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Thread: 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:

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

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

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

    CB
    Last edited by CaptainBlack; Apr 8th 2010 at 04:30 AM.
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  3. #3
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    $\displaystyle u'' + Bu' + u = \sin{\omega t}$.

    Complementary solution:

    $\displaystyle m^2 + Bm + 1 = 0$

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

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


    Therefore

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


    Particular solution:

    Guess

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

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

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


    Substituting into the DE...

    $\displaystyle u'' + Bu' + u$

    $\displaystyle = -\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} $

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


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

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

    Solve the equations simultaneously for $\displaystyle C_3$ and $\displaystyle C_4$.


    Then your solution is

    $\displaystyle u = u_c + u_p$

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

    You will need to substitute $\displaystyle C_3$ and $\displaystyle C_4$.
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