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Laplace transform of exp (-t^2)

  1. Dec 4, 2006 #1
    Hi everyone,
    I need to solve this ODE using the laplace transform:
    y" + y - exp (-t^2) = 0 , y(0) = y'(0) = 0

    My question is on how to find the laplace transform of exp (-t^2).

    I used different properties of laplace transform to solve it but I was not succesfull. I was thinking that I can find the laplace transform of exp (-t^2) by a simple transformation of u = -t^2 and then using the derivative property of laplace transform. But it didn't work.

    I really appreciate any helps to solve this problem.
  2. jcsd
  3. Dec 4, 2006 #2


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    What reason do you have to think this has a simple Laplace transform? It will clearly involve and integration of e^(-t^2)dt and that has not simple anti-derivative.
  4. Jan 1, 2008 #3
    Hi angelas,
    I have to solve that too,
    I know the answer but I have no idea how to solve. I have found the answer via Matlab.
    the bilateral Laplace transform of exp (-t^2) is pi^1/2*exp(1/4*s^2) while u have no initial condition u can use bilateral Laplace.so u must just compute inverse Laplace transform of pi^1/2*exp(1/4*s^2)/(s^2+1).
    if anybody know the way how that integral is computed leave the answer here.
  5. Jan 2, 2008 #4
    Have you tried integration by parts on deriving the result of L{exp(-t^2)} from first principles?
  6. Jan 3, 2008 #5
    Hi angelas,
    I have found the exact way to solve I hope it is not too late. ;)
    While I am using a dial up connection it is a time consuming job to type the formulas so I tried attach the file to it but I dont know why it did not work. so if you want that just mail me. :)
  7. Jan 12, 2008 #6
    This integral is actually the error function. I remembered seeing it somewhere.
    Read the following link which says that you don't have a closed form solution for this.

    The professor must have meant (exp(t))^2 not exp(t^2). If you're taking a class solving ODEs, error functions won't really come up, unless the prof just wants to go overboard. That's why I think there is a mistake in the original post.

    Last edited: Jan 12, 2008
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