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There's an example given in chapter 4 (Differential Equations) of Mathematical Tools for Physics by James Nearing:

$$m\ddot{x}+kx=F_{ext}(t) \, .$$

Obviously this is a undamped driven oscillator. The exercise is to solve the equation using the method of Green's functions. After an impulse, the motions follows $A \text{sin}(\omega _0(t-t^{'}))$. The change in momentum is $m \Delta v_x=F\Delta t'$, and $\Delta t$ is sufficiently small such that the mass is subject only to $-kx$. Just after $t=t'$, $v_x=A \omega_0=F \Delta t' / m$ $\therefore A=F \Delta t' / \omega _0$. Then the solution $x(t)$ is:

$$x(t)=\begin{cases} (F\Delta t' / m \omega_0) \text{sin}(\omega_0(t-t')) & t >t' \\ 0 & t\leq t' \end{cases}$$

The problem I've been given is to solve the equation for a damped, driven oscillator using the same method.

$$m \ddot x+b \dot x+kx=F_{ext}(t)$$

So my idea is that the change in momentum is the same, but afterward the mass is subject not only to $-kx$ but also to $-b \dot x$. So after $t=t'$,

$$v_x=A \omega _0=\frac{\Delta t'(F-b \dot x)}{m}; \therefore A=\frac{\Delta t'(F-b\dot x)}{m \omega _0} \, .$$

Is this reasonable? I just want to make sure I'm on the right track, since this is my first attempt at solving the problem in this way.

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When you have damping, the motion will not be a pure sinusoid anymore. You will have an exponential decay as well: $$x(t>t_0) = A \sin (\omega (t-t_0)) \exp(-\gamma (t-t_0)).$$ Try to compute the values of $\omega$ and $\gamma$, and you can use the Green's function technique after that.

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