Toronto Math Forum
MAT244-2014F => MAT244 Math--Tests => TT2 => Topic started by: Victor Ivrii on November 19, 2014, 08:47:30 PM
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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*}
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\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)
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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.
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Graph of #4
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Eddie, shouldn't it be asymptotically stable?
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In the case of improper node one can also use clock-wise / counter-clock-wise criteria (sign of the top-right element in the matrix $A$).
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So professor, since the top right element is positive, does that mean it has clockwise orientation, so shouldn't it be asymptotically stable?
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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 top-right element (which in the framework of the focal point/center/improper node opposite to the sign of the bottom-left)