- #1

- 1,753

- 143

^{-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) t

^{2}e

^{-2t}

But I don't know to get it.

Thanks!

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- Thread starter tony873004
- Start date

- #1

- 1,753

- 143

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) t

But I don't know to get it.

Thanks!

- #2

Homework Helper

Gold Member

- 5,000

- 7

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

- 1,753

- 143

[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

Homework Helper

Gold Member

- 5,000

- 7

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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