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Want to check my answer: What is the inverse laplace transform of 2/(s^2+4)?
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you want just the answer ?
is it sin(2t)/2?
so close ! i think 1/2(s^2+4) -----> sin(2t)
AH! I forgot to multiply back in the 2, right?
you have 2 in the denominator......you have to multiply by 4 ! no ?
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is it not a hyperbolic sin formula?
\[(1/2)*1/(s^2+4)------>\sin(2t)\]
i believe it is L^(-1){1/(s^2-B^2} = (sinh(Bt))/B
sorry, i made a mistake !
I looked again, maybe just a sin, not sinh
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yeah yeaaah your right ! it's sin(2t)
sorryy , for waisting your time !
awesome, thank you
not at all
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yeaah ! Good luck !
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