91Ó°ÊÓ

Consider the equation $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}0& 1\\ −4& 0\end{array}\right]\stackrel{→}{x}$with the initial value$\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$ .

Compare the equations $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}0& 1\\ −4& 0\end{array}\right]\stackrel{→}{x}$and$\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$ as follows.

$A=\left[\begin{array}{cc}0& 1\\ −4& 0\end{array}\right]$

Assume$\mathrm{λ}$ is an Eigen value of the matrix $\left[\begin{array}{cc}0& 1\\ −4& 0\end{array}\right]$implies$|A−\mathrm{λ}I|=0$ .

Substitute the values $\left[\begin{array}{cc}0& 1\\ −4& 0\end{array}\right]$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.

$\begin{array}{c}|A−\mathrm{λ}I|=0\\ |\left[\begin{array}{cc}0& 1\\ −4& 0\end{array}\right]−\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\end{array}$

Simplify the equation $|\left[\begin{array}{cc}0& 1\\ −4& 0\end{array}\right]−\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0$as follows.

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

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

Substitute the values$i2$ for$\mathrm{λ}$ in the equation$\left|\left[\begin{array}{cc}−\mathrm{λ}& 1\\ −4& −\mathrm{λ}\end{array}\right]\right|=0$ as follows.

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

As $E_{−1+2i}=\ker \left[\begin{array}{cc}−2i& 1\\ −4& −2i\end{array}\right]=span\left[\begin{array}{c}−2i\\ 1\end{array}\right]$, the values $\stackrel{→}{v}+i\stackrel{→}{w}$is defined as follows.

$\stackrel{→}{v}+i\stackrel{→}{w}=\left[\begin{array}{c}0\\ 1\end{array}\right]+i\left[\begin{array}{c}−2\\ 0\end{array}\right]$

Therefore, the value of $S$is$S=\left[\begin{array}{cc}−2& 0\\ 0& 1\end{array}\right]$ .

The inverse of the matrix$S=\left[\begin{array}{cc}−2& 0\\ 0& 1\end{array}\right]$ is defined as follows.

$\begin{array}{c}{S}^{−1}=\frac{1}{−2}\left[\begin{array}{cc}1& 0\\ 0& −2\end{array}\right]\\ S^{−1}=\left[\begin{array}{cc}\frac{1}{−2}& 0\\ 0& 1\end{array}\right]\end{array}$

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$\left[\begin{array}{cc}−2& 0\\ 0& 1\end{array}\right]$for$S$, $\left[\begin{array}{cc}\frac{1}{−2}& 0\\ 0& 1\end{array}\right]$for $S^{−1}$, $\left[\begin{array}{c}1\\ 0\end{array}\right]$for ${\stackrel{→}{x}}_0$,$0$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.

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

Further, simplify the equation as follows.

$\begin{array}{l}\stackrel{→}{x}\left(t\right)=\left[\begin{array}{cc}\cos \left(2t\right)& 2\sin \left(2t\right)\\ −\frac{\sin \left(2t\right)}{2}& \cos \left(2t\right)\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}\cos \left(2t\right)\\ −\frac{\sin \left(2t\right)}{2}\end{array}\right]\end{array}$

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

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

Hence, the solution of the system$\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}0& 1\\ −4& 0\end{array}\right]\stackrel{→}{x}$with the initial value$\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$is$\stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}\cos \left(2t\right)\\ −\frac{\sin \left(2t\right)}{2}\end{array}\right]$and the graph of the solution is an ellipse in counterclockwise direction.