Lyapunov exponents of a damped, driven harmonic oscillator

[Ananthan9470]

Homework Statement
I am supposed to calculate Lyapunov exponent of a damped, driven harmonic oscillator given by $\ddot{x} + 2\beta \dot{x} + \omega_0^2 x = fcos(\omega t)$
Lyapunov exponent is $\lambda$ in the equation $\delta x(t) = \delta x_0 e^{\lambda t}$

The attempt at a solution
The gerenal solution of the system is given by $Acos(\omega t - \delta) + Ce^{r_1 t} + De^{r_2 t}$

Consider two initial points 1 and 2. The solutions evolve to give $Acos(\omega t - \delta) + C_1e^{r_1 t} + D_1e^{r_2 t}$ and $Acos(\omega t - \delta) + C_2e^{r_1 t} + D_2e^{r_2 t}$ having initial points $Acos(\delta) + C_{1,2} + D_{1,2}$.

Hence we have, $\delta x(t) = C e^{r_1 t} + D e^{r_2 t}$ and $\delta x(0) = C + D$ where $C = C_1 - C_2$ and $D = D_1 - D_2$

So now my problem now comes down to being able to write $Ae^{x} + Be^{-x}$ in the form of $e^{y}(A+B)$ and figuring out y. And I don't know how I can do that. Am I doing this right? Or am I completely off track?

Ps. A and $r_1$ and $r_2$ have complex form depending on $\beta$ and $\omega$ etc.
$r_1$ and $r_2$ in the equation can be changed into the form x and -x

 

[Fred Wright]

Here's my suggestion and I'm no expert so you can take it or leave it. First, solve the differential equation explicitly. I found
r₁= -β+√(β²-ω₀²)
r₂= -β-√(β²-ω₀²)
δ=atan((ω²-ω₀²)/2ωβ).
You have δx(t) = e^-βt((C₁-C₂)e^{+√(β2-ω₀2)t} + (D₁-D₂)e^{-√(β2-ω₀2)t})
I assert that when β²-ω₀² << 1, λ≅-ω₀.