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Mathematics 13 Online
OpenStudy (anonymous):

laplace of f(t) = e^(3t) - 1/(t)

OpenStudy (anonymous):

So I thought about splitting it up to: e^(3t)/(t) - 1/(t)

OpenStudy (anonymous):

find the laplace of ea seperately?

OpenStudy (anonymous):

i dont know how though

OpenStudy (anonymous):

\[\int\limits_{0}^{\infty}f(x)e^{-sx}dx\]

hartnn (hartnn):

do u know this property ? \(\huge L[\frac{f(t)}{t}]= \int \limits_s^\infty F(s)ds\)

hartnn (hartnn):

here f(t) = e^(3t) -1

OpenStudy (anonymous):

okay but what is F(s)?

hartnn (hartnn):

F(s) is the L.T of f(t)

OpenStudy (anonymous):

okay sweet so i just take the integral and bam

hartnn (hartnn):

yes, first find Laplace of f(t) = e^(3t) -1 and then take its integral from s to infinity

OpenStudy (anonymous):

right

OpenStudy (anonymous):

thats easy enough :) thanks :)

hartnn (hartnn):

should i wait till you get the correct answer ?

OpenStudy (anonymous):

wait whats the integral of 1/(s - 3) haha

hartnn (hartnn):

ok, integral of 1/x dx is .... ?

OpenStudy (anonymous):

ln(s - 3) ?

hartnn (hartnn):

oh, yes :)

OpenStudy (anonymous):

oh, duh (i need to stop second guessing my instincts) haha and so integral of 1/s is ln(s) too! :)

hartnn (hartnn):

yes, avsolutely.

hartnn (hartnn):

so what u get finally ?

OpenStudy (abb0t):

isn't the answer just \[\frac{ 1 }{ s-3 }+1\]

OpenStudy (anonymous):

no i got ln(s) - ln(s - 3) , s > 3

OpenStudy (anonymous):

same as in the back of the book so i cant argue with that :)

hartnn (hartnn):

yeah, correct, but that can be simplified to single log.

OpenStudy (anonymous):

ln((s)/(s-3))

hartnn (hartnn):

yes. :)

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