How to solve this pair of differential equations? The equations are

$$y' = y + 3z$$$$z' = -3y + z$$I vaguely remember I need to differentiate one to get the form of the other, but by differentiating the first and subbing in the second I get$$y'' = y' + 3z'=&gt; y''- y' + 9y = 3z$$ I don't know what to do from here.

Edit:
From here I got to $y''-2y'+10y = 0$, so from $m^2-2m+10=0$ we get$$y=e^x(Acos(3x) + Bsin(3x))$$Where would I go from here?
From the first equation, $z = \frac{1}{3}(y'-y)$ and $y'' = y' + 3z'$. From the second eqution, $z' = -3y+z = -3y+\frac{1}{3}(y'-y)$. So, $$y'' = y'+ 3(-3y+\frac{1}{3}(y'-y)) \\ \Rightarrow y'' - 2y' + 10y = 0 \\ \Rightarrow y = (A\cos(3x)+B\sin(3x))\exp(x) $$

Recalling that $z = \frac{1}{3}(y'-y)$, we first calculate $$y' = y + (B\cos(3x)-A\sin(3x))\exp(x) $$

and so 

$$ z = \left(\frac{B}{3}\cos(3x)-\frac{A}{3}\sin(3x)\right)\exp(x) $$
$(y,z)'=A\cdot(y,z)$ where $A=\begin{pmatrix}1&amp;3\\-3&amp;1\end{pmatrix}$. Hence
$$
(y,z)(t) = \exp(tA)\cdot(y,z)(0)
$$