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OpenStudy (anonymous):
find the laplace transform of tcos(2t)
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hartnn (hartnn):
do u know the L.T of cos 2t ?
OpenStudy (anonymous):
yes so its s/(s^2 + 4)
OpenStudy (anonymous):
so do i have to take the integral of the function first? or...
hartnn (hartnn):
now u need to use
\(\large L[t^nf(t)]=(-1)^n\frac{d^n}{ds^n}[F(s)]\)
did u know this ?
OpenStudy (anonymous):
no i dont take the integral theres got to be an easier way
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OpenStudy (anonymous):
okay but then what is F(s) in this case?
hartnn (hartnn):
F(s) is the Laplace transfom of f(t)
in this case, f(t) is cos 2t
and F(s) is its L.T
OpenStudy (anonymous):
so it would be -1 * d/ds * F(s)
hartnn (hartnn):
thats why i first asked whether u know L.T of cos 2t
hartnn (hartnn):
YUP
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OpenStudy (anonymous):
wait so why isnt f(t) tcos(2t) and is cos(2t) haha
hartnn (hartnn):
because t^1 is multiplied by cos 2t
so, according to
t^n f(t)
f(t) is cos 2t and n=1
OpenStudy (anonymous):
oh my gosh thats why im messing up thank you so much hartnn :)
hartnn (hartnn):
welcome ^_^
i hope you can finish now....
OpenStudy (anonymous):
got it!:) thanks again
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