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Quick help with ODE

  1. May 18, 2007 #1
    Hi.

    I have this problem:
    Use the method of undetermined coefficients to find one solution of
    y'' + 2y' - 7y = [(-1t^2)+4t+7]e^(2t).
    Find a specific solution, not a general one.

    I'm having lots of trouble finding the coefficients.
    I'm assuming I need to use [(At^2)+Bt+C]*e^(2t).
    This however does not seem to work. Can anyone confirm this as the function I need to use for differentiation?

    Thank you.
     
  2. jcsd
  3. May 18, 2007 #2

    Dick

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    That's the form alright - and I have no problems getting the coefficients. So try again.
     
  4. May 18, 2007 #3
    I have [(At^2) + Bt + C]e^2t
    Y(t) = [(At^2) + Bt + C]e^2t
    Y'(t) = (2At + B)e^2t + [(At^2) + Bt + C]2e^2t
    Y''(t) = (2A)e^2t + [(At^2) + Bt + C]4e^2t

    Then [(At^2) + Bt + C]-7e^2t + (2At + B)2e^2t + [(At^2) + Bt + C]4e^2t + (2A)e^2t + [(At^2) + Bt + C]4e^2t = the right side of the ODE

    [(At^2) + Bt + C] + 2(2At + B) + 2A = (-t^2) + 4t +7

    I get that A = -1, B = 8, and C = -7. However, this is incorrect. What's wrong?

    Dick, thanks for the response by the way.
     
  5. May 18, 2007 #4

    Dick

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    Y''(t) is wrong. You didn't use the product rule enough - Y' has two products, so you should wind up with 4 terms. You are missing a (2At+B)2e^2t and a (2At+B)2e^2t.
     
    Last edited: May 18, 2007
  6. May 18, 2007 #5
    So I get,

    [(At^2) + Bt + C] + 6(At + B) + 2A = -1t^2 + 4t + 7

    With this, A = -1, B = 10, C = -51.

    My program still tells me this is wrong.
     
  7. May 18, 2007 #6

    Dick

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    It still is. Maybe you could correct your previously posted detailed solution and we could look for problems. There is nothing wrong with your approach - must just be dropping some terms.
     
  8. May 18, 2007 #7
    Are you sure it's not (At^2 + Bt + C)De^(2x)?
     
  9. May 18, 2007 #8
    Yeah I got it allowing for a D. Not quite sure if it made a difference or not or if my arithmetic just improved, but thank you very much for the help anyway.
     
  10. May 18, 2007 #9

    HallsofIvy

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    That wouldn't make much sense! It's the same as (ADt2+BDt+ CD)e2x and since A, B, C, D are all unknown constants, it is no different than (At2+ Bt+ C)e2x.
     
  11. May 18, 2007 #10

    Dick

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    D?????????????????????? I'm really curious what are the value of A,B,C and D??? turned out to be. As A=-1 was correct, D must be 1. I think the hypothesis that your arithmetic just improved is the correct one.
     
    Last edited: May 18, 2007
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