MAT244--2020F > Test 4

LEC0201-TT4-ALF-F-Q2

(1/1)

RunboZhang:
$\textbf{Problem2:}$

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

$\textbf{Solution:}$

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

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

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

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

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

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

\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}

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