91Ó°ÊÓ

Consider the linear system as follows.

$\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}-1.5& -1\\ 2& 0.5\end{array}\right]\stackrel{→}{x}\left(t\right)$

To find the eigenvalues, evaluate $\left|A-\mathrm{λ}I\right|=0$as follows.

$$\begin{gathered} \left|A-\mathrm{λ}I\right|=0 \\ \left[\begin{array}{cc}-1.5-\mathrm{λ}& -1\\ 2& 0.5-\mathrm{λ}\end{array}\right]=0 \\ \left(-1.5-\mathrm{λ}\right)\left(0.5-\mathrm{λ}\right)-2\left(-1\right)=0 \\ -0.75+1.5\mathrm{λ}-0.5\mathrm{λ}+{\mathrm{λ}}^2+2=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} {\mathrm{λ}}^2+\mathrm{λ}+1.25=0 \\ \mathrm{λ}=\frac{-1Â\pm \sqrt{{\left(1\right)}^2-4\left(1\right)\left(1.25\right)}}{2\left(1\right)} \\ \mathrm{λ}=\frac{-1Â\pm \sqrt{1-5}}{2} \\ \mathrm{λ}=\frac{-1Â\pm i\sqrt{2}}{2} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} {\mathrm{λ}}_1=\frac{-1+2i}{2} \\ {\mathrm{λ}}_1=\frac{-1}{2}+2i \\ {\mathrm{λ}}_2=\frac{-1-2i}{2} \\ {\mathrm{λ}}_2=\frac{-1}{2}-2i \end{gathered}$$

Therefore, the eigenvalues are ${\mathrm{λ}}_1=\frac{-1}{2}+2iand{\mathrm{λ}}_2=\frac{-1}{2}-2i$

The eigenvalues are ${\mathrm{λ}}_1=\frac{-1}{2}+2i$and ${\mathrm{λ}}_2=\frac{-1}{2}-2i$also in the form as follows.

$$\begin{gathered} p+iq \\ p=-\frac{1}{2} \\ q=1 \end{gathered}$$

Since, the real part of the eigenvalues are negative.

So, the phase portrait is spiral inward in clockwise directions towards the origin.

Therefore, the phase portrait is in figure 4 as follows.

Hence, the phase portrait to the linear system$\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}-1.5& -1\\ 2& 0.5\end{array}\right]\stackrel{→}{x}\left(t\right)$is IV.