### Author Topic: TT2A-P4  (Read 4585 times)

#### Victor Ivrii

• Elder Member
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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 »
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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• Physics and Astronomy Specialist '21
##### 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

Yes, I agree it should be a centre (CW). Of course, with axes: $x_1$ horizontally and $x_2$ vertically.