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\[\Large \mathcal L^{-1} \left \lbrace \frac{1}{s^3(s^2 + 1)}\right \rbrace\] i got the partial fraction to be \[-\frac 1s + \frac{1}{s^2} + \frac{s}{s^2+1}\] but i have no idea how to get the inverse of that last term
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L(cos t) = ?
you just have to recognize it
...oh yeah....
we did the proof together yesterday, now it's time to use that lol
how could i forget =_= i got so used to (s-a)^2 lol
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so it's \[-1 + t + \cos t?\]
yep
nice. thanks
http://www.wolframalpha.com/input/?i=laplace+transform+-1+%2B+t^2+%2B+cos+t
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