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ODE IVP Series Solution

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the series solution to the initial value problem.
    xy[tex]\acute{}[/tex][tex]\acute{}[/tex] + y[tex]\acute{}[/tex] + 2y = 0
    y(1) = 2
    y[tex]\acute{}[/tex](1) = 4

    2. Relevant equations


    t = (x-1), x = (t+1)

    y = [tex]\sum^{\infty}_{n=0}c_{n}t^{n}[/tex]

    y[tex]\acute{}[/tex]= [tex]\sum^{\infty}_{n=1}c_{n}(n)t^{n-1}[/tex]

    y[tex]\acute{}[/tex][tex]\acute{}[/tex]= [tex]\sum^{\infty}_{n=2}c_{n}(n)(n-1)t^{n-2}[/tex]

    3. The attempt at a solution
    I substituted the above series into the DE, adjusted the series so they all had t[tex]^{n}[/tex], and took out terms so that they all had the same starting index. By grouping the terms and the series, I got:

    (2[tex]c_{2}[/tex] + [tex]c_{1}[/tex] + 2[tex]c_{0}[/tex]) + [tex]\sum^{\infty}_{n=1}t^{n}[ (n+2)(n+1)c_{n+2} + (n+1)^{2}c_{n+1} + 2c_{n}][/tex] = 0

    Setting the terms and series equal to zero and finding several constants:

    [tex]2c_{2}[/tex] + [tex]c_{1}[/tex] + [tex]2c_{0}[/tex] = 0

    [tex](n+2)(n+1)c_{n+2}[/tex] + [tex]c_{n+1}[/tex](n+1)[tex]^{2}[/tex] + [tex]2c_{n}[/tex]] = 0

    [tex]c_{0}[/tex] = -(2c_{2}[/tex]+[tex]c_{1}[/tex])/2

    [tex]c_{1}[/tex] = -2([tex]c_{0}[/tex]+[tex]c_{2}[/tex])

    [tex]c_{2}[/tex] = -([tex]c_{1}[/tex]+[tex]2c_{0}[/tex])

    [tex]c_{3}[/tex] = [tex](2/3)c_{0}[/tex]

    [tex]c_{4}[/tex] = [tex](c1-4c0)/(3\cdot4)[/tex]

    [tex]c_{5}[/tex] = [tex]-(c1+5c0)/(3\cdot5)[/tex]

    [tex]c_{6}[/tex] = [tex](9c1+46c0)/(2\cdot3\cdot5\cdot6)[/tex]

    I don't understand how find the solution for the initial values. I can't determine a pattern for the constants for [tex]c_{n}[/tex]. Are you supposed to group these constants: y = [tex]c_{0}[/tex][ 1 + x + ...] + [tex]c_{1}[/tex][ 1 + x +...] to get the solution, and if so how do we find the constants for the initial values? Any help will be greatly appreciated.
  2. jcsd
  3. Oct 27, 2009 #2


    Staff: Mentor

    Your initial conditions y(1) = 2 and y'(1) = 4 are all in terms of x. Your power series is in terms of t = x - 1, so x = 1 corresponds to t = 0. Does that help?
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