Author Topic: LEC0201-TT4-ALF-F-Q2  (Read 334 times)

RunboZhang

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LEC0201-TT4-ALF-F-Q2
« on: December 03, 2020, 01:24:53 PM »
$\textbf{Problem2:}$

$\text{Find the general solution of }$ $$x' = \begin{bmatrix}
-3 & 25\\
-9 & 3
\end{bmatrix} x$$
$\text{classify fixed point (0,0) (type, is stable or not, orientation if applicable) and sketch trajectories.}$

$\textbf{Solution:}$

$ \text{Let }$ $$A=  \begin{bmatrix}
-3 & 25\\
-9 & 3
\end{bmatrix}$$

$\text{Then we have}$ $$det(A-\lambda I) = (-3-\lambda)(3-\lambda)-25\cdot(-9)=0$$
$\text{Solve for } \lambda ,$  $$ \lambda_1 = 6 \sqrt{6}i ,\ \lambda_2=-6\sqrt{6}i $$

$\text{Take } \lambda = 6\sqrt{6}i \text{, then }$
$$A-\lambda I = A-6\sqrt{6}iI=\begin{bmatrix}
-3-6\sqrt{6}i & 25\\
-9 & 3-6\sqrt{6}i
\end{bmatrix} \xrightarrow{\text{ref}} \begin{bmatrix}
-3-6\sqrt{6}i & 25\\
0 & 0
\end{bmatrix}$$

$\text{Hence we have }$ $$null(A-\lambda I) = \begin{bmatrix}
25 \\
3+6\sqrt{6}i
\end{bmatrix}$$

$\text{and}$ $$x=e^{6\sqrt{6}it}\begin{bmatrix}
25 \\
3+6\sqrt{6}i
\end{bmatrix}$$

$\text{Further expand it and we get }$

$
\begin{gather}
\begin{aligned}

x &= [cos(6\sqrt{6}t)+isin(6\sqrt{6}t)]\begin{bmatrix}
25 \\
3+6\sqrt{6}i
\end{bmatrix} \\\\
&=\begin{bmatrix}
25cos(6\sqrt{6}t)+i25sin(6\sqrt{6}t) \\
3cos(6\sqrt{6}t)+i3sin(6\sqrt{6}t)+i6\sqrt{6}cos(6\sqrt{6}t)-6\sqrt{6}sin(6\sqrt{6}t)
\end{bmatrix} \\\\
&=c_1 \begin{bmatrix}
25cos(6\sqrt{6}t)\\
3cos(6\sqrt{6}t)-6\sqrt{6}sin(6\sqrt{6}t)
\end{bmatrix} + c_2 \begin{bmatrix}
25sin(6\sqrt{6}t) \\
3sin(6\sqrt{6}t)+6\sqrt{6}cos(6\sqrt{6}t)
\end{bmatrix}

\end{aligned}
\end{gather}
$

$\text{(graph is attached in the pic below)}$