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Math Help - Solving DE involving dirac delta function

  1. #1
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    Solving DE involving dirac delta function

    I am trying to solve find a solution to a). i can use maple to solve it, but does anyone know how. The "sum of" part is confusing me. Thanks in advance for any help.
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    Super Member Matt Westwood's Avatar
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    The integral of \delta \left({f \left({x}\right)}\right) is just f \left({x}\right) if I remember it correctly ...

    So see which instances of the delta function are included within the limits of the integral and just add up the occurrences of f(x) at those instances.
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    Quote Originally Posted by dgmossman View Post
    I am trying to solve find a solution to a). i can use maple to solve it, but does anyone know how. The "sum of" part is confusing me. Thanks in advance for any help.
    I'd suggest using solving using the method of Laplace transforms.

    @Matt: I'm not so sure about the result you state. My memory is that \delta (g(x)) = \sum_{i} \frac{\delta (x - x_i)}{g'(x_i)} where the x_i are the real roots of g(x). Therefore the integral is \sum_{i} \frac{1}{g'(x_i)}.
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    Super Member Matt Westwood's Avatar
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    Oh yeah, I knew it was something like that, I couldn't remember for certain - apologies.
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    can someone please help me with finding the laplace transform of f(t) (shown in my first post)
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  6. #6
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    Quote Originally Posted by dgmossman View Post
    can someone please help me with finding the laplace transform of f(t) (shown in my first post)
    You should be familiar with the Dirac delta function and its properties, in particular the sifting property: \int_{-\infty}^{+\infty} \delta (t - a) \, f(t) \, dt = f(a).

    You should also be familiar withy the definition of a Laplace transform: LT[f(t)] = \int_{0}^{+\infty} e^{-st} f(t) \, dt.

    Using the above you should be able to show that LT[\delta (t - a)] = e^{-as}.

    Alternatively you can always refer to a standard table of Laplace transforms.

    Therefore the Laplace transform of your function is \sum_{k=0}^{K} \left(e^{-vs/d}\right)^k.

    Use the formula for the sum of a geometric series to evaluate this sum.
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