I am attempting to compute an approximation of the solution with the forward Euler method in [0,1] with step lengths h_1=0.2, h_2=0.1 given the initial value problem below dy/dz=1/(1+z)−y(z) y(0)=1 I am not sure what to do when I am given two step sizes instead of one. I know how to compute it if it was given with a step size. Am I supposed to find out the approximation for two different step sizes? Or is there anything I am missing?

balafiavatv

balafiavatv

Open question

2022-08-13

I am attempting to compute an approximation of the solution with the forward Euler method in [ 0 , 1 ] with step lengths h 1 = 0.2, h 2 = 0.1 given the initial value problem below
d y d z = 1 1 + z y ( z ) y ( 0 ) = 1
I am not sure what to do when I am given two step sizes instead of one. I know how to compute it if it was given with a step size. Am I supposed to find out the approximation for two different step sizes? Or is there anything I am missing?

Answer & Explanation

afinat4s

afinat4s

Beginner2022-08-14Added 13 answers

The problem asks for solving the differential equation twice. Once for the step size of h = .1 and once for the step size of h = .2 and compare the results. As you know different step sizes give you different results with the smaller step size smaller error is made .
Brooklyn Farrell

Brooklyn Farrell

Beginner2022-08-15Added 6 answers

We can apply the Euler’s method as usual using h 1 for the first solution that is
y i + 1 = y i + h 1 F ( z i , y i )
and h 2 for the second one that is
y i + 1 = y i + h 2 F ( z i , y i )
in order to compare the results since smaller isbthe step more accurate is the solution.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?