Author Topic: TT2A-P4  (Read 1809 times)

Victor Ivrii

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TT2A-P4
« on: November 20, 2018, 05:52:56 AM »
(a) Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
\hphantom{-}1 & \hphantom{-}2\\
-5 &-1\end{pmatrix}\mathbf{x}.$$
(b) Sketch trajectories. Describe the picture (stable/unstable, node, focus, center, saddle).

Samarth Agarwal

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Re: TT2A-P4
« Reply #1 on: November 20, 2018, 08:59:17 AM »
First, try to find the eigenvalues with respect to the parameter
$$ A=\begin{bmatrix} 1&2\\ -5&-1\\ \end{bmatrix} $$
$$ det(A-rI)=(1-r)(-1-r)+10=0 $$
$$ r^2 + 9 = 0 $$
$$ r = \pm 3i $$
The eigenvector is \\ \begin{bmatrix} -2\\ 1-3i \end{bmatrix}
Therefore x_1 =
$$  \begin{bmatrix} -2\\ 1-3i \end{bmatrix} (\cos3t + i\sin3t) $$
$$ = \begin{bmatrix} -2\cos3t \\ \cos3t + 3\sin3t \end{bmatrix} + i \begin{bmatrix} -2\sin3t\\ -3\cos3t + \sin3t \end{bmatrix}$$
Therefore the general solution
$$ x(t) = c_1 \begin{bmatrix} -2\cos3t \\ \cos3t + 3\sin3t \end{bmatrix} + c_2 \begin{bmatrix} -2\sin3t\\ -3\cos3t + \sin3t \end{bmatrix} $$
« Last Edit: November 20, 2018, 09:14:25 AM by Samarth Agarwal »

Mengfan Zhu

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Re: TT2A-P4
« Reply #2 on: November 22, 2018, 02:51:44 PM »
Hello, this is my answer.
To be clear, I did it step by step to get the general real solution
If there are any mistakes, please tell me below ^_^

Jingze Wang

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Re: TT2A-P4
« Reply #3 on: November 22, 2018, 03:46:14 PM »
Hello Samarth, I think your graph is not right, since the eigenvalues have no real parts, then graph should be center instead of spiral.

Michael Poon

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Re: TT2A-P4
« Reply #4 on: November 22, 2018, 05:06:53 PM »
Yes, I agree it should be a centre (CW). Of course, with axes: $x_1$ horizontally and $x_2$ vertically.

Victor Ivrii

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Re: TT2A-P4
« Reply #5 on: November 25, 2018, 12:37:04 PM »
Yes, I agree it should be a centre (CW). Of course, with axes: $x_1$ horizontally and $x_2$ vertically.
Indeed


Computer generated