Inverse Laplace transform

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  • #1
tony873004
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L-1(1/(s+2)3)
I don't see this one in the table. How do I solve the inverse Laplace transform?

I know from class notes that the answer is (1/2) t2e-2t
But I don't know to get it.
Thanks!

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The Attempt at a Solution

 

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  • #2
gabbagabbahey
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I assume that [tex]\mathcal{L}^{-1}\left[ \frac{1}{s^3} \right][/tex] is in your table?

If so, just begin by applying the frequency shift rule:

[tex]\mathcal{L}^{-1}\left[ f(s+a) \right]=e^{-at}\mathcal{L}^{-1}\left[ f(s) \right][/tex]
 
  • #3
tony873004
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ok, thanks. That gives me
[tex]\frac{1}{(e^{-2t}L^{-1}(s))^3}[/tex]

And the table gives for [tex]L^{-1}\frac{1}{s^3}[/tex] as [tex]\frac{t^2}{2}[/tex]

What do I do from here?
 
  • #4
gabbagabbahey
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ok, thanks. That gives me
[tex]\frac{1}{(e^{-2t}L^{-1}(s))^3}[/tex]
How do you get that ugly expression?

You should get

[tex]\mathcal{L}^{-1}\left[ \frac{1}{(s+2)^3} \right]=e^{-2t}\mathcal{L}^{-1}\left[ \frac{1}{s^3} \right]
[/tex]
 

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