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Laplace transform question

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Using Laplace transform to solve 4th order DE with a delta dirac forcing function. Has a tricky denominator, I just need a clue.

    [tex] y^{(4)} - y = \delta (t-2)[/tex]
    __
    IV's y'''(0)=0 , y''(0)=0 , y'(0)=0 , y(0) = 0

    2. Relevant equations

    I am asked to convert the solution to an easily invertable form, not using the integral definition to invert.

    3. The attempt at a solution

    Have the solution as --> [tex] Y(s) = \frac{e^{-2s}}{(s^4-1)} [/tex]
    looking for y(t)

    My guess is that I will use the second shift theorem to invert but I'm not sure how to break up the denominator so that it will work. Can anyone give me a clue to my next step?
     
  2. jcsd
  3. Oct 28, 2009 #2
    The inverse of 1/(s^4-1)=(1/2)sinht-(1/2)sint

    You can find it in the table of laplace transforms. If you however want to know how to find it, there are a few methods. It is good to familiarise yourself with more than one. Then if you get stuck on a test, i find it good to be able to double check my answers.

    1) use the inversion theorm, if you dont know it, basically you find all the singularities of the function, and close them in a closed contour(unless they are branch points, then you cannot enclose them, in your example they are simple poles)

    Thus 1/(s^4-1)=1/(s+1)(s-1)(s+i)(s-i)

    Evaluating all four simple poles we have:

    We multiply by e^st to "undo the laplace transform.
    then use l'hospitals rule to make it easier:

    Lim (as s approches 1) of (s-1)*f(s)e^st=e^st/4s^2=e^t/4

    similarly for -1 you get -e^-t/4

    for i you get -e^it/4i

    and for s=-i you get e^-it/4i

    then using the fact that (e^t-e^-t)/2 =sinht and -(e^it-e^-it)/2=sint

    we get f(t)=(1/2)sinht-(1/2)sint

    for e^-2s you know that this is the dirac delta function, so know you can apply the convolution theorm which is used when you have a "product " of function in s.

    so the inverse of f(s)*g(s)=f(u)*g(t-u)=∫δ(x-2)(1/2)sinh(t-u)-(1/2)sin(t-u)du

    but the dirac delta function has the property that it "picks out" the function at the value of 2, but this is a trickier example, this bit i am not 100% sure of but usually what it would give us is f(2), but the antiderivative of the dirac function is the unit step function:

    ∫δ(t-2){(1/2)sinh(t-u)-(1/2)sin(t-2)}du=u(t-2)(1/2)sinh(t-2)-u(t-2)(1/2)sin(t-2)=f(t)
     
  4. Oct 28, 2009 #3

    HallsofIvy

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    Decomposition of the denominator should be fairly straight forwared. By the "difference of squares" formula, [itex]s^4- 1= (s^2)^2- (1^2)^2 =(s^2- 1)(s^2+ 1)[/itex] and appyling "difference of squares to the first of those factors, the result is (s- 1)(s+ 1)(s^2- 1)[/itex]. [itex]s^2+ 1= 0[/itex] has only imaginary roots so this cannot be factored further in terms of real numbers.

    Use "partial fractions" to write this as
    [tex]\frac{e^{-2s}}{s^2- 4}= \frac{Ae^{-2s}}{s- 1}+ \frac{Be^{-2s}}{s+1}+ \frac{(Cxs D)e^{-2s}}{s^2+ 1}[/tex]


    Those are pretty standard formulas.
     
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