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Picard Iteration

  1. Nov 19, 2012 #1
    Use picard iteration 4 times to estimate the initial value problem solution:

    y' = 2xy thru (0,1)

    First off, when it says thru (0,1), I am assuming it means y(0) = 1. I was not sure if I was doing picard iteration right, but here it my attempt:

    dy/dx = 2xy

    dy/y = 2x dx

    dy/1 = 2x dx - since y(0) = 1

    y = x^2 + C so y = x^2 + 1. This is the first iteration

    y' = x^2 + 1. So now I just solve this

    ........ and so on 4 times.

    Or is picard iteration work like this:

    dy/dx = 2xy

    dy/y = 2x dx

    ln|y| = x^2 + C

    y = e(x^2 + C)

    y = e^(x^2)e^(C)

    y = Ce^(x^2)

    when y(0) = 1, C = 1

    So y = e^(x^2) This is the first iteration

    So then I integrate and repeat this 3 more times.

    I wasnt sure which one was the right way to do picard iteration or if neither way is right.
     
  2. jcsd
  3. Nov 19, 2012 #2

    haruspex

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    Having just read up on Picard Iteration (http://www.sosmath.com/diffeq/first/picard/picard.html) in order to respond:
    You take as first approximation y0(x) = y(0) = 1.
    Now you use y' = 2xy in the integral form to get a better approximation:
    [itex]y_1(x) = \int_{s=0}^x 2s y_0(s) ds = \int_{s=0}^x 2s (1) ds =x^2[/itex]
    and to get the next one, repeat the process:
    [itex]y_2(x) = \int_{s=0}^x 2s y_1(s) ds [/itex]
    etc.
     
  4. Nov 20, 2012 #3

    HallsofIvy

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    Your first try was correct. Picard iteration gives a power series solution.
    With y(0)= 1, y'= 2xy, then your first approximation is y(x)= 1 so that y'= 2x, y= x2+ 1, so [itex]y'= 2x(x^2+ 1)= 2x^3+ 2x[/itex] and [itex]y(x)= (1/2)x^4+ x^2+ 1[/itex], etc.
     
  5. Nov 20, 2012 #4
    Okay so let me see if I get this.

    y' = 2xy y(0) = 1

    dy/y = 2x dx

    dy = 2x dx

    y = x^2 + C

    y = x^2 + 1 (first iteration)

    integrate 2x(x^2 + 1) dx

    y = (1/2)x^4 + x^2 + C

    y = (1/2)x^4 + x^2 + 1 (second iteration)

    integrate 2x[(1/2)x^4 + x^2 + 1] dx

    y = (1/6)x^6 + (1/2)x^4 + x^2 + 1 (third iteration)

    integrate 2x[(1/6)x^6 + (1/2)x^4 + x^2 + 1] dx

    y = (1/24)x^8 + (1/6)x^6 + (1/2)x^4 + x^2 + 1
     
  6. Nov 21, 2012 #5

    haruspex

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    Looks right.
     
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