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Math Help - Euler's Method

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    Super Member Aryth's Avatar
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    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
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    Grand Panjandrum
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    Quote Originally Posted by Aryth View Post
    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
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    Super Member Aryth's Avatar
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    I am getting bigger and bigger numbers... Is it supposed to blow up like that?
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    No, it shouldn't. Please show what you have done and why you are getting "bigger and bigger numbers".
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    Grand Panjandrum
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    Quote Originally Posted by Aryth View Post
    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
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    Super Member Aryth's Avatar
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    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.
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    Grand Panjandrum
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    Quote Originally Posted by Aryth View Post
    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
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