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Picard's Method

  1. Jun 21, 2011 #1
    1. The problem statement, all variables and given/known data

    y' = -y , y(0) = 1

    2. Relevant equations

    Picard's method

    3. The attempt at a solution

    I found y1 as 1 + 1/2x2, however y1 is really 1 + x
     
    Last edited: Jun 21, 2011
  2. jcsd
  3. Jun 21, 2011 #2

    LCKurtz

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    Neither of those is correct for the equation you wrote. Show us what you did so we can find your mistake(s).
     
  4. Jun 21, 2011 #3

    lanedance

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    you should show you method, so we can see where you're going wrong - what was your first approximation? starting with a constant should integrate to a constant times x and you should have the solution after one iteration
     
  5. Jun 21, 2011 #4

    lanedance

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    also as LCKurtz points out neither is correct, you should alway check by substituting into the original DE
     
  6. Jun 22, 2011 #5
    Sorry I meant y1 is really 1 -x. It was a typo.
     
  7. Jun 22, 2011 #6

    HallsofIvy

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    I think LCKurtz and lanedance are misunderstanding your notation. Neither [itex]y= 1+ x^2/2[/itex] nor [itex]y= 1+ x[/itex] is a solution to the equation but you are not claiming it is.

    Picard's method of solving the differential equation y' = f(x,y), with initial condition [itex]y(0)= y_0[/itex] is an iterative method. Taking [itex]y_0[/itex] to be the initial value, [itex]y_1= \int f(x, y_0)dx[/itex] is the first iteration, then [itex]y_2= \int f(x, y_1(x))dx[/itex], [itex]y_3= \int f(x,y_2(x))dx[/itex], etc.

    For the problem as given, the first iteration of Picard's method is, indeed, [itex]y_1(x)= x+ 1[/itex]. But no one can show you where you went wrong until you show us what you did.
     
  8. Jun 22, 2011 #7

    LCKurtz

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    Nor are we claiming that.

    No, it isn't.
     
  9. Jun 22, 2011 #8

    HallsofIvy

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    Oh, I completely misread the equation! It is y'= -y, not y'= y as I was seeing!

    My apologies to both LCKurtz and lanedance.

    trojansc82, I still don't see how you got "1+ x^2/2". Please show us what you did.
     
  10. Jun 22, 2011 #9
    Again, I apologize, I mistyped the answer.

    The answer for y1 = 1 - x
     
  11. Jun 22, 2011 #10

    LCKurtz

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    Is that the answer you got? In your original post you said you got something else. Have you figured it out? You never did show us what you did...
     
  12. Jun 22, 2011 #11
    Within the integral I multiplied -1 (since y was -y) by t. I ended up integrating -t, which came to -1/2 x2.

    I have trouble within the integral.
     
  13. Jun 22, 2011 #12

    LCKurtz

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    What integral? We aren't mind readers. My guess is you have the integral wrong in the first place. Show us your work.
     
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