 # Let [ <mtable rowspacing="4pt" columnspacing="1em"> <mtr> <mtd> Edith Mayer 2022-05-17 Answered
Let
$\left[\begin{array}{ccc}-2& 1& 0\\ 0& -2& 1\\ 0& 0& -2\end{array}\right]$
and
$|x\left(t\right)|=\left({x}_{1}^{2}\left(t\right)+{x}_{2}^{2}\left(t\right)+{x}_{3}^{2}\left(t\right){\right)}^{1/2}$
Then any solution of the first order system of the ordinary differential equation
$\left\{\begin{array}{r}{x}^{\prime }\left(t\right)=Ax\left(t\right)\\ x\left(0\right)={x}_{0}\end{array}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}$
satisfies
1. $\underset{t\to \mathrm{\infty }|x\left(t\right)|=0}{lim}$
2. $\underset{t\to \mathrm{\infty }|x\left(t\right)|=\mathrm{\infty }}{lim}$
3. $\underset{t\to \mathrm{\infty }|x\left(t\right)|=2}{lim}$
4. $\underset{t\to \mathrm{\infty }|x\left(t\right)|=12}{lim}$
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Since −2 is the only eigenvalue of $A$, the general solution of ${x}^{\prime }\left(t\right)=Ax\left(t\right)$ has the form
$x\left(t\right)={e}^{-2t}\left({c}_{1}{v}_{1}\left(t\right)+{c}_{2}{v}_{2}\left(t\right)+{c}_{3}{v}_{3}\left(t\right)$
where ${c}_{1}$,${c}_{2}$ and ${c}_{3}$ are constants and ${v}_{j}\left(t\right)$ is a vector in ${\mathbb{R}}^{3}$ whose coordinates are polynomials with grade $\le j-1$.
Hence $|x\left(t\right)|\to 0$ for $t\to \mathrm{\infty }$.