# Green's function for a driven, damped oscillator

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.

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.