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Math Help - can't work out particular integral of this second order ODE, or figure how to rewrite

  1. #1
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    Exclamation can't work out particular integral of this second order ODE, or figure how to rewrite

    a) determine a particular integral of the second order differential equation u'' + Bu' +u = sin(wt)

    in the form u(t) = Csin(wt) + Dcos(wt) where C and D are functions of w that should be found explicitly.

    b) rewrite u(t) in the form u(t) = Asin(wt + y) where A is the amplitude of the long term response and y is a phase lag between the driving force and the response. show that A = sqrt(C^2 + D^2) and hence that

    A = 1/ (sqrt(((1-w^2)^2) + (B^2)(w^2))



    Please help!! I'm clueless and desperate
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  2. #2
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    Quote Originally Posted by thatgirlrocks View Post
    a) determine a particular integral of the second order differential equation u'' + Bu' +u = sin(wt)

    in the form u(t) = Csin(wt) + Dcos(wt) where C and D are functions of w that should be found explicitly.
    A linear system forced with a sinusoidal forcing function has a steady state sinusoidal reaponse. So for the particular integral take trial solution:

    u(t)=C \sin(\omega t) + D \cos(\omega t)

    Now plug this into the differential equation and on the left hand side collect together all the terms with \sin(\omega t) and \cos(\omega t). Equate the coefficient of the first of these to 1 and of the second to 0 and you will have a pair of simultaneous linear equations for C and D in terms of \omega and B. Solve them and you are done.

    (there is an implicit assumption that \omega is not a resonance of the system, if it is a bit more case is required)

    CB
    Last edited by CaptainBlack; April 8th 2010 at 05:30 AM.
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  3. #3
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    u'' + Bu' + u = \sin{\omega t}.

    Complementary solution:

    m^2 + Bm + 1 = 0

    m = \frac{-B \pm \sqrt{B^2 - 4(1)(1)}}{2}

     = \frac{-B \pm \sqrt{B^2 - 4}}{2}.


    Therefore

    u_c = C_1 e^{\frac{-B + \sqrt{B^2 - 4}}{2}} + C_2 e^{\frac{-B - \sqrt{B^2 - 4}}{2}}.


    Particular solution:

    Guess

    u_p = C_3\sin{\omega t} + C_4\cos{\omega t}

    u_p' = \omega C_3\cos{\omega t} - \omega C_4\sin{\omega t}

    u_p'' = -\omega^2C_3\sin{\omega t} - \omega^2C_4\cos{\omega t}.


    Substituting into the DE...

    u'' + Bu' + u

    = -\omega^2 C_3\sin{\omega t} - \omega^2 C_4\cos{\omega t} + B(\omega C_3\cos{\omega t} - \omega C_4\sin{\omega t}) + C_3\sin{\omega t} + C_4\cos{\omega t}

     = (C_3 - B\omega C_4 - \omega^2C_3)\sin{\omega t} + (C_4 + B\omega C_3 -\omega^2C_4)\cos{\omega t}.


    Since this equals \sin{\omega t}, that means

    C_3 - B\omega C_4 - \omega^2C_3 = 1 and C_4 + B\omega C_3 -\omega^2C_4 = 0.

    Solve the equations simultaneously for C_3 and C_4.


    Then your solution is

    u = u_c + u_p

    u = C_1 e^{\frac{-B + \sqrt{B^2 - 4}}{2}} + C_2 e^{\frac{-B - \sqrt{B^2 -  4}}{2}} + C_3\sin{\omega t} + C_4\cos{\omega t}.

    You will need to substitute C_3 and C_4.
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