Derivative of transforms

[Lorenz]

Can anyone help me do this? I can't figure it out.

Evaluate the given Laplace transform

L{t^2sinh3t}

My attempt at it:

sinhkt = k/s^2-k^2

f(t) = sinh3t
F(s) = L{f(t)}
=L{sinh3t}

= 3/s^2-3^2
= 3/s^2-9

L{t^n f(t)} = (-1)^n d^n/ds^n F(s)

so L{t^2sinh3t} = (-1)^2 d^2/ds^2 (3/s^2-9)

and I get stuck there.

but heres my attempt to continue it
(-1)^2 d^2/ds^2 (3/s^2-9) = d^2/ds^2 (3/s^2-9)
=d/ds (3/(s^2-9)^2) = 3(-1/(s^2-9)^4)(2s)
=3(-2s/(s^2-9)^4) = -6/(s^2-9)^4)

Solution: L{t^2sinh3t} = -6/(s^2-9)^4

My second attempt to continue it is pretty long but here was my solution to it

(s^2-9)^2 (-2) - (-6s) 2(s^2-9) (2s) / (s^2-9)^4
(s^2-9) [ (s^2-9) (-2) - (-6s) (2) (2s) / (s^2-9)^3 ]
..
skip a few steps..
..
and heres what I ended up with
18+22s^2/(s^2-9)^3

Please help!!!

 

[galactus]

\(\displaystyle L(t^{2}sinh(3t))=\frac{1}{(s-3)^{3}}-\frac{1}{(s+3)^{3}}\)

 

[Lorenz]

how did you get that?

 

[galactus]

By using:

\(\displaystyle \int_{0}^{\infty}e^{-st}\cdot t^{2}sinh(3t)dt\)

To integrate, try using the identity for sinh(3t)

\(\displaystyle \frac{1}{2}e^{3t}-\frac{1}{2}e^{-3t}\)

This gives:

\(\displaystyle \frac{1}{2}\int\left[t^{2}e^{-st}(e^{3t}-e^{-3t})\right]dt=\frac{-18(s^{2}+3)}{(s-3)^{3}(s+3)^{3}}\)

Now, use the limits of integration.