Take the system
$\begin{cases} w=v'\\ v''-\mu v'+v=0 \end{cases}$
for $\mu \in \mathbb{R}$
Find a constant matrix $A^{\mu}$ such that the above system is equivalent to
$\begin{bmatrix}v'\\w'\end{bmatrix}=A^{\mu}\begin{bmatrix}v'\\w'\end{bmatrix}$
My first thought was to find a solution to $v''-\mu v'+v=0$. Taking the characteristic polynomial as $r^2-\mu r+1=0$, I get $r=\frac{1}{2}(\mu \pm\sqrt{\mu^2-4})$. Not really sure what to do at this point, however. Another way to solve this might be first to replace $v'$ in the second equation with $w$ to get $v''-\mu w+v=0$ but I'm not sure where to go from there either.
(I'm assuming here that you meant $\begin{bmatrix}v'\\w'\end{bmatrix}=A^{\mu}\begin{bmatrix}v\\w\end{bmatrix}$ instead of $\begin{bmatrix}v'\\w'\end{bmatrix}=A^{\mu}\begin{bmatrix}v'\\w'\end{bmatrix}$; in the latter case, $A^\mu$ would just be the identity matrix.)
If you replace $v'$ with $w$ in the second line of your equation and isolate each equation for $v'$ and $w'$, you get:
$\begin{cases} v'=w\\ w'=-v + \mu w \end{cases}$
Now all you have to do is find a matrix of coefficients $a$ through $d$ such that:
$$\begin{bmatrix}v'\\w'\end{bmatrix}=\begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}v\\w\end{bmatrix}$$
For this matrix to correspond to your set of equations, it must be the case that $a = 0$, $b = 1$, $c = -1$, and $d = \mu$.