Question

\[L^{-1} (\frac{2s^3}{s^2-81})\]


The actual problem in the problem set is
\[L^{-1} (\frac{2s^3}{s^4-81})\]

Answer 1

i like Laplace

Answer 2

\[L^{-1} (\frac{2s^3}{s^2-81})=L^{-1} (2s+\frac{81}{s-9}+\frac{81}{s+9})\]

Answer 3

I don't like Laplace /_\

Answer 4

i assume, thats correct, so whats the problem ?

Answer 5

\[L^{-1}(2s)=trouble\]
Btw.. I... suddenly... discovered that... I read the problem wrong :\

Answer 6

But how to find \(L^{-1}(2s)\)?

Answer 7

For the correct question in the problem set:
\[L^{-1} (\frac{2s^3}{s^4-81})\]\[=L^{-1} (\frac{2s^3}{s^4-81})\]\[=L^{-1} (\frac{s}{s^2+9}+\frac{1}{2(s-3)}+\frac{1}{2(s+3)})\]\[=cos(3t)+cosh(3t)\]

Answer 8

sorry, my computer restarted....
to get L^{-1} (2s), u just replace 's' by 2s ....

Answer 9

L^{-1} (s) also looks weird to me...
Usually, it's a/s^n which I can do the inverse easily (easier, I mean)..

Answer 10

ohh...i read that wrong .. u know whats L^{-1} s ?

Answer 11

its \(\delta'(t)\)
so, \(L^{-1}(2s) = 2\delta'(t)\)

Answer 12

What is \(\delta'(t)\)?

Answer 13

u know what is \(\delta(t) ?\)

Answer 14

Dirac delta function..
and ' means its derivative

Answer 15

I.. don't remember I've learnt such function...

Answer 16

really ?
http://en.wikipedia.org/wiki/Dirac_delta_function
0 everywhere except origin.....like an impulse.

Answer 17

I think I've only learnt int. from 0 to infty and from s to infty case, not the -ve infty to +ve infty...
Thanks for introducing a new friend to me though :)

Answer 18

hmm... welcome ^_^

Answer 19

I want to get back my words. I don't know the name of this new friend, but he's just an old friend of mine :'(
I'm sorry!!