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Y''+9y = 6sin(3t); Diff EQ, weee! where did i go wrong?

  1. Feb 23, 2006 #1
    Ello ello!
    I'm looking at an example problem my professor did which is very similar to the one that i am doing, the only difference is his problem is:
    y''+y = sin(t)
    mine is:
    y''+9y = 6sin(3t);

    THe problem says:
    Find a particular solution to
    y'' + 9 y = 6sin(3 t) .

    y_{p} =
    so here is what i did:
    [​IMG]

    I basically copied what my professor did, but instead of equating sin(t) to 1, i equated it to 6.

    Any ideas on what i'm doing wrong? THanks!

    Looking at it again should i have started out with the guess of:
    y =A*t*sin(3t) + B*t*cos(3t) ?
     
    Last edited: Feb 23, 2006
  2. jcsd
  3. Feb 23, 2006 #2

    arildno

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    Dearly Missed

    You have forgotten that when differentiating the term Btcos(3t), a 3 jumps out every time you differentiate the cos-factor.
     
  4. Feb 23, 2006 #3
    Thanks,
    Okay i did it this time, but still got it wrong:
    [​IMG]
    The answwer i submitted was:
    [​IMG]
    i just let c1 and c2 = 1, they give no inital condition! so what am i suppose to do with that c1 and c2?
     
  5. Feb 23, 2006 #4

    arildno

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    You must have -tcos(3t) as your particular solution, not -tcos(t).
     
  6. Feb 23, 2006 #5
    Thanks a ton arildno, worked great!! :D
     
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