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Summer Bradford 2022-06-26 Answered
Consider the IVP
y = y
for t 0, and y ( 0 ) = 1, y ( 0 ) = 2.

I have rewritten this differential equation as a system of first-order ODE's such that
u = v v = u
with u ( 0 ) = 1 , v ( 0 ) = 2.

The solution is y = 2 sin ( t ) + cos ( t ), y = 2 cos ( t ) sin ( t )

I am asked to perform one step of Euler's method with h = 0.5, and determine if Euler's method is stable for this problem.

For the first part, I find that one step of Euler's method yields
y 1 = y 0 + h f ( t 0 , y 0 ) = 1 + ( 0.5 ) ( 2 cos ( 0 ) sin ( 0 ) ) = 2.
But how do I determine if Euler's method is stable? I know that for the equation y = λ y, Euler's method is stable for | 1 + h λ | 1, but since this problem is in a different form I'm not sure what I need to do here.
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Answers (2)

Bornejecbo
Answered 2022-06-27 Author has 19 answers
Applying the Euler iteration procedure we have
{ u k = u k 1 + h v k 1 v k = v k 1 h u k 1
or
( u k v k ) = ( 1 h h 1 ) ( u k 1 v k 1 )
or
U k = M k U 0
this sequence converges as long as the eigenvalues of M have absolute value less than 1. Here the M eigenvalues are 1 ± i h with absolute value 1 + h 2 > 1 so the Euler procedure diverges.
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Layla Velazquez
Answered 2022-06-28 Author has 11 answers
You apply the Euler step to the first-order system.
u 1 = u 0 + h u 0 = u 0 + h v 0 , v 1 = v 0 + h v 0 = v 0 h u 0 .
As the Euler step is tangential to the convex solution curve, it will always move outwards, away from the center of the concentric exact solution curves.
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I have been studying the Backward Euler Method, and I am have having some of problems representing the stability of the method of some particular ODEs.
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Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized:
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