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Math Help - solving the differential equation

  1. #1
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    solving the differential equation

    tihs is going to sound like a stupid question, but how do i do it when the right side is like this?

    x''+4x=1+t+sin(t) with initial conditions x(0)=0 and x'(0)=0
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  2. #2
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    Differential equation

    Hello gambler84
    Quote Originally Posted by gambler84 View Post
    tihs is going to sound like a stupid question, but how do i do it when the right side is like this?

    x''+4x=1+t+sin(t) with initial conditions x(0)=0 and x'(0)=0
    Since you talk about the right-hand-side, I assume you know how to solve the Complementary Function to get

    x = A\cos 2t + B \sin 2t

    Next, you need to find the Particular Integral, by saying.

    Let x = P +Qt +R\cos t + S\sin t

    Then \dot{x} = Q - R\sin t + S\cos t

    \Rightarrow \ddot{x} = -R\cos t - S\sin t

    \Rightarrow -R\cos t - S\sin t + 4(P +Qt +R\cos t + S\sin t) \equiv 1 + t + \sin t

    Comparing coefficients gives

    P=Q= \tfrac{1}{4}, R = 0, S = \tfrac{1}{3}

    So the solution to the PI is

    x =\tfrac{1}{4} + \tfrac{1}{4}t +\tfrac{1}{3}\sin t

    The general solution to your equation is the sum of the solutions to the PI and the CF:

    x = A\cos 2t + B \sin 2t + \tfrac{1}{4} + \tfrac{1}{4}t +\tfrac{1}{3}\sin t

    Plug in your initial conditions and you're done.

    Grandad
    Last edited by Grandad; February 5th 2009 at 02:13 AM. Reason: Typo
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