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Differential Equations and Power Series

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the differential equation f' = 2xf2 with the initial condition f(0)=1 in the following way:
    i) First, assume that there is a solution given by a power series
    f(x) = [​IMG]
    with a positive radius of convergence. SUbstitude this into the differential equation and figure out the coefficients an. (it is enough to guess a pattern - you do not have to prove that your guess is correct)

    3. The attempt at a solution
    I know f' = sigma(from n=1 to infinity)nanxn-1.
    So sigma(from n=1 to infinity)nanxn-1 = 2x[​IMG]^2
    I substituded x=0 into f(x) and found that a0=1
    I don't really know where to go from here :S. How do I figure out the coefficients??
     
    Last edited: Sep 13, 2008
  2. jcsd
  3. Sep 13, 2008 #2

    Dick

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    Let's skip the infinite series notation, it will just get in the way here. This problem is tough to do this way because of the f^2. You have f=1+a1*x+a2*x^2+... so f'=a1+2*a2*x+... That's equal to 2*x*(1+a1*x+a2*x^2+...)^2. You want to equate equal powers of x on either side. So you have to square out that expression to get the coefficients of the power x^k up to whatever k you feel you need to solve for.
     
  4. Sep 13, 2008 #3
    Thanks! That really helped :). Hm the pattern Im getting is 1, 0, 1, 0, 1.... What function gives you alternating 0 and 1's?? :S
    Would cos2(n pi/2) work?
     
    Last edited: Sep 13, 2008
  5. Sep 13, 2008 #4

    Dick

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    I'm getting 1,0,1,0,1... too. It's not a cos, it's a geometric series. Add it up or try solving the original differential equation by separation of variables. Either way, you find it's non-series form is 1/(1-x^2).
     
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