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2nd order non-homogenous differentila equation

  1. Oct 7, 2008 #1
    I have to start from a simple 2nd order ordinary deifferential equation as:

    y’’+2ξωny’+ω2y = F

    The solution should be of the form

    y = ∫F(Ω) G(t - Ω) dΩ (integral from 0-t)


    where

    G(t) = 1/ω * e^(-ξωnt)sin(ξt) for ξ<1

    G(t) = e^(-ωnt) for ξ=1

    G(t) = 1/ω * e^(-ξωnt)sinh(ξt) for ξ>1

    and

    ω = ωn√|1-ξ^2|

    n in ωn is subscript.

    Any help in this regard will be highly gratified.

    Thanks in advance.
     
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2

    HallsofIvy

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    Science Advisor

    What exactly do you want to do?
     
  4. Oct 8, 2008 #3
    Re: 2nd order non-homogenous differential equation

    I have to start from a simple 2nd order ordinary deifferential equation as:

    y’’+2ξωy’+ω[tex]^{2}_{n}[/tex]y = F


    The solution should be

    y = ∫F(Ω) G(t - Ω) dΩ (integral from 0-t)


    G(t) and ω is given in my first post.

    Can u help. If u can give me ur email i will email u the problem in a better readable manner.

    Thanks a lot for your concern.
     
    Last edited: Oct 8, 2008
  5. Oct 8, 2008 #4
    I'm not sure what you are trying to do but I'm pretty sure the problem is not properly stated.
    By convolution, G(t) must be the impulse response of the second order LTI system which implies [tex]G^{''}+2 \zeta \omega _{n} G^{'} + \omega _{n}^{2} G=\delta (t)[/tex]

    The stated G(t) are wrong for *all* cases (underdamped, critically-damped, and overdamped). i.e for the underdamped case, the damped rad freq should be omega_d, not zeta; in critically-damped case, the response is not a monotone function.

    Solve the G(t) correctly first. It can be done by finding the step response first then take a time derivative of the step response.

    If F(t) is LapLaceable, using Laplace transform to find y(t) is usually easier.
     
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