Question

laplace transform f(t) = t^2 cosh^2 4t

Answer 1

cosh^2 4t can be written as (1+cosh8x)/2
(1/2)L((1+cosh8x))
=(1/2)((1/s)+(s/s^2-8^2)

t^2 cosh^2 4t=(1/2)d^2/ds^2)((1/s)+(s/s^2-8^2)
Take two times derivative to get answer

Source: http://www.mathskey.com/question2answer/ask

Answer 2

Another method: Use the definition of \(\cosh t\) into its exponentials:

\[\large \begin{align} \mathcal{L}\left\{ t^2 \cosh^2(4t)\right\}&=\mathcal{L}\left\{t^2\left( \frac{e^{4t}+e^{-4t}}{2}\right)^2\right\} \\&=\frac{1}{4}\left[\mathcal{L}\left\{ t^2(e^{8t}+e^{-8t}+2)\right\} \right]\\&=\frac{1}{4}\left[ \mathcal{L}\left\{t^2 e^{8t} \right\}+\mathcal{L}\left\{ t^2e^{-8t}\right\}+2\mathcal{L}\left\{t^2 \right\}\right] \\ =& \frac{1}{4}\left[\frac{2}{(s-8)^3} +\frac{2}{(s+8)^3}+\frac{4}{s^3}\right]\end{align}\] using # 23 and #3 from: http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx#LT_30

Answer 3

the answer of the textbook is \[-2\left[ \frac{ S^{3}-12S }{ (S^{2}+4)^{3} } \right]\], not sure how this answer is derived