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Ordinary Differential Equation Series Solution

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data
    y' = [tex]\sqrt{(1-y^2)
    }[/tex]
    Initial condition y(0) = 0
    a) Show y = sinx is a solution of the initial value problem.
    b) Look for a solution of the initial value problem in the form of a power series about x = 0. Find coefficients up to the term in x^3 in this series.

    2. Relevant equations

    part a) was Ok.

    3. The attempt at a solution
    This is for my part b attempt.

    (y')^2 + (y)^2 = 1

    (cosx)^2 + (sinx)^2 = 1

    [tex]\sum(-1)^ {2n} * (x^ {4n} )}/(2n!)^2[/tex] + [tex]\sum (-1)^{2n} * (x^ {4n+2})/((2n+1)!)^2[/tex] = 1

    I had also tried the general solution series for y and y'. y = a0 + a1x + ... + an*x^n
    y' = a1 + ... + n*an x^n
    Please note the values such as a0, a1, ...an have the nth term as a subscript.

    Maybe I should not be using part a where y=sinx and y'=cosx because I will have even powers when squared always...?

    What series will allow me to evaluate at the x^3 term?

    Thanks,
    -Adam
     
    Last edited: Nov 17, 2009
  2. jcsd
  3. Nov 17, 2009 #2

    Mark44

    Staff: Mentor

    Double posted.
     
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