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Second Order ODE

  1. Feb 5, 2012 #1
    I'm not sure exactly how to solve this ODE. (dx^2)/(dt^2) + (w^2)x = Fsinwt, where x(0) = 0 and X'(0) = 0.
    What I've got so far is:
    x'' + w^2x = Fsinwt --> x(homogenous) = Acoswt + Bsinwt

    I know I have to find a particular solution but I keep getting zero as a result which I know won't solve the ODE.

    Also, I know that the answer is (F/2w^2)sinwt - (F/2w)tcoswt
     
  2. jcsd
  3. Feb 5, 2012 #2

    vela

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    Show us how you solved for the particular solution.
     
  4. Feb 5, 2012 #3
    For the particular solution I set x = Ccoswt + Dsinwt where C and D are arbitrary constants, so x' = -wCsinwt + wDcoswt, x'' = -w^2Ccoswt - w^2Dsinwt. so for the original equation
    x'' + w^2x = Fsinwt, I have (-w^2Ccoswt - w^2Dsinwt) + w^2(Ccoswt + Dsinwt) = Fsinwt
    but that reduces to 0 = Fsinwt which doesn't tell me anything about the particular solution because I have no values for C and D. I tried making x = tCcoswt + tDsinwt among others and even when I find some value for the particular solution, once I combine it with the homogeneous and try to solve for A and B I run into the same problem.
     
  5. Feb 5, 2012 #4

    vela

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    Oh, okay. Your problem is in your guess for the particular solution. You should note that it's the same as the homogeneous solution, so of course when you plug it into the differential equation, you get 0.

    The problem is that the forcing function looks like the homogeneous solution. That is, both contain the term ##\sin \omega t##. When this happens, you need to multiply your guess for the particular solution by t to get solutions to the differential equation independent of the homogeneous solution. So try ##x_p(t) = Ct\cos \omega t + Dt\sin \omega t##.
     
  6. Feb 5, 2012 #5
    I did that before and still did not end up with the correct answer, but I just tried it again right now and now I have the right answer, so I guess I just made a mistake before. Thank you for answering.
     
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