Removing Free Oscillations A forced oscillator can be described by the equation $\ddot{x} + \omega^2x = F(t)$. The solution of this equation will have a free and forced solution, with the free solution being just $x(t) = Ae^{i\omega t + \phi}$ and $A$ and $\phi$ are constants.
I was told that a "slow-start" to the system can "remove" the free oscillation and isolate the forced solution. By "slow-start", I mean $F(t)$ is increased exponentially from zero. I don't understand how this works.
 A: I am going to translate your problem into the terminology I am familiar with.  Then I will give an example and hope that you can translate it back to your application.  When you apply a step function, the signal overshoot its steady state, reverses and undershoots the steady state.  This behavior continues for a specified time until the resistance of the circuit dampens the "spurious" oscillations and the circuit reaches steady state.  The signal then reverses and the process repeats.
To eliminate (or reduce) the "spurious" oscillations, you can use: higher damping resistance, "slower" application of the signal (not a step function), and filtering.
An example of slowing the signal is to use a "large enough" input capacitor to slow down the leading and trailing edges of the signal.
A mathematical capacitor might be setting F(t) =$ ke^{jw} $ (or something like it).
For filtering, one uses a bandpass filter to eliminate the "spurious" frequency components on the output.
A "mathematical filter" might be one that removes the frequency range of the spurious components.
Hope this helps.     
