To be precise for the Forward Euler scheme, can we find \( C \) to make \( R \) $\Oof{\Delta t^2}$?
$$ \begin{equation} \lbrack D_t^+ \uex + a\uex = C + R \rbrack^n\tp \tag{30} \end{equation} $$
$$ \half\uex''(t_n)\Delta t - \frac{1}{6}\uex'''(t_n)\Delta t^2 + \Oof{\Delta t^3} = C^n + R^n\tp$$ Choosing
$$ C^n = \half\uex''(t_n)\Delta t,$$ makes
$$ R^n = \frac{1}{6}\uex'''(t_n)\Delta t^2 + \Oof{\Delta t^3}\tp$$