Laplace transform of exp (-t^2)

[angelas]

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.

 

[HallsofIvy]

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.

 

[HoneyMoon]

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.
Thanx

 

[unplebeian]

Have you tried integration by parts on deriving the result of L{exp(-t^2)} from first principles?

 

[HoneyMoon]

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 don't know why it did not work. so if you want that just mail me. :)
hell_iut@yahoo.com

 

[unplebeian]

Hi angelas,
I have found the exact way to solve I hope it is not too late. ;)

hell_iut@yahoo.com

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.

http://en.wikipedia.org/wiki/Error_function