Help with Laplace Transform of (t+2)sinh2t

[cabellos]

My understanding of the laplace trasnform isn't so great so i would appreciate some help with this question please:

find the laplace transform of (t+2)sinh2t

now i know the laplace transform of sinh2t is 2/(s^2 -4) as this is a standard rule...

looking through textbooks they show the multiplication by t^n rule is needed and i found that the laplace transform of t (sin kt) = 2ks/(s^2 + k^2) ^2

how do i apply this to my equation...

 

[arildno]

Do not double post!

 

[Karlisbad]

the Laplace transform of (t+2) is [tex] 1/s^{2}+2/s [/tex]

If you multiply f(t) by exp(-at) then there's a shift so F(s+a) and

[tex] 2sinh(ax)=e^{xa}+e^{-ax} [/tex]

then next is just hand-work...

 

[cabellos]

thanks for the tips,

can u find the LT of 2sinh2t and the tsinh2t and add them together which gives 4/(s^2 - 4) + 4s/(s^2 + 4) ^2

is this correct?

 

[HallsofIvy]

Yes. The definition of the Laplace tranform is:
[tex]L(f(t))= \int_0^\infty f(s)e^{-st}dt[/tex]
Since
[tex]\int (f(x)+ g(x))dx= \int f(x)dx+ \int g(x)dx[/tex]
It follows that you can add Laplace transforms.

It should be easy to integrate
[tex]\int_0^\infty (t+2)sinh t dt[/tex]
(Break it into two integrals and use integration by parts)
as an exercise.