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Finding the general solution of a system of differential equations

by chris_0101
Tags: differential, equations, solution
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chris_0101
#1
Apr6-11, 12:49 AM
P: 71
1. The problem statement, all variables and given/known data
The question is:
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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,

Thanks
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HallsofIvy
#2
Apr6-11, 05:13 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,348
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>.
chris_0101
#3
Apr6-11, 09:02 AM
P: 71
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,
Thanks


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