the backward Euler method
/ OY-ler /
Forward Euler asks 'which way is the slope HERE?' and walks that direction. Backward Euler is cannier: it asks 'where could I land so that the slope AT THE DESTINATION points back along the path I just took?' Instead of trusting the slope where you start, it insists the step be consistent with the slope where you finish. That small twist costs more work each step but buys remarkable stability.
The rule is y_{n+1} = y_n + h * f(t_{n+1}, y_{n+1}). Notice the unknown y_{n+1} appears on BOTH sides — on the right it is buried inside f at the new time. So you cannot just evaluate; you must SOLVE for y_{n+1}, which makes the method IMPLICIT. If f is linear in y the equation is a simple rearrangement; if f is nonlinear you solve it each step with Newton's method or a fixed-point iteration. For example, on y' = -2y the step y_{n+1} = y_n - 2h*y_{n+1} rearranges to y_{n+1} = y_n / (1 + 2h), a clean closed form. Geometrically you draw the tangent at the END of the step and require the step to lie along it.
Why pay for the extra solve? Stability. Backward Euler is A-stable: on the test problem y' = lambda*y with any decaying mode (real part of lambda negative), the computed solution decays for EVERY step size h, never blowing up. This makes it the workhorse for STIFF problems, where forward Euler would demand absurdly tiny steps just to avoid exploding. The honest trade-offs: it is still only first order (low accuracy, like forward Euler), each step costs a nonlinear solve, and it is actually OVER-damped — it can smear out sharp features and kill oscillations that should persist. It is chosen for robustness, not precision.
On the stiff decay y' = -100y, y(0) = 1, forward Euler needs h < 0.02 or it explodes. Backward Euler with a huge h = 1 still behaves: y_1 = 1 / (1 + 100*1) = 1/101 = 0.0099, a small positive number that sensibly tracks the true (vanishingly small) e^(-100) = 4e-44 — wrong in detail but stable and bounded, never blowing up.
Implicit: solve for the new value using the slope at the destination — stable even with large steps.
A-stability does not mean accuracy: backward Euler can return a stable but quite wrong answer with a big step. Its strong damping is a feature for stiff decay but a bug for problems with genuine oscillations you want to preserve — there a symplectic or higher-order implicit method is better.