Finding the Laplace Transformation of a Piecewise Function

Northbysouth
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Homework Statement


Obtain the Laplace transformation of the function defined by

f(t) = 0 t<0

= t2e-at t>=0

Homework Equations


The Attempt at a Solution



I'm a little unsure of what I'm doing here, so bear with me.

L {t2e-at} = ∫inf0 t2e-at dt

= ∫0inf t2e-(a+s)tdt

How do I integrate this? I tried using my TI-89 but it told me it is undefined. Any help would be greatly appreciated.

EDIT: The complete solution is required, which I think means I can't just take it out of the Laplace tables
 
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Northbysouth said:

Homework Statement


Obtain the Laplace transformation of the function defined by

f(t) = 0 t<0

= t2e-at t>=0


Homework Equations





The Attempt at a Solution



I'm a little unsure of what I'm doing here, so bear with me.

L {t2e-at} = ∫inf0 t2e-at dt

= ∫0inf t2e-(a+s)tdt

How do I integrate this? I tried using my TI-89 but it told me it is undefined. Any help would be greatly appreciated.

EDIT: The complete solution is required, which I think means I can't just take it out of the Laplace tables

First of all: never, never write what you did above, which was
\int_0^{\infty} t^2 e^{-at} \, dt = \int_0^{\infty} t^2 e^{-(a+s)t} \, dt.
This is only true if s= 0.

Anyway, for your final (correct) integral, change variables from t to x = (a+s)t (assuming s > -a).

I strongly recommend that you put away the TI-89 until after you have learned how to do these problems, or at least restrict its use to numerical computation.
 
Sorry, I missed out a part.

There should have been an e-st

But I don't understand your point about changing the variable from t to x. Could you explain further please?
 
Northbysouth said:
Sorry, I missed out a part.

There should have been an e-st

But I don't understand your point about changing the variable from t to x. Could you explain further please?

I assume you have taken integration already. If so, you already studied change-of-variable methods (or so I hope).

If you have not done integration yet you would need a longer explanation than I am prepared to give here: we would need to go over material that takes several weeks to present in coursework! However, such information is widely available in books and on-line.
 
Ray Vickson said:
I assume you have taken integration already. If so, you already studied change-of-variable methods (or so I hope).

If you have not done integration yet you would need a longer explanation than I am prepared to give here: we would need to go over material that takes several weeks to present in coursework! However, such information is widely available in books and on-line.

Yes, it's just integration. But the trick you really need to handle the t^2 factor is integration by parts. By all means, review integration.
 
It's much simpler to evaluate
\tilde{g}(s)=\int_0^{\infty} \mathrm{d} t \exp[-(a+s) t]
and then taking the 2nd derivative wrt. s. It's very easy to see that
\tilde{f}(x)=\tilde{g}&#039;&#039;(s).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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