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Math Help - 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
    x'=\begin{bmatrix}<br />
4 & 4\\ <br />
3 & 5<br />
\end{bmatrix}x
    which satifies the initial conditions
    x(0)=\begin{bmatrix}<br />
-4\\ <br />
2<br />
\end{bmatrix}

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

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

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

    -4a+b=-4,

    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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    Joined
    Feb 2010
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    Oh, yes, you're right. I got:
    (-4)6/7e^t-4/7e^{8t}
    and
    (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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