Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Nonhomogeneous Power Series Solution

  1. Mar 15, 2012 #1
    Hi. I have to solve: [tex]y''+xy'-2y=e^x[/tex]
    Using series. So, this is what I did:
    [tex]y(x)=\sum_0^{\infty}a_n x^n[/tex]
    [tex]y'(x)=\sum_1^{\infty}n a_n x^{n-1}[/tex]
    [tex]y''(x)=\sum_2^{\infty}n(n-1) a_n x^{n-1}[/tex]
    And [tex]e^x=\sum_0^{\infty}\frac{x^n}{n!}[/tex]
    Then, using that m=n-2 for y'' and then replacing in the diff. eq:
    [tex]\sum_0^{\infty}(n+2)(n+1)a_{n+2} x^n+x\sum_1^{\infty}n a_n x^{n-1}-2\sum_0^{\infty}a_n x^n=\sum_0^{\infty}\frac{x^n}{n!}[/tex]
    [tex]2a_2-2a_0-1+\sum_1^{\infty}\left [(n+2)(n+1)a_{n+2}+n a_n -2a_n -\frac{1}{n!} \right ] x^n=0[/tex]
    Then: [tex]2a_2-2a_0-1=0\rightarrow a_2=1/2+a_0[/tex]
    And: [tex](n+2)(n+1)a_{n+2}+n a_n -2a_n -\frac{1}{n!}=0\rightarrow a_{n+2}=\frac{a_n(2-n)+\frac{1}{n!}}{(n+2)(n+1)}[/tex]
    Now I have to find the recurrence relation for [tex]a_n[/tex]
    Is this ok?
    Now, I've tried some terms to find the recurrence relation, but I couldn't find it yet.

    Last edited: Mar 15, 2012
  2. jcsd
  3. Mar 15, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi Telemachus! :smile:
    Multiply throughout by n! ? :wink:
  4. Mar 16, 2012 #3
    Oh, I think thats a good idea. Thank you tiny-tim.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook