$\textbf{Problem2:}$
$\text{Find the general solution of }$ $$x' = \begin{bmatrix}
-3 & 25\\
-9 & 3
\end{bmatrix} x$$
$\text{classify fixed point (0,0) (type, is stable or not, orientation if applicable) and sketch trajectories.}$
$\textbf{Solution:}$
$ \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}
$
$\text{(graph is attached in the pic below)}$
