# Lyapunov exponents of a damped, driven harmonic oscillator

I am supposed to calculate Lyapunov exponent of a damped, driven harmonic oscillator given by $\ddot{x} + 2\beta \dot{x} + \omega_0^2 x = f\cos(\omega t)$

Lyapunov exponent is $\lambda$ in $\delta x(t) = \delta x_0 e^{\lambda t}$

The general solution of the system is given by $A\cos(\omega t - \delta) + Ce^{r_1 t} + De^{r_2 t}$

Consider two initial points 1 and 2. The solutions evolve to give $A\cos(\omega t - \delta) + C_1e^{r_1 t} + D_1e^{r_2 t}$ and $A\cos(\omega t - \delta) + C_2e^{r_1 t} + D_2e^{r_2 t}$. 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 $e^{y}(A+B)$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 form depending on $\beta$, $\omega$ etc. $r_1$ and $r_2$ can be in the equation can be changed into the form $x$ and $-x$