# Curious issue with an inverse Laplace Transform...

• Jul 17th 2011, 04:00 PM
Lancet
Curious issue with an inverse Laplace Transform...
In the course of working a 2nd order ODE with initial conditions, I ended up with this part of what I needed to get the inverse Laplace transform of:

$\frac{1}{(s - 2)^2}$

Now, the textbook did this:

$t^n e^{at} = \frac{n!}{(s - a)^{n + 1}}$

$n = 1, a = 2$

$L^{-1} = te^{2t}$

But this is what I did:

$e^{at} = \frac{1}{s - a}$

$(e^{at})^2 = \left[\frac{1}{s - a}\right]^2 = \frac{1}{(s - a)^2}$

$L^{-1} = e^{4t}$

These two answers don't look equivalent, but what I did looks like it should work. So, why is it wrong?
• Jul 17th 2011, 11:14 PM
chisigma
Re: Curious issue with an inverse Laplace Transform...
Quote:

Originally Posted by Lancet
In the course of working a 2nd order ODE with initial conditions, I ended up with this part of what I needed to get the inverse Laplace transform of:

$\frac{1}{(s - 2)^2}$

Now, the textbook did this:

$t^n e^{at} = \frac{n!}{(s - a)^{n + 1}}$

$n = 1, a = 2$

$L^{-1} = te^{2t}$

But this is what I did:

$e^{at} = \frac{1}{s - a}$

$(e^{at})^2 = \left[\frac{1}{s - a}\right]^2 = \frac{1}{(s - a)^2}$

$L^{-1} = e^{4t}$

These two answers don't look equivalent, but what I did looks like it should work. So, why is it wrong?

I wonder in what textbook is written that if $\mathcal {L} \{f(t)\} = F(s)$ then $\mathcal {L} \{f^{2}(t)\} = F^{2}(s)$...

Kind regards

$\chi$ $\sigma$
• Jul 18th 2011, 03:18 AM
Lancet
Re: Curious issue with an inverse Laplace Transform...
Quote:

Originally Posted by chisigma
I wonder in what textbook is written that if $\mathcal {L} \{f(t)\} = F(s)$ then $\mathcal {L} \{f^{2}(t)\} = F^{2}(s)$...

Kind regards

$\chi$ $\sigma$

How many textbooks on higher math do you come across that are *clearly* written in the first place? Most books on higher math seem to be written in such a way that you already have to know the subject to begin with just to comprehend their explanation - which is rather pointless.

With everything that we do to manipulate the resulting fraction in order to find a set of inverse transforms, this seemed like a natural - and frankly, mild - step.
• Jul 18th 2011, 03:55 AM
chisigma
Re: Curious issue with an inverse Laplace Transform...
Quote:

Originally Posted by Lancet
... but this s is what I did:

$e^{at} = \frac{1}{s - a}$

$(e^{at})^2 = \left[\frac{1}{s - a}\right]^2 = \frac{1}{(s - a)^2}$

$L^{-1} = e^{4t}$

These two answers don't look equivalent, but what I did looks like it should work. So, why is it wrong?

May be that my observation hasn't been 'full clear'. What I understand in what You have written is that...

$\mathcal {L} \{e^{a t}\} = \frac{1}{s-a} \implies \mathcal {L} \{e^{2 a t}\} = \frac{1}{(s-a)^{2}}$

... that of course is a nonsense...

Kind regards

$\chi$ $\sigma$
• Jul 18th 2011, 04:05 AM
Lancet
Re: Curious issue with an inverse Laplace Transform...
Quote:

Originally Posted by chisigma
May be that my observation hasn't been 'full clear'. What I understand in what You have written is that...

$\mathcal {L} \{e^{a t}\} = \frac{1}{s-a} \implies \mathcal {L} \{e^{2 a t}\} = \frac{1}{(s-a)^{2}}$

... that of course is a nonsense...

Kind regards

$\chi$ $\sigma$

No, you made it clear the first time that it couldn't be done. The problem was that I didn't know it couldn't be done, it was never explicitly mentioned in regards to this type of situation, and it *appears* that we do a lot worse when we manipulate the fractions the way we normally do, so I never knew this was an issue.

Thank you for helping to clarify.