91影视

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

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

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

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

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

$\begin{array}{c}|A鈭�位I|=0\\ |\left[\begin{array}{cc}鈭�1& 1\\ 鈭�2& 1\end{array}\right]鈭�位\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\end{array}$

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

$\begin{array}{c}|\left[\begin{array}{cc}鈭�1& 1\\ 鈭�2& 1\end{array}\right]鈭�位\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\\ |\left[\begin{array}{cc}鈭�1& 1\\ 鈭�2& 1\end{array}\right]鈭�\left[\begin{array}{cc}位& 0\\ 0& 位\end{array}\right]|=0\\ \left|\left[\begin{array}{cc}鈭�1鈭�位& 1\\ 鈭�2& 1鈭�位\end{array}\right]\right|=0\\ (鈭�1鈭�位)(1鈭�位)+2=0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{c}(鈭�1鈭�位)(1鈭�位)+2=0\\ 鈭�1+{位}^2+2=0\\ {位}^2+1=0\\ {位}^2=鈭�1\end{array}$

Therefore, the Eigen values of $A$are$位=卤i$ .

Substitute the values$i$ for$位$ in the equation $\left|\left[\begin{array}{cc}鈭�1鈭�位& 1\\ 鈭�2& 1鈭�位\end{array}\right]\right|=0$as follows.

$\begin{array}{c}\left|\left[\begin{array}{cc}鈭�1鈭�位& 1\\ 鈭�2& 1鈭�位\end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}鈭�1鈭�i& 1\\ 鈭�2& 1鈭�i\end{array}\right]\right|=0\end{array}$

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

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

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

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

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

As $\stackrel{鈫�}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& 鈭�\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{鈭�1}{\stackrel{鈫�}{x}}_0$, Substitute the value$\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$for $S$, $\left[\begin{array}{cc}鈭�1& 1\\ 1& 0\end{array}\right]$for ,$S^{鈭�1}$$\left[\begin{array}{c}0\\ 1\end{array}\right]$for ,${\stackrel{鈫�}{x}}_0$

$0$for$p$ and$1$ for $q$in the equation $\stackrel{鈫�}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& 鈭�\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{鈭�1}{\stackrel{鈫�}{x}}_0$as follows.

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

Further, simplify the equation as follows.

$\begin{array}{l}\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\sin \left(t\right)& \cos \left(t\right)\\ \cos \left(t\right)+\sin \left(t\right)& 鈭�\sin \left(t\right)+\cos \left(t\right)\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{c}\sin \left(t\right)\\ \sin \left(t\right)+\cos \left(t\right)\end{array}\right]\end{array}$

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

As${位}_{1,2}=卤i$, draw the graph of the solution $\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{c}\sin \left(t\right)\\ \sin \left(t\right)+\cos \left(t\right)\end{array}\right]$as follows

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