Question

Laplace transform for f(t)=tcosh(3t)?

Answer 1

I'm actually not sure how to do this.

Answer 2

\[
\mathcal L\{tf(t)\} = - \frac {d}{ds}\mathcal L\{f(t)\}
\]

Answer 3

\[
\mathcal L\{f(at)\} = \frac 1{|a|} \mathcal L\{f(t)\} \left(\frac sa\right)
\]

Answer 4

Laplace of f(t) deals with the del operator. Essentially, it is taking the partial derivative of the function, twice. Unfortunately, the equation editor does not have that capability. however, I'll try to be as clear as possible. Here goes: \[del^2 (t \cosh(3t))\] is the function that we need to differentiate twice using partial derivatives. Using the product rule we get \[= 3t \sinh(3t) + \cosh (3t)\] for the first derivative. now, we need to differentiate it again to get \[= 3(3t \cosh(3t) + \sinh(3t)) + 3\sinh(3t)\] simplifying further we get \[= 9t \cosh(3t) + 3\sinh(3t) + 3 \sinh(3t)\] further simplification gives us the final answer \[= 9t \cosh(3t) + 6\cosh(3t)\]

Answer 5

let me know if that is what you are looking for

Answer 6

I made a mistake on my part...I gave you the Laplacian vice the transform

Answer 7

The transform looks something like this \[F(s) = \int\limits_{-\infty}^{\infty} e ^{-st} t \cosh(3t) dt\] it's gonna be messy cause you'll be doing integration by parts and simplifying them then applying the limits. However, if you use your table of Lapalce transforms you would get the following \[ = (s^2 + 9)/(s^2 - 9)^2\]