Laplace transform of exp (-t^2)

In summary, the conversation revolves around finding the Laplace transform of exp (-t^2) and using it to solve the given ODE. There are different proposed methods, but ultimately it is determined that there is no simple closed form solution for this integral.
  • #1
angelas
8
0
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.
 
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  • #2
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.
 
  • #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.
Thanx
 
  • #4
Have you tried integration by parts on deriving the result of L{exp(-t^2)} from first principles?
 
  • #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 don't know why it did not work. so if you want that just mail me. :)
hell_iut@yahoo.com
 
  • #6
HoneyMoon said:
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
 
Last edited:

Related to Laplace transform of exp (-t^2)

What is Laplace transform?

Laplace transform is a mathematical operation used to convert a function from the time domain to the frequency domain. It is commonly used in engineering and physics to solve differential equations and analyze systems.

What is the Laplace transform of exp(-t^2)?

The Laplace transform of exp(-t^2) is √π/s, where s is the complex variable in the frequency domain.

Why is the Laplace transform of exp(-t^2) important?

The Laplace transform of exp(-t^2) is important because it is a key component in solving many differential equations that arise in various fields of science and engineering. It also has many applications in signal processing and control systems.

How is the Laplace transform of exp(-t^2) calculated?

The Laplace transform of exp(-t^2) is calculated using the definition of Laplace transform, which involves integration of the function from 0 to infinity. However, there are also tables and software programs available to calculate the transform.

What are the properties of the Laplace transform of exp(-t^2)?

Some properties of the Laplace transform of exp(-t^2) include: it is an even function, it is a decreasing function, and it has a maximum value of √π/s at s=0. These properties can be useful in solving differential equations and analyzing systems.

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