Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 2 => Topic started by: Victor Ivrii on November 20, 2018, 05:46:55 AM

(a) Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
\hphantom{}5 & \hphantom{}5\\
5 &1\end{pmatrix}\mathbf{x}.$$
(b) Sketch trajectories. Describe the picture (stable/unstable, node, focus, center, saddle).

First, try to find the eigenvalues with respect to the parameter
$A=\begin{bmatrix}
5&5\\
5&1\\
\end{bmatrix}$
$det(ArI)=(5r)(1r)+25=0$
$r^24r+20=0$
$r=\frac{4\pm\sqrt{64}}{2}$
$r=2\pm4i$
Since they are complex conjugates
Then just use one of the eigenvector to find real solution
Use eigenvalue $r=3+4i$ to find its corresponding eigenvector
\begin{bmatrix}
34i&5\\
5&34i\\
\end{bmatrix}
The eigenvector is
$\begin{bmatrix}
5\\
4i3
\end{bmatrix}$
$X=e^{2+4i}\begin{bmatrix}5\\4i3\end{bmatrix}\cos4t+i\sin4t$
Rearrange this, we get $U=e^{2t}\begin{bmatrix}
\cos4t\\
3cos4t4\sin4t
\end{bmatrix}$
Also, $V=e^{2t}\begin{bmatrix}
5\sin4t\\
4\cos4t3\sin4t
\end{bmatrix}$
And they are real valued solutions
Since 5<0, it is clockwise
Also real parts is 2>0, it is unstable spiral

Also, I am wondering can I just say it is spiral instead of focus :)

Hello, this is my answer.
To be clear, I did it step by step to get the general real solution ^_^

Computergenerated