MAT244--2019F > Term Test 2

Problem 4 (noon)

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Changhao Jiang:
To find eigenvalues, $det(A-\lambda x)=(1-\lambda)(-3-\lambda)-(-2)(3)=0$,we can get $\lambda = -1-\sqrt{2}$ or $\lambda = -1+\sqrt{2}$
To find eigenvectors, when $\lambda=-1-\sqrt{2}$,
$\begin{pmatrix}
2+\sqrt{2} & 3 \\
-2 & -2+\sqrt{2}
\end{pmatrix}
~
\begin{pmatrix}
2 & 2-\sqrt{2} \\
0 & 0
\end{pmatrix}$
the eigenvector is $\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}$
so $e^{(-1-\sqrt{2})t}\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}
= e^{-t}(\cos\sqrt{2}t-i\sin\sqrt{2}t)\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}
= e^{-t} (\begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + i\begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix})$
therefore, the general solution is $x(t)=c_1 e^{-t} \begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix}$

NANAC:
Please see the attachment for the answer

baixiaox:
answer for tt2 question4

Mingdi Xie:
This is my solution. :)

Victor Ivrii:
What everybody is missing

we see that characteristic roots $k_{1,2}=-1\pm \sqrt{2}i$ are complex, with negative real part. So, it is  stable focus  and with  clock-wise  orientation  since the bottom-left element is negative.

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