How Can I Solve a Linear System of Differential Equations Without Determinants?

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SUMMARY

This discussion focuses on solving a linear system of differential equations represented by the matrix equation \(\begin{pmatrix}\dot{x}_1 \\ \dot{x}_2\end{pmatrix} = \begin{pmatrix}-2 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}\) without using determinants. The eigenvalues identified are \(\lambda = -1\) and \(\lambda = 3\). Feedback indicates that the eigenvectors presented contain an unnecessary factor of \(t\) and that one eigenvector is a multiple of another, suggesting a need for clarification in the solution process.

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mrxtothaz
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I am solving for a linear system of differential equations. Since LaTeX never works out for me on this forum, I have written out my problem and my solution on LaTeX and made an image of it: http://i.imgur.com/ncra.jpg .

I basically have to solve the system of equations and express it as a diagonal matrix. For this, I use the concept of eigenvectors (this is done without determinants). I ran into some difficulties and I am not confident in my solution, so I would really appreciate any feedback on my solution.

Thanks in advance.
 
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There's a bug in the forum software where incorrect images sometimes appear for the rendered LaTeX in previews. The work-around is to reload the page. The correct images will show up then.

Your problem is to solve the system

[tex]\begin{pmatrix}\dot{x}_1 \\ \dot{x}_2\end{pmatrix} = \begin{pmatrix}-2 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}[/tex]

You found the eigenvalues are [itex]\lambda = -1[/itex] and [itex]\lambda=3[/itex]. So far, so good.

Then you wrote the eigenvector for [itex]\lambda = -1[/itex] is

[tex]\vec{v}_1 = t\begin{pmatrix}1 \\ 1\end{pmatrix}[/tex]

The factor of t shouldn't be there.

Next you said the eigenvectors for [itex]\lambda = 3[/itex] are

[tex]\vec{v}_2 = t\begin{pmatrix}-1 \\ 1\end{pmatrix} \hspace{0.5in} \vec{v}_3 = t\begin{pmatrix}1 \\ -1\end{pmatrix}[/tex]

Again, the factor of t shouldn't be there. Also, note that the first vector is just a multiple of the second vector, so they're essentially the same eigenvector.
 

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