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Help with implicit euler method

  1. May 27, 2014 #1
    Hey
    I don't understand this backward euler solution, in particular why the f(y,t+h) is equal to y(t+2)
     

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  2. jcsd
  3. Jun 11, 2014 #2
    Because it's wrong.

    It should be:

    y(t+2)=y(t)+y'(t+2)h

    Since, in this problem y'=-y, you have:

    y(t+2)=y(t)-y(t+2)h

    Solving for y(t+2) gives:
    [tex]y(t+2)=\frac{y(t)}{1+2h}[/tex]
    Chet
     
  4. Jun 11, 2014 #3

    D H

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    I agree with you there!

    Actually, it should be y(t+h)=y(t)+h*y'(t+h). Given that y'(t)=-y(t), this becomes y(t+h)=y(t)-h*y(t+h), or y(t+h)=y(t)/(1+h).

    Better: ##y(t+2)=\frac {y(t)}{3}##.

    You've already set h=2.
     
  5. Jun 11, 2014 #4
    Thanks DH. I'm usually more careful about checking over what I've written before I submit my replies. I hope I didn't confuse the OP too much.

    Chet
     
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