# Euler's Method

• Mar 3rd 2009, 07:29 PM
Aryth
Euler's Method
Use Euler's Method with $h = 0.2$ to approximate the solution to the initial value problem:

$y' = -20y$
$y(1) = 3$

at

$x = 1, 1.2, 1.4$
• Mar 4th 2009, 04:08 AM
CaptainBlack
Quote:

Originally Posted by Aryth
Use Euler's Method with $h = 0.2$ to approximate the solution to the initial value problem:

$y' = -20y$
$y(1) = 3$

at

$x = 1, 1.2, 1.4$

Euler's method for the numerical integration of an ODE IVP uses the stepping formula:

$y(t+ \delta t)=y(t)+\delta t\times y'(t)$

You are given $y(1)$ and that $\delta t =0.2$, so immediately the stepping formula will tell you the value at $t=1.2$, then applying it again from $t=1.2$ and the value of $y(1.2)$ you found from the first step, it will give the value at $t=1.4$.

CB
• Mar 4th 2009, 06:25 AM
Aryth
I am getting bigger and bigger numbers... Is it supposed to blow up like that?
• Mar 4th 2009, 06:29 AM
HallsofIvy
No, it shouldn't. Please show what you have done and why you are getting "bigger and bigger numbers".
• Mar 4th 2009, 06:36 AM
CaptainBlack
Quote:

Originally Posted by Aryth
I am getting bigger and bigger numbers... Is it supposed to blow up like that?

This equation is integrable, and so you should know what the solution is analyticaly.

You are being given a step size for which the Euler method is unstable, probably to show/teach you something!

CB
• Mar 4th 2009, 06:48 AM
Aryth
Yeah. I knew the solution to the equation already.

Thanks for the help. I appreciate it. And this was a test question... I just wanted to get clarification on the answer.

But the next question was the same thing with the Improved Euler's method and I got a much better answer, which would mean that bigger values of h are allowed in the Improved method as opposed to the classical method.
• Mar 4th 2009, 07:19 AM
CaptainBlack
Quote:

Originally Posted by Aryth
Yeah. I knew the solution to the equation already.

Thanks for the help. I appreciate it. And this was a test question... I just wanted to get clarification on the answer.

But the next question was the same thing with the Improved Euler's method and I got a much better answer, which would mean that bigger values of h are allowed in the Improved method as opposed to the classical method.

The Improved Euler method (which if it is what I think it is - a predictor corrector using Euler steps) then it is unconditionally stable.

CB