The Backward Euler scheme

Inserting the finite difference approximation in $u'(t_n)=-au(t_n)$ yields the Backward Euler (BE) scheme: $$\begin{equation} \frac{u^{n}-u^{n-1}}{t_{n}-t_{n-1}} = -a u^n \tag{8} \end{equation}$$ Solve with respect to the unknown $u^{n+1}$: $$\begin{equation} u^{n+1} = \frac{1}{1+ a(t_{n+1}-t_n)} u^n \tag{9} \end{equation}$$