MAT244--2018F > Term Test 2

TT2-P4

(1/1)

Victor Ivrii:
(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).

Jingze Wang:
First, try to find the eigenvalues with respect to the parameter

$A=\begin{bmatrix} 5&5\\ -5&-1\\ \end{bmatrix}$

$det(A-rI)=(5-r)(-1-r)+25=0$

$r^2-4r+20=0$

$r=\frac{4\pm\sqrt{-64}}{2}$

$r=2\pm4i$

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}
3-4i&5\\
-5&-3-4i\\
\end{bmatrix}

The eigenvector is
$\begin{bmatrix} 5\\ 4i-3 \end{bmatrix}$

$X=e^{2+4i}\begin{bmatrix}5\\4i-3\end{bmatrix}\cos4t+i\sin4t$

Rearrange this, we get $U=e^{2t}\begin{bmatrix} \cos4t\\ -3cos4t-4\sin4t \end{bmatrix}$

Also, $V=e^{2t}\begin{bmatrix} 5\sin4t\\ 4\cos4t-3\sin4t \end{bmatrix}$

And they are real valued solutions

Since -5<0, it is clockwise

Also real parts is 2>0, it is unstable spiral

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

Mengfan Zhu:
To be clear, I did it step by step to get the general real solution ^_^

Victor Ivrii:
Computer-generated