# Implicit Euler method for linear first order ODE's

We have the linear first order ODE $y'(t) = m(t) y(t) + n(t)$ where $m(t), n(t)$ are given functions. We need to show that after applying the implicit Euler method we receive a linear system of equations for the variable $y_{k+1}$.

Let $h>0$ be the step size, $t_0 \in \mathbb{R}$ such that the ODE suffices the initial value condition $y(t_0) = y^0$. The implicit Euler method in our case is given by $y_0 = y^0$ and $y_{k+1} = y_k + h (a(t_{k+1}) y_{k+1} + b(t_{k+1}))$.

This leads to $y_{k+1} = \frac{y_k+h n(t_{k+1})}{1-h m(t_{k+1})}$.

My questions:

• How does this make a linear system of equations or what do we mean with that? Here we have only one equation and in what sense is this equation linear? Usually I would associate a system of linear equations with something like $Ax = b$.
• We use the implicit Euler method. But is it correct if the result is an explicit equation? That confuses me a bit.

Any help is very appreciated.

The equation you get from the implicit method (whose solution you correctly give in the question) is: $$\frac{y_{k+1}-y_k}{h}=m(t_{k+1})\,y_{k+1}+n(t_{k+1}).$$ This expression is linear in $y_{k+1}$ and $y_k$. Since the ODE is linear, the equation coming from the implicit Euler method turns out to be linear. And since this equation is linear, it is possible to solve explicitly for $y_{k+1}$ in terms of $y_k$, even if the method is implicit. There is no contradiction here.