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?
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
$$ X = Q^{-1}Y = \left( \begin{array}{c} \frac{2}{5} e^{3 t} c_1+\frac{1}{5} e^{-2 t} c_2 \\ \frac{1}{5} e^{-2 t} c_2-\frac{3}{5} e^{3 t} c_1 \\ \end{array} \right) $$