Author Topic: TT2--P4M  (Read 5039 times)

Victor Ivrii

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TT2--P4M
« on: March 21, 2018, 03:06:11 PM »
Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} 1 & -2\\ 2 &1\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

Jas Kainth

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Re: TT2--P4M
« Reply #1 on: March 21, 2018, 04:55:40 PM »
The picture was too large to upload so I have a link where you can view it from:
https://gyazo.com/7c7ba97b6a8bf65ea4c6434cf7eb04e7

Meng Wu

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• MAT3342018F
Re: TT2--P4M
« Reply #2 on: March 21, 2018, 05:06:42 PM »
The picture was too large to upload so I have a link where you can view it from:
https://gyazo.com/7c7ba97b6a8bf65ea4c6434cf7eb04e7

You have to type it out.
Prof. Victor will tell you the same

Syed Hasnain

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• mat244h1s-winter2018
Re: TT2--P4M
« Reply #3 on: March 21, 2018, 05:16:19 PM »
Here is my solution.....
It differs from the one mentioned above

Victor Ivrii

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Re: TT2--P4M
« Reply #4 on: March 24, 2018, 11:51:27 AM »
No point in two "pictures"
$\renewcommand{\Re}{\operatorname{Re}}$

Characteristic equation
$$\left|\begin{matrix} 1-k & -2\\ 2 &1-k\end{matrix}\right|=(k-1)^2+4\implies k_{1,2}=1\pm 2i.$$
Finding eigenvectors
$$k_1=1+2i\implies \begin{pmatrix} -2i & -2\\ 2 &-2i \end{pmatrix} \begin{pmatrix} \alpha\\ \beta \end{pmatrix}= \begin{pmatrix} 0\\ 0 \end{pmatrix}\implies \beta=-i\alpha\implies \mathbf{e}=\begin{pmatrix} 1\\ -i \end{pmatrix}$$
and  $k_2$ and $\mathbf{e}_2$ are complex conjugate. Then
\begin{align*}
\mathbf{x}=\Re \Bigl[(C_1+iC_2)\begin{pmatrix}
1\\ -i
\end{pmatrix}
e^{(1+2i)t}\Bigr]=
&e^{t}\Re \Bigl[(C_1+iC_2)\begin{pmatrix}
1\\ -i
\end{pmatrix}
(\cos(2t)+i\sin(3t))\Bigr]=\\
&e^{t} \begin{pmatrix}
C_1\cos(2t)-C_2\sin(2t)\\
C_1\sin(2t)+C_2\cos (2t)
\end{pmatrix}
\end{align*}
« Last Edit: March 24, 2018, 11:57:48 AM by Victor Ivrii »