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Non-homogeneous ODE via Power Series

  1. Mar 5, 2009 #1
    Hey there! I'm new here and I just want to ask anyone willing how to solve this problem using power series:

    y"+3y'+2y= sin x

    y(0)=0
    y'(0)=1

    Evaluate y(0.1)

    Thanks! :smile:
     
  2. jcsd
  3. Mar 5, 2009 #2

    HallsofIvy

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    It is impossible to explain something like that without knowing what you already understand about differential equations and power series.

    It would be better if you what you understand about this problem and where you have trouble. What have you tried?

    (I suppose it would be tacky of me to suggest that this looks like a homework problem. You wouldn't do that, would you?)
     
  4. Mar 5, 2009 #3
    You are aware this can be solved analytically, yes? Just making sure.
     
  5. Mar 5, 2009 #4
    Yes, I'm aware it could be solved analytically. My problem lies with the fact that after having changed the entire expression into power series, I'm totally blank on what to do next. The power series for the sine function is complicated relative to the one for y and its derivatives. I'm looking for some way to enable the sine function to be workable. Do you have any hints at least? Or am I looking at an inappropriate way to solve the ODE?
     
  6. Mar 5, 2009 #5

    HallsofIvy

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    You say you have "changed the entire expression into power series". Please show what you have. The power series for sin x is relatively simple and I would like to see what exactly is giving you trouble. I suspect I know what your difficulty is but I don't want to go into a long-winded explanation if I am wrong.
     
  7. Mar 5, 2009 #6
    Okay, so here's the equation:

    ∑_(i=2)^∞▒〖i(i-1) a_i x^(i-2)+3∑_(i=1)^∞▒〖ia_i x^(i-1)+2∑_(i=0)^∞▒〖a_i x^i=∑_(i=0)^∞▒(〖(-1)〗^i x^(2i+1))/(2i+1)!〗〗〗
     
  8. Mar 5, 2009 #7
    Here's another way to put it upon expansion:

    (2a_2+6a_3 x+12a_4 x^2+⋯)+(3a_1+6a_2 x+9a_3 x^2+⋯)+(2a_0+2a_1 x+2a_2 x^2+⋯)=x-x^3/3!+x^5/5!+⋯
     
  9. Mar 5, 2009 #8
    After expansion, I don't know how to factor x^3 and x^5 terms.
     
  10. Mar 5, 2009 #9
    Oh, everything's okay now, I was able to get the answer. Haha, I'm an idiot. My problem's easily remedied by just expanding the series further. Thanks anyway :smile:
     
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