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Finding the general solution of a system of differential equations

  1. Apr 6, 2011 #1
    1. The problem statement, all variables and given/known data
    The question is:
    question 7.JPG

    2. Relevant equations
    I really don't know what to put here but my method is:
    -Find det(A-[tex]\lambda[/tex]I)
    -Find the roots of the determinant - which are the eigenvalues
    -Solve for (X -([tex]\lambda[/tex])I)

    I am stuck at this part

    3. The attempt at a solution

    So I managed to find the determinant, which is:
    [tex]\lambda[/tex]^3 -5[tex]\lambda[/tex]^2 -3[tex]\lambda[/tex] +9

    The Eigenvalues are:
    [tex]\lambda[/tex]1 = 1
    [tex]\lambda[/tex]2 = -3
    [tex]\lambda[/tex]3 = -3

    Now I'm trying to solve for the eigenvector of eigenvalue 1, however manipulating the matrix does not yield anything that I can use to solve for k1 k2 and k3

    Any help with this will be greatly appreciated,

  2. jcsd
  3. Apr 6, 2011 #2


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    Staff Emeritus
    Science Advisor

    An eigenvector [tex]\begin{pmatrix}x \\ y \\ z\end{pmatrix}[/tex] corresponding to eigenvalue 1 must, of course, satisfy
    [tex]\begin{pmatrix}1 & -1 & 0 \\ 0 & -1 & -2 \\ 0 & 2 & -5\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y \\ z\end{pmatrix}[/tex]

    [tex]\begin{pmatrix}x - y \\ -y- 2z \\ 2y- 5z \end{pmatrix}= \begin{pmatrix}x \\ y \\ z\end{pmatrix}[/tex]
    Which means we must solve x- y= x, -y- 2z= y, 2y- 5z= z. From the first, equation, subtracting x from both sides, -y= 0 so y= 0. Putting that into both of the other equations, z= 0. Therefore, any eigenvector must be of the form <x, 0, 0>.
  4. Apr 6, 2011 #3
    Hey, thanks for the response, I now understand what you have mentioned. Now I have run into another issue, finding the third eigenvector. I managed to find the second eigenvector for eigenvalue2, which is <1,4,4>. Now the third eigenvalue is the same as the second so I would assume that the eigenvector will be the same however this is not the case. Using Wolfram alpha, the third eigenvector is <0,0,0> and I don't know how to get this eigenvector. Any help would be great,
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