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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 \( u^{n+1} \): $$ \begin{equation} u^{n+1} = \frac{1}{1+ a(t_{n+1}-t_n)} u^n \tag{9} \end{equation} $$
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