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Solving a differential equation using Picard's iteration method

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Use Picard's iteration method to solve the initial value problem y' = t + y, y(0) = 0. Determine [itex]\phi_{n}(t)[/itex] for an arbitrary value of n, and take the limit as n goes to infinity. Of course, you know the series for e^t, so you will recognize the limit function [itex]\phi_{n}(t)[/itex]. Also evaluate [itex]\phi_{3}(t) [/itex] at t=0, 0.1, and 0.2.

    2. Relevant equations

    none

    3. The attempt at a solution

    I set up the problem by letting [itex]\phi_{n+1}(t) = \int^{0}_{t}f(s, \phi_{n}(s))ds[/itex]

    What I get is [itex]\phi_{0}(t) = 0[/itex], [itex]\phi_{1}(t) = \frac{t^{2}}{2}[/itex], [itex]\phi_{2}(t) = \frac{t^{2}}{2} + \frac{t^{3}}{6}[/itex], [itex]\phi_{3}(t) = \frac{t^{2}}{2} + \frac{t^{3}}{6} + \frac{t^{4}}{24}[/itex]. So clearly,

    [itex]\phi_{n}(t) = \frac{t^{2}}{2!} + \frac{t^{3}}{3!} + \frac{t^{4}}{4!} + ... + \frac{t^{n}}{n!} = e^{t} - e -1[/itex].

    But I know that the differential equation y' = t + y has the unique solution [itex]y = \frac{-te^{-t} - e^{-t} +1}{e^{-t}}[/itex]:

    Rewrite the de as follows:
    [itex]y' + (-1)y = t[/itex]
    [itex]\mu(t) = e^{\int(-1)dt} = e^{-t}[/itex]
    [itex]e^{-t}y' + e^{-t}(-1)y = e^{-t}t[/itex]
    [itex]\frac{d}{dt}[ye^{-t}] = e^{-t}t[/itex]
    [itex]ye^{-t} = -te^{-t} + \int e^{-t} = -te^{-t} - e^{-t}[/itex]
    [itex]y =\frac{-te^{-t} - e^{-t} + C}{e^{-t}}[/itex]

    And we can solve for C easily...but i'll save you that and tell you C = 1 (I think)

    These two are not equal...

    I'm really confused...as for the evaluation part, I can do that once I've solved the DE.
     
    Last edited: Feb 27, 2012
  2. jcsd
  3. Feb 27, 2012 #2

    kai_sikorski

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    Gold Member

    Series for et is
    [tex]e^t =1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+...[/tex]
    So your series is
    [itex]e^t - 1 - t[/itex]

    Thats the same thing as your second solution.
     
  4. Feb 27, 2012 #3
    So [itex]y = \frac{-te^{-t} - e^{-t} +1}{e^{-t}} = e^{t} - e -1[/itex]?
     
  5. Feb 27, 2012 #4

    kai_sikorski

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    Gold Member


    No, where do you keep getting this -e from?

    [itex]y = \frac{-te^{-t} - e^{-t} +1}{e^{-t}} = e^{t} - t -1[/itex]
     
  6. Feb 27, 2012 #5
    Right. I kept thinking e^1. But it's t^1. Thank you.
     
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