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Finding the general solution of a system of differential equations 
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#1
Apr611, 12:49 AM

#2
Apr611, 05:13 AM

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PF Gold
P: 39,682

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>. 


#3
Apr611, 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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