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Thread: 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:

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

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

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

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

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

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

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

    Taking the nth power of a diagonal matrix is the same as taking the nth power of the diagonal elements, component-wise. Solving $\displaystyle A=PDP^{-1}$ 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 $\displaystyle D$, and the eigenvectors make up the columns of $\displaystyle P$. Hence the importance of the eigenvalues for a system of ODE's.

    Now, to find the eigenvalues, you have to set $\displaystyle \det(A-\lambda I)=0,$ not $\displaystyle \det(A-\lambda^{2}I)=0.$ 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

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

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

    Therefore, your invertible $\displaystyle P$ can be written as

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

    To form your final solution, you write

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

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

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

    Can you finish from here?
    Last edited by Ackbeet; Aug 20th 2010 at 12:33 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 $\displaystyle \vec{x}$ is an eigenvector of the matrix $\displaystyle A$ if and only if $\displaystyle \vec{x}\not=0$ and $\displaystyle A\vec{x}=\lambda \vec{x}$ for some scalar $\displaystyle \lambda,$ which is called the eigenvalue corresponding to $\displaystyle \vec{x}$. So, suppose that $\displaystyle \vec{x}$ is an eigenvector of the matrix $\displaystyle A$ with corresponding eigenvalue $\displaystyle \lambda$. Let $\displaystyle y$ be a nonzero scalar. Then $\displaystyle A(y\vec{x})=yA\vec{x}=y\lambda\vec{x}=\lambda(y\ve c{x}).$ Since $\displaystyle A(y\vec{x})=\lambda(y\vec{x}),$ and $\displaystyle y\vec{x}\not=0$ (since $\displaystyle y\not=0$ and $\displaystyle \vec{x}\not=0$), it follows by definition that $\displaystyle y\vec{x}$ is an eigenvector of matrix $\displaystyle A$.

    Now, if you look at my $\displaystyle \vec{x}_{-2},$ and multiply by $\displaystyle 7$, 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 $\displaystyle e^{Dt}.$

    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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