TT2A-P4

[Victor Ivrii]

(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]

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
$$ x(t) = c_1 \begin{bmatrix} -2\cos3t \\ \cos3t + 3\sin3t \end{bmatrix} + c_2 \begin{bmatrix} -2\sin3t\\ -3\cos3t + \sin3t \end{bmatrix} $$

[Mengfan Zhu]

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]

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]

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.

Indeed


Computer generated