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Thread: Simple eigenvalue problem (x'=[2x2]x)

  1. #1
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    Simple eigenvalue problem (x'=[2x2]x)

    Hi, I've got a homework problem that should be very simple, the previous problem was identical (different numbers) and was a breeze, but this one is giving me problems. Here's what we're asked for:
    Apply the eigenvalue method to find the particular solution to the system of differential equations
    $\displaystyle x'=\begin{bmatrix}
    4 & 4\\
    3 & 5
    \end{bmatrix}x$
    which satifies the initial conditions
    $\displaystyle x(0)=\begin{bmatrix}
    -4\\
    2
    \end{bmatrix}$

    So I begin by finding the eigenvalues from the characteristic equation:
    $\displaystyle (4-\lambda)(5-\lambda)-12=(\lambda^2-9\lambda+8)=(\lambda-1)(\lambda-8)$
    $\displaystyle \lambda=1,8$
    To get the matrices:
    $\displaystyle \begin{bmatrix}
    3 & 4\\
    3 & 4
    \end{bmatrix}\begin{bmatrix}
    a\\
    b
    \end{bmatrix}=0$
    and
    $\displaystyle \begin{bmatrix}
    -4 & 4\\
    3 & -3
    \end{bmatrix}\begin{bmatrix}
    a\\
    b
    \end{bmatrix}=0$
    To get
    $\displaystyle \begin{bmatrix}
    a\\
    b
    \end{bmatrix}=\begin{bmatrix}
    -4\\
    3
    \end{bmatrix}$ and $\displaystyle \begin{bmatrix}
    a\\
    b
    \end{bmatrix}=\begin{bmatrix}
    1\\
    1
    \end{bmatrix}$

    Which gives the particular solution $\displaystyle x_p=a\begin{bmatrix}
    -4\\
    3
    \end{bmatrix}e^t+b\begin{bmatrix}
    1\\
    1
    \end{bmatrix}e^{8t}$

    But at this point, no values for a and/or b will satisfy $\displaystyle x(0)=\begin{bmatrix}
    -4\\
    2
    \end{bmatrix}$

    Can anyone help me out, please?
    Thanks!
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  2. #2
    A Plied Mathematician
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    Don't you just need to solve the system

    $\displaystyle -4a+b=-4,$

    $\displaystyle 3a+b=2?$

    That system has one unique solution, as far as I can tell.
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  3. #3
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    Oh, yes, you're right. I got:
    $\displaystyle (-4)6/7e^t-4/7e^{8t}$
    and
    $\displaystyle (3)6/7e^t-4/7e^{8t}$
    Which are both correct.

    Thank you!
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  4. #4
    A Plied Mathematician
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    You're very welcome. Have a good one!
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  5. #5
    MHF Contributor

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    there HAS to be a solution. at t = 0, the exponential functions are both 1. and since (-4,3) and (1,1) are linearly independent, they span a 2-dimensional subspace of R^2, which MUST be R^2 itself.

    hint: a and b are not integers.
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