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\( u^n = ct_n+I \) fulfills the discrete equations!
First, $$ \begin{align} \lbrack D_t^+ t\rbrack^n &= \frac{t_{n+1}-t_n}{\Delta t}=1, \tag{34}\\ \lbrack D_t^- t\rbrack^n &= \frac{t_{n}-t_{n-1}}{\Delta t}=1, \tag{35}\\ \lbrack D_t t\rbrack^n &= \frac{t_{n+\half}-t_{n-\half}}{\Delta t}=\frac{(n+\half)\Delta t - (n-\half)\Delta t}{\Delta t}=1\tag{36} \end{align} $$
Forward Euler: $$ [D^+ u = -au + b]^n $$
\( a^n=a(t_n) \), \( b^n=c + a(t_n)(ct_n + I) \), and \( u^n=ct_n + I \) results in $$ c = -a(t_n)(ct_n+I) + c + a(t_n)(ct_n + I) = c $$
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