Steady Solution of ODE: Exploring Time Independence with Small Perturbations

[KFC]

I have a set of ODE of the following form

$$\begin{cases} \displaystype{\frac{dx(t)}{dt}} = F(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm] \displaystype{\frac{dy(t)}{dt}} = G(x, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t})\\[4mm] \displaystype{\frac{dz(t)}{dt}} = H(z, y, z; \delta e^{i\omega t}, \Delta e^{-i\omega t}) \end{cases}$$

where $$\delta, \Delta, \omega$$ are constants.

If only concern about the steady solution, can I conclue that the solution must be time-independent?

The equations is quite complicate so one must consider the small pertubration ($$\delta, \Delta$$ are very small number. So when $$\delta \to 0$$ and $$\Delta \to 0$$, the steady solutions are $$x^{(0)}, y^{(0)}, z^{(0)}$$. Take x as example, the first order corrections of the steady solution is of the form

$$x = x^{(0)} + y^{(1)} \delta e^{i\omega t} + z^{(1)} \Delta e^{-i\omega t}$$

I wonder why the above steady solution is time dependent? In this sense, can I conclude that $$y^{(1)}, z^{(1)}$$ are time independent?

 

[HallsofIvy]

A "steady state solution" is by definition a solution constant in time. Yes, it is independent of time.

Again, the definition of "steady state solution" is that it is time independent!