Toronto Math Forum

\begin{equation*} \textbf{x}'=\begin{pmatrix}\hphantom{-}-6 & 5\\\hphantom{-}-5 &4 \end{pmatrix}\textbf{x}\ . \end{equation*}

find eigenvalues

\begin{equation*} \det (A - rI) = \left|\begin{matrix}-6 - r &5\\-5&  4 - r\end{matrix}\right| =  r^2+ 2r + 1 = 0\implies r_1=r_2=-1\end{equation*}

then, find eigenvectors

\begin{equation*} \begin{pmatrix} -6 - r & \hphantom{-}5\\  \hphantom{-}-5 &4 -r\end{pmatrix}\begin{pmatrix}\mathbf{\xi}_1\\\mathbf{\xi}_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}

then

\begin{equation*}\mathbf{\xi}^1 =\begin{pmatrix}1\\1\end{pmatrix}\end{equation*}

generalized eigenvector

\begin{equation*} \begin{pmatrix} -6 - r & \hphantom{-}5\\  \hphantom{-}-5 &4 -r\end{pmatrix}\mathbf{\xi}^2=\mathbf{\xi}^1 \end{equation*}
\begin{equation*} \mathbf{\xi}^2=\begin{pmatrix}0\\1/5\end{pmatrix}\end{equation*}

so

\begin{equation*}\mathbf{x}(t)= C_1e^{-t}\begin{pmatrix}1\\1\end{pmatrix}+ C_2e^{-t}\left(  t \begin{pmatrix}1\\1\end{pmatrix} + \begin{pmatrix}0\\1/5\end{pmatrix}\right)\end{equation*}

phase portrait(improper node, stable)

\begin{equation*} \textbf{x}'=\begin{pmatrix} -2 & \hphantom{-}1\\  \hphantom{-}1 &-2 \end{pmatrix}\textbf{x}\ . \end{equation*}
find eigenvalues
\begin{equation*} \det (A - rI) = \left|\begin{matrix}-2 - r &1\\1&  - 2 - r\end{matrix}\right| =  r^2+ 4r + 3 = 0\implies r_1=-3, r_2=-1\end{equation*}
then, find eigenvectors
\begin{equation*} \begin{pmatrix} -2 - r & \hphantom{-}1\\  \hphantom{-}1 &-2 -r\end{pmatrix}\begin{pmatrix}\mathbf{\xi}_1\\\mathbf{\xi}_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}
then
\begin{equation*}\mathbf{\xi}^1 =\begin{pmatrix}1\\-1\end{pmatrix} , \mathbf{\xi}^2 =\begin{pmatrix}1\\1\end{pmatrix}\end{equation*}
so
\begin{equation*}\mathbf{x}= C_1e^{-3t}\begin{pmatrix}1\\-1\end{pmatrix}+ C_2e^{-t}\begin{pmatrix}1\\1\end{pmatrix}\end{equation*}
plot(stable):