 # Section 7.8 – Repeated Eigenvalues

Directions. The following are review problems for the section. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it.
1. Find the general solution of the system and the fundamental matrix. Classify the type of the critical point (0,0), and determine whether it is stable or unstable. Sketch the phase portrait.${\bf x}'=\begin{pmatrix} 3&-4\\1&-1\end{pmatrix}{\bf x}$

The general solution is $\mathbf{x}(t)=C_1e^{t}\begin{pmatrix}2\\1\\\end{pmatrix}+C_2e^{t}\left[t\begin{pmatrix}2\\1\end{pmatrix}+\begin{pmatrix}1\\0\end{pmatrix}\right]$ and the fundamental matrix is $\begin{pmatrix}2e^{t}&(2t+1)e^{t}\\e^{t}&te^{t}\end{pmatrix}$ The critical point $$(0,0)$$ is an improper node and unstable. Refer to the video for a sketch of the phase portrait.

To see the full video page and find related videos, click the following link.

2. Classify the type and stability of the equilibrium point(s) of the system
${\bf x}'=\begin{pmatrix} \alpha-1 &\alpha+1\\-2/3&1/3\end{pmatrix}{\bf x}$
for different values of the parameter $$\alpha.$$

• If $$\alpha<\alpha_1=-\dfrac{1}{3}$$ $$\implies$$ $$(0,0)$$ is a saddle point and is unstable.
• If $$\alpha_1=-\dfrac{1}{3}<\alpha<\alpha_2=\dfrac{8}{3}-2\sqrt{2}$$ $$\implies$$ $$(0,0)$$ is an asymptotically stable sink (nodal sink).
• If $$\alpha_2=\dfrac{8}{3}-2\sqrt{2}<\alpha<\alpha_3=\dfrac{2}{3}$$ $$\implies$$ $$(0,0)$$ is an asymptotically stable spiral sink.
• If $$\alpha_3=\dfrac{2}{3}<\alpha<\alpha_4=\dfrac{8}{3}+2\sqrt{2}$$ $$\implies$$ $$(0,0)$$ is an unstable spiral source.
• f $$\alpha>\alpha_4=\dfrac{8}{3}+2\sqrt{2}$$ $$\implies$$ $$(0,0)$$ is an unstable source (nodal source).
• If $$\alpha=\alpha_1$$, we have a line of equilibria which is asymptotically stable.
• If $$\alpha=\alpha_2 \implies$$ $$(0,0)$$ is an asymptotically stable improper node.
• If $$\alpha=\alpha_3\implies$$ $$(0,0)$$ is a stable center.
• If $$\alpha=\alpha_4\implies$$ $$(0,0)$$ is unstable improper node.

To see the full video page and find related videos, click the following link.