While solving a differential equation like $(D^2+4)x=A\cos\omega t$, for example, the complex gain is said to be $\dfrac{1}{4-\omega^2}A$. Is there any deeper reason why it is defined this way, as opposed to simply $\dfrac{1}{4-\omega^2}$, which is the actual ratio of the solution $x(t)$ to the input signal $A\cos\omega t$?
Or is the input signal itself defined as $\cos\omega t$, rather than as $A\cos\omega t$? If so, why?
A possible explanation that occured to me -- correct me if I'm wrong -- is that the amplitude $A$ isn't necessarily a scalar value, but could be another polynomial $Q(D)$, in which case the solution to the differential equation would take the form:
$$x(t)=\Re\left[\frac{Q(i\omega)}{P(i\omega)}\right]\cos(\omega t)$$
And it would "make sense" to consider $Q(i\omega)/P(i\omega)$ as the complex gain (and its real part as the gain), because $Q(D)$ can be considered part of the "processing" done on the input signal $\cos\omega t$.
Another thought: one could divide both sides of the initial equation by $A$, and the differential equation would remain the same. However, while the gain $\frac1{P(i\omega)}$ would not remain invariant, as $P$ has changed, $\frac{A}{P(i\omega)}$ would. In other words, the gain is $\frac{A}{P(i\omega)}$, because only that is invariant under a scaling transformation.