Question

How do I get the inverse laplace transform of the following? I am completely lost...

6/(s+1)^3

Answer 1

I'm tagging you again :-P @zepdrix

Answer 2

@amistre64

Answer 3

hmm, if i could recall the tables, it would help

Answer 4

I have a table of transforms, but it does not have any helpful info. Well, not to me, anyway

Answer 5

\[\frac{t^k}{k!}\to\frac{1}{s^{k+1}}\]

Answer 6

in this case, you might wasnt to say u = (s+1)

i dont think a partial fraction decomp would be useful ... but its been awhile

Answer 7

Decomp didn't help.. I was stuck with a ()^2 and ()^3 again.
Brb, going to try substitution

Answer 8

recall the 6 = 3!

Answer 9

#4 on the table seems to address it nicely

Answer 10

no partial,
that table doesn't have the required theorem

Answer 11

hard to keep a focus when you have to flip between screens ....

Answer 12

Okay I tried, but then we would totally ignore whatever is in the (), since the inverste transforms has no s's in. Or where are we going to substitute back again?

Answer 13

\[\frac{6}{(s+1)^3}=3\frac{2!}{(s+1)^3}\]

Answer 14

should I then just write (t+1) instead of t?

Answer 15

consider: the laplace of say

i believe so

Answer 16

So, 3(t+1)^3?

Answer 17

thats looking better yes; try ^2 tho

Answer 18

http://www.wolframalpha.com/input/?i=inverse+laplace+(6%2F(s%2B1)%5E3)

Answer 19

use this :
\(\huge L^{-1}F(s-a)=e^{at}f(t)\)

Answer 20

yeah, theres in e in there

Answer 21

so, from your Q,
\(\huge e^{-t}[L^{-1}\dfrac{6}{s^3}]\)
now use the table.

Answer 22

So it's\[e^{-t}3t^2\]?

Answer 23

yes.

Answer 24

Thanks!
I'll do some more problems like these, I'm really struggling

Answer 25

practice makes the man perfect! :)

Answer 26

Thanks a lot Goku!

Answer 27

welcome ^_^