{x}=y\lambda\vec{x}=\lambda(y\ve  c{x}). Since A(y\vec{x})=\lambda(y\vec{x}), and y\vec{x}\not=0 (since y\not=0 and \vec{x}\not=0), it follows by definition that y\vec{x} is an eigenvector of matrix A.

    Now, if you look at my \vec{x}_{-2}, and multiply by 7, you might get a more recognizable eigenvector. Does that make sense?

    I can see that D and e^Dt are from the two eigenvalues; so do you then work the determinant for that out?
    I don't compute the determinant, I just compute the exponential e^{Dt}.

    I'm not sure I understand what you're not understanding. The solution to this problem: pepper me with questions. Anything at all you don't understand, ask about. Ok?
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