Toronto Math Forum
MAT244-2014F => MAT244 Math--Tests => TT2 => Topic started by: Victor Ivrii on November 19, 2014, 08:46:44 PM
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Find the general solution, sketch the phase portrait and determine the type of behavior near the origin of the the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t= -x - 4y\ , \\
&y'_t=\ \ x -\ y .
\end{aligned}\right.
\end{equation*}
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x′=(−1 −4)(x) .
y'=(1 -1) (y)
find eigenvalues
det(A−rI)=r2+2r+5=0
r1=-1+2i
r2=-1-2i
then, find eigenvectors
(-2i -4) v1=(2i)
(1 -2i) (1)
(2i -4) v2=(2i)
(1 2i) (-1)
x(t)=C1e(−1+2i)t(2i)+C2e−(-1-2i)t(2i)
(1) (-1)
stable spiral point
-4<0, counterclockwise
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#3
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Picture
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first write the equation in matrix form:
\begin{equation*}\textbf{x}'=\begin{pmatrix}\hphantom{-}-1 & -4\\\hphantom{-}1 &-1\end{pmatrix}\textbf{x}\ . \end{equation*}
find eigenvalues:
\begin{equation*} r^2 - trace(A) + (ad - bc)= r^2+ 2r + 5 = 0\implies r_1= -1 + 2i, r_2=-1 -2i\end{equation*}
then, find eigenvectors, which are conjugated
\begin{equation*} \begin{pmatrix} -1 - r & \hphantom{-}-4\\ \hphantom{-}1 &-1 -r\end{pmatrix}\begin{pmatrix}\mathbf{\xi}_1\\\mathbf{\xi}_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}
find the two conjugate eigenvectors
\begin{equation*}\mathbf{\xi}^1 =\begin{pmatrix}2\\-i\end{pmatrix}
\mathbf{\xi}^2 =\begin{pmatrix}2\\i\end{pmatrix}\end{equation*}
therefore
\begin{equation*}\mathbf{x}(t)= C_1e^{-t}\begin{pmatrix}2\cos(2t)\\\sin(t) \end{pmatrix}+ C_2e^{-t}\begin{pmatrix}-2\sin(2t)\\\cos(t) \end{pmatrix} \end{equation*}
the attachment is the phase portrait generated by PPLANE
(spiral point, stable)
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Despite some glitches (one should write \sin (t) resulting in upright $\sin $ rather than sin (t) in LaTeX) the last solution is the only one I can read