91Ó°ÊÓ

Consider the equation $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]\stackrel{→}{x}$with the initial value$\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}1\\ −1\end{array}\right]$ .

Compare the equations$\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]\stackrel{→}{x}$ and $\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$as follows.

$A=\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]$

Assume $\mathrm{λ}$is an Eigen value of the matrix $\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]$ implies $|A−\mathrm{λ}I|=0$.

Substitute the values$\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]$ for $A$and$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for$I$ in the equation$|A−\mathrm{λ}I|=0$ as follows.

$\begin{array}{l}|A−\mathrm{λ}I|=0\\ |\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]−\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\end{array}$

Simplify the equation$|\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]−\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0$ as follows.

$\begin{array}{c}|\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]−\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\\ |\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]−\left[\begin{array}{cc}\mathrm{λ}& 0\\ 0& \mathrm{λ}\end{array}\right]|=0\\ \left|\left[\begin{array}{cc}−1−\mathrm{λ}& −2\\ 2& −1−\mathrm{λ}\end{array}\right]\right|=0\\ (−1−\mathrm{λ})(−1−\mathrm{λ})+4=0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{c}(−1−\mathrm{λ})(−1−\mathrm{λ})+4=0\\ 1+2\mathrm{λ}+{\mathrm{λ}}^2+4=0\\ {\mathrm{λ}}^2+2\mathrm{λ}+5=0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{c}\mathrm{λ}=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}\\ \mathrm{λ}=\frac{−\left(2\right)Â\pm \sqrt{{\left(2\right)}^2−4\left(1\right)\left(5\right)}}{2\left(1\right)}\\ \mathrm{λ}=\frac{−2Â\pm \sqrt{4−20}}{2}\\ \mathrm{λ}=\frac{−2Â\pm 4i}{2}\end{array}$

Therefore, the Eigen values of $A$are $\mathrm{λ}=−1Â\pm i2$.

Substitute the values$−1+i2$ for$\mathrm{λ}$ in the equation $\left|\left[\begin{array}{cc}−1−\mathrm{λ}& −2\\ 2& −1−\mathrm{λ}\end{array}\right]\right|=0$as follows.

$\begin{array}{c}\left|\left[\begin{array}{cc}−1−\mathrm{λ}& −2\\ 2& −1−\mathrm{λ}\end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}−1−(−1+2i)& −2\\ 2& −1−(−1+2i)\end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}−1+1−2i& −2\\ 2& −1+1−2i\end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}−2i& −2\\ 2& −2i\end{array}\right]\right|=0\end{array}$

As$E_{−1+2i}=\ker \left[\begin{array}{cc}−2i& −2\\ 2& −2i\end{array}\right]=span\left[\begin{array}{c}i\\ 1\end{array}\right]$ , the values $\stackrel{→}{v}+i\stackrel{→}{w}$is defined as follows.

$\stackrel{→}{v}+i\stackrel{→}{w}=\left[\begin{array}{c}0\\ 1\end{array}\right]+i\left[\begin{array}{c}1\\ 0\end{array}\right]$

Therefore, the value of$S$ is .$S=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

The inverse of the matrix$S=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ is defined as follows.

$S^{−1}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

As$\stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& −\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{−1}{\stackrel{→}{x}}_0$ , Substitute the value$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for$S$ , $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for$S^{−1}$ , $\left[\begin{array}{c}1\\ −1\end{array}\right]$for ${\stackrel{→}{x}}_0$, $−1$for $p$and$2$ for$q$ in the equation$\stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& −\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{−1}{\stackrel{→}{x}}_0$ as follows.

$\begin{array}{c}\stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& −\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{−1}{\stackrel{→}{x}}_0\\ \stackrel{→}{x}\left(t\right)=e^{−t}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\cos \left(2t\right)& −\sin \left(2t\right)\\ \sin \left(2t\right)& \cos \left(2t\right)\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}1\\ −1\end{array}\right]\\ \stackrel{→}{x}\left(t\right)=e^{−t}\left[\begin{array}{c}\cos \left(2t\right)+\sin \left(2t\right)\\ \sin \left(2t\right)−\cos \left(2t\right)\end{array}\right]\end{array}$

Therefore, the solution of the system is $\stackrel{→}{x}\left(t\right)=e^{−t}\left[\begin{array}{c}\cos \left(2t\right)+\sin \left(2t\right)\\ \sin \left(2t\right)−\cos \left(2t\right)\end{array}\right]$.

As ${\mathrm{λ}}_{1,2}=−1Â\pm 2i$and$−1<0$, draw the graph of the solution$\stackrel{→}{x}\left(t\right)=e^{−t}\left[\begin{array}{c}\cos \left(2t\right)+\sin \left(2t\right)\\ \sin \left(2t\right)−\cos \left(2t\right)\end{array}\right]$as follows.

Hence the solution of the system $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}−1& −2\\ 2& −1\end{array}\right]\stackrel{→}{x}$with the initial value$\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}1\\ −1\end{array}\right]$is $\stackrel{→}{x}\left(t\right)=e^{−t}\left[\begin{array}{c}\cos \left(2t\right)+\sin \left(2t\right)\\ \sin \left(2t\right)−\cos \left(2t\right)\end{array}\right]$and the graph of the solution is a spirals in counterclockwise direction toward the origin.