Toronto Math Forum
MAT244--2018F => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 20, 2018, 05:46:55 AM
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(a) Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
\hphantom{-}5 & \hphantom{-}5\\
-5 &-1\end{pmatrix}\mathbf{x}.$$
(b) Sketch trajectories. Describe the picture (stable/unstable, node, focus, center, saddle).
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First, try to find the eigenvalues with respect to the parameter
$A=\begin{bmatrix}
5&5\\
-5&-1\\
\end{bmatrix}$
$det(A-rI)=(5-r)(-1-r)+25=0$
$r^2-4r+20=0$
$r=\frac{4\pm\sqrt{-64}}{2}$
$r=2\pm4i$
Since they are complex conjugates
Then just use one of the eigenvector to find real solution
Use eigenvalue $r=3+4i$ to find its corresponding eigenvector
\begin{bmatrix}
3-4i&5\\
-5&-3-4i\\
\end{bmatrix}
The eigenvector is
$\begin{bmatrix}
5\\
4i-3
\end{bmatrix}$
$X=e^{2+4i}\begin{bmatrix}5\\4i-3\end{bmatrix}\cos4t+i\sin4t$
Rearrange this, we get $U=e^{2t}\begin{bmatrix}
\cos4t\\
-3cos4t-4\sin4t
\end{bmatrix}$
Also, $V=e^{2t}\begin{bmatrix}
5\sin4t\\
4\cos4t-3\sin4t
\end{bmatrix}$
And they are real valued solutions
Since -5<0, it is clockwise
Also real parts is 2>0, it is unstable spiral
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Also, I am wondering can I just say it is spiral instead of focus :)
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Hello, this is my answer.
To be clear, I did it step by step to get the general real solution ^_^
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Computer-generated