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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]
    So:
    [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.
    Even:
    [tex]a_2=a_{0+2}=1/2+a_0[/tex]
    [tex]a_4=a_{2+2}=\frac{a_2(0)+1/2}{4.3}=1/4![/tex]
    [tex]a_6=a_{4+2}=\frac{a_4(-2)+1/4!}{6.5}=-\frac{1}{6!}[/tex]
    [tex]a_8=a_{6+2}=\frac{a_6(-4)+1/6!}{8.7}=\frac{5}{8!}[/tex]
    [tex]a_{10}=a_{8+2}=\frac{a_8(-6)+1/8!}{10.9}=\frac{-29}{10!}[/tex]

    Odd:
    [tex]a_3=a_{1+2}=\frac{a_1+1}{6}=\frac{a_1+1}{3!}[/tex]
    [tex]a_5=a_{3+2}=\frac{a_3(-1)+1/3!}{5.4}=-a_3+1/5!=-\frac{a_1+1}{5!}+1/5!=-\frac{a_1}{5!}[/tex]
    [tex]a_7=a_{5+2}=\frac{a_5(-2)+1/5!}{7.6}=\frac{-3a_1+1}{7!}[/tex]
    [tex]a_9=a_{7+2}=\frac{a_7(-5)+1/7!}{9.8}=\frac{15a_1-4}{9!}[/tex]
     
    Last edited: Mar 15, 2012
  2. jcsd
  3. Mar 15, 2012 #2

    tiny-tim

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    Hi Telemachus! :smile:
    Multiply throughout by n! ? :wink:
     
  4. Mar 16, 2012 #3
    Oh, I think thats a good idea. Thank you tiny-tim.
     
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