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Find the equilibrium solution and eigenvalues and eigenvectors of system?

  1. Mar 6, 2012 #1
    Hey guys, I need to find the equilibrium solution (critical point) for the given system. Also I need to take the homogeneous equation x' = Ax (matrix notation) and find the eigenvalues and eigenvectors.

    system: x' = -x - 4y - 4
    y' = x - y - 6

    Can you help?

    Last edited: Mar 6, 2012
  2. jcsd
  3. Mar 6, 2012 #2


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    Hi Norm!!! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :wink:

    (start with the eigenvalues and eigenvectors)
  4. Mar 6, 2012 #3
    Okay so, the critical points are when x' and y' equal zero I believe, so adding the two equations gets -5y - 10 = 0 => y = -2, x = 4. So the critical point is (x,y) = (4,-2).

    Now for writing the homogeneous equation in matrix form, by using change of variables:

    x_1 = x
    x_2 = y
    x_2 = y'

    Gives equations:

    (x_1)' = -x_1 - 4x_2 - 4
    (x_2)' = x_1 - x_2 - 6

    And that gives the matrix form x' = Ax + b, which would be (follow link)

    but we want the homogeneous, so we have x' = Ax, which would be (follow link)

    Now to find eigenvalues, det(A - λI) = λ^2 + 2λ + 5.

    This is where i'm confused so far. I can do the quadratic but we haven't had to do quadratic equation yet so I want to make sure i have everything correct so far?

  5. Mar 7, 2012 #4


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    Hi Norm850! :smile:

    (just got up :zzz:)
    Fine so far. :smile:
  6. Mar 7, 2012 #5


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    You will get a homogeneous set of equations if you let [itex]x_1= x- 4[/itex] and [itex]x_2= y+ 2[/itex]. Of course, [itex]x_1'= x'[/itex] and [itex]x_2'= y'[/itex] and [itex]x= x_1+ 4[/itex], [itex]y= x_2- 2[/itex]. Putting those into the equation.

    [itex]x_1'= -(x_1+ 4)-4(x_2- 2)- 4= -x_1- 4x_2[/itex]
    [itex]x_2'= (x_1+ 4)- (x_2- 2)- 6= x_1- x_2[/itex]

    Determine the eigenvalues of
    [tex]\begin{bmatrix}-1 & 4 \\ 1 & -1\end{bmatrix}[/tex]
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