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Laplace of the function

  1. May 19, 2004 #1
    hi how do u find the laplace of the function below:


    x'(t) = kx(t) -hH(44-t)

    k is the rate which is >0, x(t) is the actual population, h is the harvesting rate
     
  2. jcsd
  3. May 19, 2004 #2
    Do you mean

    [tex]x'(t)=kx(t)-hH(44-t)[/tex]

    where k and h are constants? If so, what is H(t)? Is it the Heaviside unit step function?

    Edited for LaTeX errors.
     
  4. May 19, 2004 #3
    hi,

    yes H is the heaviside unit step.
     
  5. May 19, 2004 #4

    arildno

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    I suggest you use Laplace transform
     
  6. May 19, 2004 #5
    thats the method i am using, but im not sure how to apply it to this equation.
     
  7. May 19, 2004 #6
    Rewriting:

    x' - kx + hH(44 - t) = 0

    You probably already know how to find the Laplace transform of x' - kx . I'm betting that the problem is with the step function. Do you know how to transform a step function when it is written in the standard form of H(t - a) ? I think it's usually covered in an introductory chapter to Laplace transforms, along with the Dirac delta function.

    If hH(44 - t) is giving you the trouble, try rewriting it like this: hH(44 - t) = h - hH(t - 44). You should see that they both have the same shape: h when t < 44 ; 0 when t > 44 . With this modification, you have the Heaviside function in the right form to perform a Laplace transform on it.

    If you haven't got any idea what to do with any step function at all, let me know and I will try to help a little more. Also, you should include the initial condition x(0) = ? for a Laplace transform problem.
     
  8. May 19, 2004 #7
    ok, i know how to find the laplace of h(t-a) the answer is:

    e^-as/s

    ok, i understand how u got the values but what im really confused on is to apply the laplace on everything, even the first bit.

    i know what the answer should be, its:

    x(t) = (x0 - h/k)*e^kt + h/k * H(44-t) + h/k * (e^k(t-44))H(t-44)

    but i havent got a clue how to get to the answer.

    I can either prove it going forward to the laplace or going from the laplace to the actual x'(t).

    Please help
     
    Last edited: May 19, 2004
  9. May 19, 2004 #8
    initila condition is not provided, but you can assume it to be 16.
     
  10. May 19, 2004 #9
    May I ask if this question is from a certain course, or are you doing this on your own?

    The differential equation again:

    [tex]x' - kx + h - hH(t-44) = 0[/tex]

    You can perform a Laplace transform on each of the terms of the equation separately and then add them together:

    [tex]\mathcal{L}\{x' - kx + h - hH(t-44)\} = \mathcal{L}\{x'\} - \mathcal{L}\{kx\}
    + \mathcal{L}\{h\} - \mathcal{L}\{hH(t-44)\}[/tex]
    [tex]\mathcal{L}\{x'\} = sX - x(0)[/tex]
    [tex]\mathcal{L}\{kx\} = kX[/tex]
    [tex]\mathcal{L}\{h\} = h/s[/tex]
    [tex]\mathcal{L}\{hH(t-44)\} = he^{-44s}/s[/tex]

    Therefore:

    [tex]\mathcal{L}\{\mathrm{d.e.}\} = sX - x_0 - kX + h/s -he^{-44s}/s = 0[/tex]

    Does that help?

    Edited, because LaTeX is hard and whatnot.
     
    Last edited: May 19, 2004
  11. May 19, 2004 #10
    thanks alot
     
    Last edited: May 19, 2004
  12. May 19, 2004 #11
    Sorry. Reload the thread, I fixed it.
     
  13. May 19, 2004 #12
    its working now, thanks
     
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