Solving a system involving a matrix? I have this question:
Solve the system 
$$
\frac{dx}{dt} = \begin{bmatrix} -2 & -5 \\ 5 & -2 \end{bmatrix} x
$$ 
with 
$$x(0) = \begin{bmatrix} 1 \\ -3 \end{bmatrix}.
$$
From my notes I figured out it has a spiral trajectory, spinning counter clockwise.
I also need to find $x_1$ and $x_2$ but I can't figure it out.
Note, those equations include matrices, that I wasn't sure how to draw in the question
 A: The general solution to $\dot x = Ax$ is 
$$
x(t) = e^{At}x(0) = e^{-2t}\begin{bmatrix}\cos{5t}&-\sin{5t}\\\sin{5t}&\cos{5t} \end{bmatrix}\begin{bmatrix}1\\-3\end{bmatrix}
$$
The tedious part is calculating the matrix exponential and you can consult any textbook or online source for it. 
Well, ..., this was already given in the comments anyways here is the computed version
A: It is : 
$$\Bigg(\begin{matrix} x_1'\\ x_2' \end{matrix}\Bigg) = \begin{bmatrix} -2 & -5 \\ 5 & -2 \end{bmatrix} \Bigg(\begin{matrix} x_1 \\ x_2 \end{matrix} \Bigg)$$
So :
$$x_1' = -2x_1 -5x_2$$
$$x_2' = 5x_1 -2x_2 $$
It's easy to solve :
$$x_1' = -2x_1 - 5x_2 \Rightarrow \dots \Rightarrow x_1 =c_1e^{-2x_2} + \frac{5x_2}{2} - \frac{5}{4}$$
So, finally, you'll have to solve : 
$$x_2' = 5c_1e^{-2x_2} + \frac{25x_2}{2} - \frac{25}{4} - 2x_2 \Rightarrow \dots$$
Can you solve this one and yield your final result ?
Remember that you can simplify $x_1$ and thus $x_2$ by your initial values : 
$$x(0) = \begin{bmatrix} 1 \\ -3 \end{bmatrix} = \begin{cases} x_1(0) = 1 \\ x_2(0) = -3\end{cases}$$
A: If you change the variables
$$
\begin{bmatrix}z_1\\z_2\end{bmatrix}=e^{2t}\begin{bmatrix}x_1\\x_2\end{bmatrix}
$$
then you will get
$$
\begin{cases}
z_1'=-5z_2,\\
z_2'=5z_1.
\end{cases}
$$
If you differentiate it once again, you may notice that both $z_k$ are solutions to $y''+25y=0$. Solve it and adapt the initial values to find $z_k$ and after that $x_k$.
