Toronto Math Forum
MAT2442014F => MAT244 MathTests => TT2 => Topic started by: Victor Ivrii on November 19, 2014, 08:47:30 PM

Find the general solution and determine the type of behavior near the origin of the the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t= 6x + 5y\ , \\
&y'_t= 5x + 4y .
\end{aligned}\right.
\end{equation*}

\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)

I just showed the steps for finding the second eigenvector..
After that, you would proceed to plug it into the standard general equation for a repeated eigenvalue question.

Graph of #4

Eddie, shouldn't it be asymptotically stable?

In the case of improper node one can also use clockwise / counterclockwise criteria (sign of the topright element in the matrix $A$).

So professor, since the top right element is positive, does that mean it has clockwise orientation, so shouldn't it be asymptotically stable?

Completely correct solution was provided by Shuyang Wang. Orientation and stability are different questions: stability follows from the sign of the eigenvalue ($r<0$; $r>0$ means instability) and orientation from the sign of the topright element (which in the framework of the focal point/center/improper node opposite to the sign of the bottomleft)