Author Topic: TT2-P4  (Read 1042 times)

Victor Ivrii

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TT2-P4
« on: November 20, 2018, 05:46:55 AM »
(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).

Jingze Wang

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Re: TT2-P4
« Reply #1 on: November 20, 2018, 08:38:46 AM »
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


« Last Edit: November 20, 2018, 02:30:25 PM by Jingze Wang »

Jingze Wang

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Re: TT2-P4
« Reply #2 on: November 20, 2018, 08:53:07 AM »
Also, I am wondering can I just say it is spiral instead of focus :)

Mengfan Zhu

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Re: TT2-P4
« Reply #3 on: November 22, 2018, 02:49:36 PM »
Hello, this is my answer.
To be clear, I did it step by step to get the general real solution ^_^

Victor Ivrii

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Re: TT2-P4
« Reply #4 on: November 25, 2018, 11:17:02 AM »
Computer-generated