\( u(0)=I \), \( u'(0)=V \): $$ \begin{align*} \lbrack u &=I\rbrack^0\quad\Rightarrow\quad u^0=I\\ \lbrack D_{2t}u &=V\rbrack^0\quad\Rightarrow\quad u^{-1} = u^{1} - 2\Delta t V \end{align*} $$ End result: $$ u^1 = u^0 + \Delta t\, V + \frac{\Delta t^2}{2m}(-bV - s(u^0) + F^0) $$ Same formula for \( u^1 \) as when using a centered scheme for \( u''+\omega u=0 \).