Carrying out the truncation error analysis

Using (15)-(16) and (5)-(6) we get $$\begin{align*} \lbrack mD_tD_t \uex + \beta D_{2t} \uex\rbrack^n &= m\uex''(t_n) + \beta\uex'(t_n) + \\ &\quad \left(\frac{m}{12}\uex''''(t_n) + \frac{\beta}{6}\uex'''(t_n)\right)\Delta t^2 + \Oof{\Delta t^4} \end{align*}$$ The terms $$m\uex''(t_n) + \beta\uex'(t_n) + \omega^2\uex(t_n) + s(\uex(t_n)) - F^n,$$ correspond to the ODE (= zero).

Result: accuracy of $\Oof{\Delta t^2}$ since $$\begin{equation} R^n = \left(\frac{m}{12}\uex''''(t_n) + \frac{\beta}{6}\uex'''(t_n)\right)\Delta t^2 + \Oof{\Delta t^4}, \tag{39} \end{equation}$$

Correction terms: complicated when the ODE has many terms...