# Inverse Laplace Transforn

• Sep 24th 2006, 02:43 AM
jmytil
Inverse Laplace Transforn
Hello!
Could anyone help me solve the problem below? I've tried it all day but I have stuck. Thanks a lot!

Problem
Find the inverse Laplace transform of {(e^(-a/s))/Sqrt[s]}, a>0, knowing that InverseLaplaceTransform{(e^(-1/s))/Sqrt[s]} = (cos(2Sqrt[t]))/Sqrt[Pi*t]
• Sep 24th 2006, 03:46 AM
CaptainBlack
Quote:

Originally Posted by jmytil
Hello!
Could anyone help me solve the problem below? I've tried it all day but I have stuck. Thanks a lot!

Problem
Find the inverse Laplace transform of {(e^(-a/s))/Sqrt[s]}, a>0, knowing that InverseLaplaceTransform{(e^(-1/s))/Sqrt[s]} = (cos(2Sqrt[t]))/Sqrt[Pi*t]

You have:

L {cos(2 sqrt(t))/sqrt(pi t)} = exp(-1/s)/sqrt(s) ... (1)

Now we also know that:

L uF(ut) = f(s/u) ... (2)

Now let u=a and use (2) in (1):

L {a cos(2 sqrt(at))/sqrt(pi at)} = exp(-a/s)/sqrt(s/a)
............................................= sqrt(a) exp(-a/s)/sqrt(s).

Hence:

L {sqrt(a) cos(2 sqrt(at))/sqrt(pi at)} = exp(-a/s)/sqrt(s).

So the inverse LT of exp(-a/s)/sqrt(s) is sqrt(a) cos(2 sqrt(at))/sqrt(pi at).

RonL
• Sep 24th 2006, 08:21 AM
jmytil
Quote:

Originally Posted by CaptainBlack
You have:

L {cos(2 sqrt(t))/sqrt(pi t)} = exp(-1/s)/sqrt(s) ... (1)

Now we also know that:

L uF(ut) = f(s/u) ... (2)

Now let u=a and use (2) in (1):

L {a cos(2 sqrt(at))/sqrt(pi at)} = exp(-a/s)/sqrt(s/a)
............................................= sqrt(a) exp(-a/s)/sqrt(s).

Hence:

L {sqrt(a) cos(2 sqrt(at))/sqrt(pi at)} = exp(-a/s)/sqrt(s).

So the inverse LT of exp(-a/s)/sqrt(s) is sqrt(a) cos(2 sqrt(at))/sqrt(pi at).

RonL

THANK YOU VERY MUCH!!!!!