1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential Equations - Power Series problem with e^t

  1. Jul 15, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem is to solve:

    y''+ty'+e[itex]^{t}[/itex]y=0, y(0)=0 and y'(0)=-1

    2. Relevant equations



    3. The attempt at a solution

    My main issue is the following: I normally find the recursion relation, and then factor out the t[itex]^{whatever}[/itex] and I know that the coefficient to this must equal 0. However, the addition of the e[itex]^{t}[/itex] means I can't factor out ALL of the t[itex]^{whatever}[/itex] so I can't exactly equate the coefficient to 0.

    So specifically with the question above, I get the following:

    ty'=[itex]\sum[/itex][itex]^{inf}_{k=1}[/itex]ka[itex]_{k}[/itex]t[itex]^{k}[/itex]

    e[itex]^{t}[/itex]=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{k}[/itex]/k!

    e[itex]^{t}[/itex]y=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{2k}[/itex]a[itex]_{k}[/itex]/k!

    y''=[itex]\sum[/itex][itex]^{inf}_{k=2}[/itex]k(k-1)a[itex]_{k}[/itex]t[itex]^{k-2}[/itex]

    Then I plug those into the DE, write out the first few terms so that they all start at k=2, and put them all under the same sum to get as follows:

    a[itex]_{0}[/itex]+a[itex]_{1}[/itex]t+a[itex]_{1}[/itex]t[itex]^{2}[/itex]+2a[itex]_{2}[/itex]+6a[itex]_{3}[/itex]t+[itex]\sum[/itex][itex]^{inf}_{k=2}[/itex][(k+2)(k+1)a[itex]_{k+2}[/itex]+ka[itex]_{k}[/itex]+a[itex]_{k}[/itex]t[itex]^{k}[/itex]]t[itex]^{k}[/itex]=0

    From this you find that a[itex]_{2}[/itex]=-a[itex]_{0}[/itex]/2

    and

    a[itex]_{1}[/itex]=0

    and

    a[itex]_{3}[/itex]=0


    Like I said, normally at this point I find the recursion relation and factor out a t[itex]^{k}[/itex] so that I know that the entire coefficient is 0. But in this case, after factoring out the t[itex]^{k}[/itex] one of the terms in the coefficient still have a t[itex]^{k}[/itex], so I don't know if I can't just equate this whole thing to 0.

    Can anyone help?

    Thank-you
     
  2. jcsd
  3. Jul 15, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You don't need to put the itex tags around every item. Just put them around the whole equation. Right click on the above expression and see how I changed it for you.

    You don't multiply two series like that. You need the Cauchy Product. See:
    http://planetmath.org/cauchyproduct
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Differential Equations - Power Series problem with e^t
Loading...