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Power series solution to a second order o.d.e.

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the terms up to x^5 in the power series solution of the following equation


    2. Relevant equations

    Power series, sum from 0 to infinity

    [tex]y=\sum a_{n}x^{n}[/tex]

    3. The attempt at a solution

    At first I just differentiated each term separately and then equated coefficients,


    After equating and rearranging i ended up with,


    Is this correct?

    Also I was wondering how I would go about doing this for the general case, using complete sums and then equating coefficients,

    [tex]y=\sum a_{n}x^{n}[/tex]

    [tex]y'=\sum na_{n}x^{n-1}[/tex]

    [tex]y''=\sum n(n-1)a_{n}x^{n-2}[/tex]

    [tex]x^{2}y=\sum a_{n}x^{n+2}[/tex]

    Subbing back into the original,

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

    Equating coefficients,


    But no matter what I do with this relation I can't get it into a helpful form that agrees with my above result???

    Please can someone help?
    Last edited: Dec 1, 2008
  2. jcsd
  3. Dec 1, 2008 #2


    Staff: Mentor

    I don't see that you multiplied your series for y by (1 + x^2). After you do that, equate your series for y'' with the series for (1 + x^2)y, and compare like powers of x to get a relationship involving your series coefficients a_0, a_1, etc.
  4. Dec 1, 2008 #3
    That was a silly mistake to make :blushing:, thankyou for pointing it out :smile:

    [tex]2a_{2}+6a_{3}x+12a_{4}x^{2}+20a_{5}x^{3}+...=(a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_ {5}x^{5}+...)+(a_{0}x^{2}+a_{1}x^{3}+a_{2}x^{4}+a_{3}x^{5}+a_{4}x^{6}+a_ {5}x^{7}+...)[/tex]



    and so on which leads to the relation,


    which agrees with my other relation (with some rearranging).

    Now I'm confused as to how this is helpful. What am I supposed to with this relation now that I have found it?
  5. Dec 1, 2008 #4


    Staff: Mentor

    Use the relation you found plus the others you showed to get the coefficients a_0 through a_5. You're not given any initial conditions, so that means that a_0 and a_1 will be arbitrary and the other coefficients will be in terms of those two.
  6. Dec 1, 2008 #5
    I'm completley at a loss as to how to get the coefficients in terms of a_0 and a_1. Sorry if I'm being dumb, but I all I've got so far is,

    Using these relations


    [tex]a_{2}=12a_{3}-a_{0}[/tex] 1
    [tex]a_{3}=20a_{4}-a_{1}[/tex] 2
    [tex]a_{4}=30a_{5}-a_{2}[/tex] 3
    [tex]a_{5}=42a_{6}-a_{3}[/tex] 4



    [tex]a_{4}=\frac{a_{2}+a_{0}}{12}[/tex] 5
    [tex]a_{5}=\frac{a_{3}+a_{1}}{20}[/tex] 6

    By substituting 2 into 1 and then the result of that into 5 I got,


    But this seems to be a circular way of doing it and sure enough I get, upon subs,


    I think I'm going the wrong way about this...please could you guide me as to how I can find the coefficients in terms of a_0 and a_1?
  7. Dec 1, 2008 #6


    Staff: Mentor

    Equating the series for y'' with x^2 * y + y, I got:
    2a_2 = a_0
    6a_3 = a_1
    12a_4 = a_0 + a_2
    20a_5 = a_1 + a_3
    and so on.
    You always want to solve for the higher-index coefficients in terms of those with lower indexes. In this problem, you'll eventually get back to a_0 and a_1. For example, a_4 is in terms of a_0 and a_2, but a previous equation relates a_2 and a_0.

    All you need are a_0 through a_5.
  8. Dec 3, 2008 #7
    With your assistance I now get,

    [tex]y=a_{0}(1+\frac{x^{2}}{2}+\frac{x^{4}}{8}+...)+a_{1}(x+\frac{x^{3} }{6}+\frac{7x^{5}}{120}+...)[/tex]

    Which seems more reasonable and my realtions all agree. Does this look okay to you?

    Thanks a million for your help, I would still be stuck without it! :smile:
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