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This is my first time solving a problem like this and I just wanted to make sure if what I did was correct.

\begin{equation} x'_1=5x_1 + 2 x_2 - x_3 \\ x'_2=-2x_1 + x_2 - 2x_3 \\ x'_3=-6x_1 - 6 x_2 \end{equation}

Using the eigenvalues and eigenvectors, I found the matrix $P$ such that $P^{-1} AP$ is a diagonal matrix where $A$ is the coefficient matrix of the system above.

$P=$ \begin{pmatrix} 0&-1&-1\\ 1&1&0\\ 2&0&1 \end{pmatrix}

$P^{-1} AP=$ \begin{pmatrix} -3&0&0\\ 0&3&0\\ 0&0&6 \end{pmatrix}


I am sure everything above is correct. So assuming it is correct, is the following process correct?

$U'=(P^{-1} AP)U$

\begin{equation} u'_1=-3u_1 \\ u'_2=3u_2 \\ u'_3=6u_3 \end{equation}

\begin{equation} u_1=c_1e^{-3x} \\ u_2=c_2e^{3x} \\ u_3=c_3e^{6x} \end{equation}

$U=$ \begin{pmatrix} c_1e^{-3x} \\ c_2e^{3x} \\ c_3e^{6x} \end{pmatrix}

$X=PU$

I understand this is pretty long. I'd appreciate if someone can just give a quick look at the process and the substitutions I made (with U and U'). Thanks.

EDIT Comment below helped me answer my own question.

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The easiest verification is seeing if your solution satisfies the original system of differential equations. Does it? –  kigen Dec 10 '12 at 7:32
    
Did not realize that. I tried it and it does! Thank you. So I guess this question is closed. –  Jey Dec 10 '12 at 7:39
1  
No, this question isn't closed. It will remain unanswered and will forever periodically take up space on the main page unless you do something about it. You can either, preferably, write an answer yourself and accept it, or, if you don't want to invest that effort, you could delete the question. –  joriki Dec 10 '12 at 8:07
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