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:

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

Now, the textbook did this:

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

$\displaystyle n = 1, a = 2$

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

But this is what I did:

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

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

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

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

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:

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

Now, the textbook did this:

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

$\displaystyle n = 1, a = 2$

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

But this is what I did:

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

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

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

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 $\displaystyle \mathcal {L} \{f(t)\} = F(s)$ then $\displaystyle \mathcal {L} \{f^{2}(t)\} = F^{2}(s)$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Curious issue with an inverse Laplace Transform...

Quote:

Originally Posted by

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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.

Re: Curious issue with an inverse Laplace Transform...

Quote:

Originally Posted by

**Lancet** ... but this s is what I did:

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

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

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

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...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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.