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Power Series Solution to a Diff EQ

  1. Mar 24, 2008 #1
    [SOLVED] !!!Power Series Solution to a Diff EQ!!!

    1. The problem statement, all variables and given/known data
    Find the first 5 term of a Power series solution of

    [tex]y'+2xy=0[/tex] (1)

    Missed this class, so please bear with my attempt here.

    3. The attempt at a solution

    Assuming that y takes the form


    Then (1) can be written:


    Re-written 'in phase' and with the same indices (in terms of k):


    [tex]\Rightarrow c_1+\sum_{k=1}^{\infty}[(k+1)c_{k+1}+2c_{k-1}]x^k=0[/tex]

    Now invoking the identity property, I can say that all coefficients of powers of x are equal to zero (including [itex]c_1*x^0[/itex])

    So I can write:

    [itex]c_1=0[/itex] and


    Now I am stuck (I know I am almost there though!)

    Should I just start plugging in numbers for k=1,2,3,4,5 ? Will this generate enough 'recursiveness' to solve for the 1st five terms?

    Is that the correct approach?

  2. jcsd
  3. Mar 24, 2008 #2
    Don't worry guys, I got it. And for those who might make future use of this thread, my approach was correct. Solving


    For k=1,2....,9 generates enough coefficients to write out the first five nonzero terms of the solution by plugging them back into

    Last edited: Mar 24, 2008
  4. Mar 24, 2008 #3
    I think you need one more initial condition on your series (example [itex] y(0)=x_0[/itex]). You need to define [itex] c_0[/itex] (or you can just leave [itex] c_0 [/itex] as the 'integration' constant of the ODE). However, the recursion formula becomes pretty obvious once you have that. If you can't see it right away, try plugging in a few values. Since [itex] c_1=0[/itex] what can we say about all the odd labelled coefficients? You should get something that looks like the series of an exponential function. In fact, the series will have a closed form solution if you can see how your answer relates to the exponential series.
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