Solve the initial value problem x'=Ax $A =
\begin{array}{cc}
-2 & 1 \\
5 & -4 \\
\end{array}$
$x(0) = 
\begin{array}{cc}
1\\
3\\
\end{array}$
I undertsand that this is an eigenvector problem, and I got the values of $ -3+ \sqrt{6}$ and $ -3- \sqrt{6}$. I am unable to calculate the eigenvectors from here.
 A: Using the eigenvalue $\lambda = -3 - \sqrt{6}$, we find $[A-\lambda I]v_1 = 0$ and arrive at a RREF of:
$$\left(
\begin{array}{cc}
 1 & \frac{1}{5} \left(-1+\sqrt{6}\right) \\
 0 & 0 \\
\end{array}
\right)v_1 = 0$$
We choose $b = 1, a = \dfrac{1}{5} \left(1-\sqrt{6}\right)$, so:
$$v_1 = \left(\dfrac{1}{5} \left(1-\sqrt{6}\right), 1\right)$$
Since we have conjugate eigenvalues, we can write the eigenvector for the second eigenvalue as:
$$v_2 = \left(\frac{1}{5} \left(1+\sqrt{6}\right),1\right)$$
You can now write:
$$x(t) = c_1 ~e^{\lambda_1 t}~v_1 + c_2~ e^{\lambda_2 t}~v_2$$
Use the IC to find the constants.
Your final solution should be:

$x(t) = \dfrac{-4 e^{\left(-3-\sqrt{6}\right) t}+\sqrt{6} e^{\left(-3-\sqrt{6}\right) t}+4 e^{\left(-3+\sqrt{6}\right) t}+\sqrt{6} e^{\left(-3+\sqrt{6}\right) t}}{2 \sqrt{6}} =\dfrac{1}{3} e^{-3 t} \left(2 \sqrt{6} \sinh \left(\sqrt{6} t\right)+3 \cosh \left(\sqrt{6} t\right)\right)  \\y(t) =\dfrac{-2 e^{\left(-3-\sqrt{6}\right) t}+3 \sqrt{6} e^{\left(-3-\sqrt{6}\right) t}+2 e^{\left(-3+\sqrt{6}\right) t}+3 \sqrt{6} e^{\left(-3+\sqrt{6}\right) t}}{2 \sqrt{6}} = \dfrac{1}{3} e^{-3 t} \left(\sqrt{6} \sinh \left(\sqrt{6} t\right)+9 \cosh \left(\sqrt{6} t\right)\right)$

