Truncation error of the Forward Euler scheme

The Forward Euler scheme: $$\begin{equation} \lbrack D_t^+ u = -au \rbrack^n \tag{25} \tp \end{equation}$$ Definition of the truncation error $R^n$: $$\begin{equation} \lbrack D_t^+ \uex + a\uex = R \rbrack^n \tag{26} \tp \end{equation}$$ From (9)-(10): $$[D_t^+ \uex]^n = \uex'(t_n) + \half\uex''(t_n)\Delta t + \Oof{\Delta t^2}\tp$$ Inserted in (26): $$\uex'(t_n) + \half\uex''(t_n)\Delta t + \Oof{\Delta t^2} + a\uex(t_n) = R^n \tp$$ Note: $\uex'(t_n) + a\uex^n = 0$ since $\uex$ solves the ODE. Then $$\begin{equation} R^n = \half\uex''(t_n)\Delta t + \Oof{\Delta t^2} \tag{27} \tp \end{equation}$$