Question

Please help me find the inverse Laplace transform of -2/(s+2)^2

Answer 1

To clarify, I'm trying to find the inverse Laplace transform of...

\[\frac{ -2 }{ (s+2)^{2} }\]

Answer 2

My answer is...

\[-2e ^{-4t}\]
but it is incorrect.

Answer 3

On the table of laplace transform, formula 23 tells you that inverse Laplace transform of
\(\dfrac{n!}{(s-a)^{n+1}}\) is \(t^ne^{at}\)

Your function is \(\dfrac{-2}{(s+2)^2}\)
so, your n+1 = 2 or n =1, and a =(-2)

It becomes: \(\dfrac{1!}{(s-(-2))^2}\) Hence Inverse of it will be \(t^1 e^{-2t}=te^{-2t}\)

Answer 4

Sorry, I forgot -2 on numerator, so, the answer should be \(-2te^{-2t}\)

Answer 5

My table has 7 transforms to choose from, I suppose I'll need to find another table.

Answer 6

Why can't I take the transform for 1/(s+a) and square it to get the answer? This is what I did to get my original answer.

Answer 7

@Loser66

Answer 8

Because you don't have that form.

Answer 9

Go through the definition of Laplace transform to see why you can't use it.

Answer 10

\[\frac{ 1 }{ s+a } \times \frac{ 1 }{ s+a }\]

Answer 11

You don't have it

Answer 12

How do I "not have it"?

Answer 13

What do you want me to explain? from formula or from definition or properties of Laplace ?

Answer 14

Idk, forget it

Answer 15

ok, I go from definition of Laplace
\[\mathcal{L}(1)=\int_0^{\infty}e^{-st}dt=\dfrac{1}{s}\]

Now, \[\mathcal{L}(f(t))=\int_0^\infty e^{-st}f(t)dt\]

So if
\[\mathcal{L}(te^{-2t})=\int_0^\infty e^{-st}te^{-2t}dt=\int_0^\infty te^{-t(s+2)}dt\]

let u =t, du =dt,
\(dv=e^{-t(s+2)}dt\rightarrow v=\dfrac{-e^{-t(s+2)}}{s+2}\)

so the integral is
\[t\dfrac{-e^{-t(s+2)}}{s+2}(from~0 ~to~\infty)+\int_0^{\infty} \frac {e^{-t(s+2)}}{s+2}dt\]

Answer 16

the first guy gives you 0, the second guys give you \(\dfrac{-1}{(s+2)^2}\)

Answer 17

If you apply definition for your answer, the Laplace transform is not that
Thanks for asking. I have a good chance to review.