Mathematics
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OpenStudy (anonymous):
laplace of f(t) = e^(3t) - 1/(t)
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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 ?
\(\huge L[\frac{f(t)}{t}]= \int \limits_s^\infty F(s)ds\)
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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
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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 .... ?
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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 ?
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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))
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hartnn (hartnn):
yes. :)