Why do we say that the roots of the characteristic polynomial of an ODE, are the eigenvalues? Suppose I have a differential equation, $$\ddot{x} -  \dot{x}-6x = 0$$
I've read two ways to obtain the solution, 

$i)$ Assume $x = e^{\lambda t}$, then I obtain the characteristic equation,  $$\lambda^2 -  \lambda -6 =0 $$Now I obtain the values for $\lambda$, namely $\lambda_1 =  3$ and $\lambda_2=-2$. Now the solution is $x(t) = c_1 e^{3t} +c_2e^{-2t}$

$ii)$ Form  a system of equations by introducing dummy variables, $$\dot{x} = v\\ \dot{v} =  6x+v$$
$$\begin{bmatrix}
\dot{x}\\ 
\dot{y}
\end{bmatrix}=\begin{bmatrix}
0 & 1\\ 
6 & 1
\end{bmatrix}\begin{bmatrix}
x\\ 
v
\end{bmatrix}$$
$$\dot{X}=AX$$
Find the eigen values and vectors of $A$, $\lambda_1 = 3$ and $\xi_1 =[1\,\,\,3]$, $\lambda_2=-2$ and $\xi_2=[1\,\,\,-0.5]$
$$\begin{bmatrix}
x(t)\\ 
v(t)
\end{bmatrix}=\begin{bmatrix}
1 & 1\\ 
3 & -0.5
\end{bmatrix}\begin{bmatrix}
e^{3t} & 0\\ 
0 & e^{-2t}
\end{bmatrix}\begin{bmatrix}
1 & 1\\ 
3 & -0.5
\end{bmatrix}^{-1}\begin{bmatrix}
x(0)\\ 
y(0)
\end{bmatrix}$$
$$X(t)=TDT^{-1}X(0)$$
Finally, it also yields 
$x(t) = c_1 e^{3t} +c_2e^{-2t}$

My Question is:
Why do we get the characteristic equation while solving for the eigen values? What is the relation between these two methods?
 A: Considering the equivalent system
$$
\cases{
\dot x_1 = x_2\\
\dot x_2 = x_2 + 6x_1
},\ \ \ \text{or}\ \ \ \dot X = A X
$$
supposing we can factorize $A = Q^{-1}\Lambda Q$ we have then
$$
\dot X = Q^{-1}\Lambda Q X
$$
and now introducing the change of variables $Y = Q X$ we follow with
$$
\dot Y = \Lambda Y
$$
so concluding, if we can do $A = Q^{-1}\Lambda Q$ then there exist a reference coordinate system such that the solutions are associated to the $A$ eigenvalues.
NOTE
Here $\Lambda$ is the diagonal eigenvalues matrix and $Q$ the eigenvectors matrix, both associated to $A$ matrix.
$$
Q = \left(
\begin{array}{cc}
 1 & -1 \\
 3 & 2 \\
\end{array}
\right),\ \ \ \Lambda = \left(
\begin{array}{cc}
 3 & 0 \\
 0 & -2 \\
\end{array}
\right)
$$
The solution for $\dot Y = \Lambda Y$ is 
$$
\cases{
y_1 = c_1 e^{3t}\\
y_2 = c_2 e^{-2t}
}
$$
and to recover the initial coordinates we should do
$$
X = Q^{-1}Y = \left(
\begin{array}{c}
 \frac{2}{5} e^{3 t} c_1+\frac{1}{5} e^{-2 t} c_2 \\
 \frac{1}{5} e^{-2 t} c_2-\frac{3}{5} e^{3 t} c_1 \\
\end{array}
\right)
$$
