To begin,we assume $\frac{du}{dt}=u$ and $\frac{du}{dt}=\lambda u$ are solved by exponentials:
$$ ⁍ $$
when we substitute $e^{\lambda t}x(u)$,it becomes $\frac{du}{dt}=\lambda e^{\lambda t}x=Au$,and we solve it by use undetermined coefficients like, $u(t)=c_1e^{\lambda_1 t}x_1+c_2e^{\lambda_2 t}x_2$
There may be a question when two $\lambda$ are equal(2x2 matrix) its solution need to add $te^{\lambda t}x$
let me prove it a simple way:
exp.
$\frac{d^2u}{dt}-6\frac{du}{dt}+9u=r^2-6r+9=(r-3)^2=0\to r=3$
as we see 3 is a multiple root and we might write the answer as $u(t)=c_1e^{3t}+c_2te^{3t}$
Observe the equation we can find $(D-3)(D-3)u=0\quad D$ is derivative
look the formula $(D-3)u=Du-3u=\frac{du}{dt}-3u$ ,follow it we can get the original formular
Now let $z=(D-3)u$, then we find $\frac{dz}{dt}=3z$.It can be solved by exponentials:$z(t)=Ae^{3t}\\\to u''-3u'=Ae^{3t}\\\to e^{-3t}u'-3e^{-3t}u=A\\\to u'e^{-3t}+u(e^{-3t})'=A\\\to (ue^{-3t})'=A \\\to ue^{-3t}=At+B\\\to u=Ate^{3t}+Be^{3t}$
$p(t)=c_0+c_1t+...+c_{n-1}t^{n-1}+t^n$
$C(p)=\begin{bmatrix}0\quad0\quad ...\quad0\quad-c_0\,\\ 1\quad0\quad ...\quad 0\quad-c_1\,\\ 0\quad1\quad ...\quad 0\quad-c_2\,\\...\\\quad0\quad0\quad ...\quad 1\quad-c_{n-1}\end{bmatrix}$
A is stable and u(t) $\to$ 0 when all eigenvalues $\lambda$ have negative real parts.