Toronto Math Forum
MAT2442018S => MAT244Tests => Term Test 2 => Topic started by: Victor Ivrii 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.

The picture was too large to upload so I have a link where you can view it from:
https://gyazo.com/7c7ba97b6a8bf65ea4c6434cf7eb04e7

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 ;D

Here is my solution.....
It differs from the one mentioned above

No point in two "pictures"
$\renewcommand{\Re}{\operatorname{Re}}$
Characteristic equation
$$
\left\begin{matrix}
1k & 2\\ 2 &1k\end{matrix}\right=(k1)^2+4\implies k_{1,2}=1\pm 2i.
$$
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*}