Resonance is explained here as the matching of the frequency of the system with the drive in the RHS of the equation, so that the solution of the ODE grows to infinity:
$$u'' + 9u = 2 \sin 3t$$
with $$u=-\frac 1 3 t\cos 3t + C_1 \sin 3t + C_2 \cos 3t$$
The question is
Shouldn't this resonant frequencies be identified by the poles of the Laplace transform?
If I get the LT of $u$ (ignoring the null space) I get
$$\mathcal L\Big [-\frac 1 3 t \cos 3t \Big](s)= -\frac{s^2 - 9}{3(s^2+9)^2}$$
with roots $s=3$ and $s=-3,$ which seem to point to the resonant frequency, but no poles!
I thought that the roots of the LT carried information on the amplitude of the solution equation.
I see after the answer that I was taking the LT of the actual answer... So perhaps I should be reading the tea leaves on the LT of the initial equation, which according to Wolfram Alpha is:
$$ \frac{6+ 9 u'(0) + 9s u(0) + s^2 u'(0) + s^3 u(0)}{(s^2 +9)^2}$$
Analyzing the process
$$ u'' + 9 u = v $$
and assuming null initial conditions, the LT gives
$$ U(s)=\frac{1}{s^2+9}V(s) $$
The resonance is associated to the transfer function modulus behavior with input frequencies or
$$ |U(\omega)| = \left|\frac{1}{9-\omega^2}\right||V(\omega)| $$
where $\omega$ represents the angular frequency at the input $v(t)$. If in $v(t)$ appears a carrier with frequency $\omega = 3$ then the output amplitude at $U(3)$ goes to infinite.
Note that if instead we have
$$ u'' + 9 u' = v $$
then in this case
$$ U(s)=\frac{1}{s(s+9)}V(s) $$
and
$$ |U(\omega)| = \left|\frac{1}{9\omega\sqrt{1+\left(\frac{\omega}{9}\right)^2}}\right||V(\omega)| $$
In this case if $v(t)$ has a carrier with frequency $\omega = 3$ the output will be limited without resonance.
NOTE
Given $G(s) = \frac{1}{s(s+9)}$ we obtain the module $|G(j\omega)|$ as follows:
Making $s = j\omega$
$$ G(j\omega) = \frac{1}{j\omega(j\omega+9)} $$
and
$$ |G(j\omega)|=\sqrt{G(j\omega)G(-j\omega)} = \frac{1}{9\omega\sqrt{1+\left(\frac{\omega}{9}\right)^2}} $$