Does improved Euler method fail when we increase step size?meaning does it give worse answer than standard Euler method when our step size is big?

Question

ohhappyday890b · Accepted Answer

Let&#039;s do the first step. For the ODE       y    ′    =  −  y, for which the solution is monotically decreasing towards 0, the transition from       y    0   to       y    1   with step size   h isEuler forwardy      y    1    =      y    0    +  h  f  (      y    0    )  =  (  1  −  h  )      y    0   which becomes unstable for   h  ≥  2.Improved Euler      y    1    =      y    0    +  h  ·  f  (      y    0    +  h      /    2  ·  f  (      y    0    )  )  =      y    0    −  h  ·  (      y    0    −  h      /    2  ·      y    0    )  =  (  1  −  h  +      h    2        /    2  )  ·      y    0  Now explore when       |    1  −  h  +      h    2        /    2      |    ≥  1

Does improved Euler method fail when we increase step size?meaning does it give worse answer than standard Euler method when our step size is big?

Answered question

Answer & Explanation

New Questions in Integral Calculus