I am solving an atypical problem. The problem is of forced undamped oscillations: $$x''(t) + 9 x(t) = 5\sin(2t) − 6 \cos(4t)$$ with initial conditions $$x(0) = 0$$ $$x'(0) = 4$$ The solution is clear to me: $$-\frac{6}{7}\cos(3t)+\frac{2}{3}\sin(3t)+sin(2t)+\frac{6}{7}\cos(4t)$$ My question arises when analyzing the amplitude and frequency of the solution. Since it is not possible to arrive at a single answer for the amplitude and frequency (not even a steady state because of the undamped nature), what should I reason them to be? One way to put it would be to say that the amplitude of the function is the difference $$max(x(t))-min(x(t))$$ and frequency is a value k that satisfies $$min\{k:\frac{k}{3},\frac{k}{2},\frac{k}{4}\in\mathbb{N}\}$$ On the other hand it would seem reasonable to simply state that the oscillations are a combination of three harmonic oscillations: $$x(t)=\frac{2\sqrt{130}}{21}\cos(3t-\delta)+sin(2t)+\frac{6}{7}\cos(4t)$$ where $\delta=\arctan(\frac{14}{9})$ and we have three amplitudes for the three seperate oscillations: $A_0=\frac{2\sqrt{130}}{21}$, $A_1=1$, $A_2=\frac{6}{7}$. We have three frequencies as well: $\omega_0=3$, $\omega_1=2$, $\omega_2=4$ accordingly.
Even though I consider the problem to be formulated incorrectly at the point where it is being asked for an amplitude of forced undamped oscillations, which line of reasoning would make more sense in your opinion?