Author Topic: LEC5101 Quiz 6  (Read 737 times)

LLY

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LEC5101 Quiz 6
« on: November 17, 2019, 03:39:57 AM »
\begin{equation}
\mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {2+i} \\ {-i} & {-1-i}\end{array}\right) \mathbf{x}
\end{equation}
\begin{equation}
\left(\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right) r \xi e^{n}=\left(\begin{array}{cc}{2} & {2+i} \\ {-i} & {-1-i}\end{array}\right) \xi e^{n}
\end{equation}
\begin{equation}
\left(\begin{array}{ll}{r} & {0} \\ {0} & {r}\end{array}\right) \xi e^{n}=\left(\begin{array}{cc}{2} & {2+i} \\ {-i} & {-1-i}\end{array}\right) \xi e^{n}
\end{equation}
\begin{equation}
\left(\begin{array}{cc}{2-r} & {2+i} \\ {-1} & {-1-i-r}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)
\end{equation}
\begin{equation}
\begin{array}{cc}{2-r} & {2+i} \\ {-1} & {-1-i-r |=0} \\ {r^{2}-(1-i) r-i} & {=0} \\ {r^{2}-(1-i) r-i} & {=0} \\ {r} & {=1,-i}\end{array}
\end{equation}
\begin{equation}
r_{1}=1 \text { and } r_{2}=-i
\end{equation}
\begin{equation}
\text { For } r=r_{1}=1
\end{equation}
\begin{equation}
\begin{array}{l}{\left(\begin{array}{cc}{2-1} & {2+i} \\ {-1} & {-1-i-1}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)} \\ {\left(\begin{array}{cc}{1} & {2+i} \\ {-1} & {-(2+i)}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)}\end{array}
\end{equation}
\begin{equation}
\xi_{1}=-(2+i) \xi_{2}
\end{equation}
\begin{equation}
\xi^{(1)}=\left(\begin{array}{c}{2+i} \\ {-1}\end{array}\right)
\end{equation}
\begin{equation}
\text { for } r=r_{2}=-i
\end{equation}
\begin{equation}
\left(\begin{array}{cc}{2-(-i)} & {2+i} \\ {-1} & {-1-i-(-i)}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)
\end{equation}
\begin{equation}
\left(\begin{array}{cc}{2+i} & {2+i} \\ {-1} & {-1}\end{array}\right)\left(\begin{array}{l}{\xi_{1}} \\ {\xi_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)
\end{equation}
\begin{equation}
\xi_{1}=-\xi_{2}
\end{equation}
\begin{equation}
\xi^{(2)}=\left(\begin{array}{c}{1} \\ {-1}\end{array}\right)
\end{equation}
\begin{equation}
\mathbf{x}^{(1)}(t)=\left(\begin{array}{c}{2+i} \\ {-1}\end{array}\right) e^{\prime}, \mathbf{x}^{(2)}(t)=\left(\begin{array}{c}{1} \\ {-1}\end{array}\right) e^{-i t}
\end{equation}
\begin{equation}
\begin{aligned} \mathbf{x} &=c_{1} \mathbf{x}^{(1)}(t)+c_{2} \mathbf{x}^{(2)}(t) \\ &=c_{1}\left(\begin{array}{c}{2+i} \\ {-1}\end{array}\right) e^{\prime}+c_{2}\left(\begin{array}{c}{1} \\ {-1}\end{array}\right) e^{-i t} \end{aligned}
\end{equation}