vagnhestagn

2022-09-07

I know the difference between the forward (explicit) Euler method and backward (implicit) Euler method in the context of approaching solutions to ODEs.
However, I have a doubt regarding the following method:
${y}_{n+1}={y}_{n}+\frac{h}{2}\left(f\left({t}_{n},{y}_{n}\right)+f\left({t}_{n+1},{y}_{n+1}\right)\right)$
I am not sure whether it is an implicit or an explicit method.
I tend to say it is implicit one step methode because we have to solve an equation to calculate ${y}_{n+1}.$

Jase Powell

You're quite right.
An explicit method is given by ${y}_{n+1}=F\left({t}_{n},{y}_{n}\right)$, with no mention of ${y}_{n+1}$ on the RHS, and you can compute ${y}_{n+1}$ directly.
An implicit method is given by $F\left({t}_{n},{y}_{n},{y}_{n+1}\right)=0$, and you have to solve an equation to find ${y}_{n+1}$.
Both these definitions are for one-step methods: ${y}_{n+1}$ depends directly or indirectly just on ${t}_{n}$ and ${y}_{n}$, not on any ${y}_{k}$ for $k.

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