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Picard's Iteration

  1. Aug 21, 2005 #1
    Sorry for making another thread,but this problem is really penetrating for me!

    dy/dx=y^2 with initial condition y(0)=1

    I have reached upto
    5/9 + 4/9((1+x) + (1+x)^4/4 + (1+x)^7/7 + ........)
    And ahead of that i have no clue,!!
     
  2. jcsd
  3. Aug 21, 2005 #2

    Hurkyl

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    That one's actually pretty easy: just use separation of variables.
     
  4. Aug 21, 2005 #3
    the question has to be done by Picard's iteration,not by seperation of variables!
     
  5. Aug 22, 2005 #4
    Is this really penetrating question!
     
  6. Aug 22, 2005 #5

    HallsofIvy

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    WHY do you have it in terms of (1+ x) ? Your initial value is given at x= 0.

    For those of you who don't know, Picard's iteration is this:

    Given the intial value problem, y'= f(x,y), y(x0)= y0, imagine that we know y as a function of x and integrate both sides:
    [tex]y(x)= y_0+ \int_{x_0}^x f(t,y(t))dt[/tex]
    The initial value problem has a solution if and only if that integral equation has a solution. The integral equation can be thought of as a "fixed value" problem and, since Banach's fixed value theorem holds (see thread on "existance and uniqueness"), we can do it by iteration. Let Y be any function. The constant y(x)= y_0 works nicely. Plug that into the righthand side and integrate. Use the value of y(x) you get to repeat.

    In this case, the initial value problem is y'= y2, y(0)= 1. That converts to the integral equation [itex]y(x)= 1+ \int_0^x (y(t))^2 dt[/itex].

    Taking y(t)= 1 we get the new solution
    [tex]y(x)= 1+ \int_0^x (1)^2 dt= 1+ x[/tex].
    Taking y(t)= 1+ t, we get
    [tex]y(x)= 1+ \int_0^x(1+ t)^2 dt= 1+ x+ x^2+ (1/3)x^3[/tex].
    Continue until you think you see a pattern (or until you are exhausted).
     
    Last edited: Aug 22, 2005
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