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2nd-Order Non-homogeneous Differential Equations

  1. Feb 18, 2010 #1
    Hello,
    I would like some help solving the following differential equation:
    [tex] \frac{d^2 y(t)}{dt^2} + \lambda ^2 y(t) = F(t)[/tex]

    In my document, it is solved in this form, but I do not understand how or why:
    [tex] y(t) = A cos(\lambda t) + B sin(\lambda t) + \int_0^t F(t) sin\lambda (t-\tau ) \,d\tau. [/tex]


    I can understand how to solve for the first two terms (A and B), using the initial conditions. But where in the world is the third term from? In fact in the text they write,

    [tex] \frac{F(t) \ast sin\lambda }{\lambda} [/tex] where the asterisk, I suppose, means the integration??
     
    Last edited: Feb 18, 2010
  2. jcsd
  3. Feb 18, 2010 #2

    LCKurtz

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    I think the * is the notation for the convolution of F and the sine function, which is what that integral is. If you solve the equation by LaPlace transforms you will probably see where it comes from.
     
  4. Feb 18, 2010 #3
    I see. So now I'm trying to find the inverse transform of the following:

    L{F(t)} / (s^2 + d^2)

    where d is a constant.

    How do I go about solving this and obtaining the convolution solution?
     
    Last edited: Feb 18, 2010
  5. Feb 18, 2010 #4

    LCKurtz

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    Just call if f(s). Remember that when you take inverses, if you have a product like f(s)φ(s), the inverse will be a convolution of the inverse of f(s), which is F(t) and the inverse of φ(s), whatever that is in your equation.

    I haven't actually worked it but, hey, it is my best guess and it's your problem :wink:
     
  6. Feb 18, 2010 #5
    Oh I see! Thank you very much.

    And I suppose the mathematical expansion of "convolution" is the integral mentioned in the first post.


    Thank you!
     
  7. Feb 18, 2010 #6

    LCKurtz

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    That's what I would expect. Good luck, sack time here.
     
  8. Feb 18, 2010 #7

    vela

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    You might also want to look up Green's functions.

    Your convolution integral is a little bit off. The convolution of two functions f and g is

    [tex] f*g=\int_{-\infty}^\infty f(\tau)g(\tau-t) d\tau[/tex]

    The limits in your integral are probably due to boundary conditions, but the arguments of the functions in the integrand are definitely wrong.
     
  9. Feb 18, 2010 #8
    Yes just caught that. Thanks also.

    Will Green's functions offer a better alternative to this solution method?
     
  10. Feb 19, 2010 #9

    vela

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    This method is the Green's function method.

    Conceptually, the Green's function is the response of the system to a unit impulse. The forcing function F(t) can be thought of as a train of scaled impulses, each resulting in a response. The convolution integral sums all of the responses. By the principle of superposition, the sum is the response of the system to F(t).
     
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