1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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


    User Avatar
    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook