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

Differential Equations
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i like Laplace
\[L^{-1} (\frac{2s^3}{s^2-81})=L^{-1} (2s+\frac{81}{s-9}+\frac{81}{s+9})\]
I don't like Laplace /_\

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Other answers:

i assume, thats correct, so whats the problem ?
\[L^{-1}(2s)=trouble\] Btw.. I... suddenly... discovered that... I read the problem wrong :\
But how to find \(L^{-1}(2s)\)?
For the correct question in the problem set: \[L^{-1} (\frac{2s^3}{s^4-81})\]\[=L^{-1} (\frac{2s^3}{s^4-81})\]\[=L^{-1} (\frac{s}{s^2+9}+\frac{1}{2(s-3)}+\frac{1}{2(s+3)})\]\[=cos(3t)+cosh(3t)\]
sorry, my computer restarted.... to get L^{-1} (2s), u just replace 's' by 2s ....
L^{-1} (s) also looks weird to me... Usually, it's a/s^n which I can do the inverse easily (easier, I mean)..
ohh...i read that wrong .. u know whats L^{-1} s ?
its \(\delta'(t)\) so, \(L^{-1}(2s) = 2\delta'(t)\)
What is \(\delta'(t)\)?
u know what is \(\delta(t) ?\)
Dirac delta function.. and ' means its derivative
I.. don't remember I've learnt such function...
really ? http://en.wikipedia.org/wiki/Dirac_delta_function 0 everywhere except origin.....like an impulse.
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 :)
hmm... welcome ^_^
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!!

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