Question

Laplace of cosh^2(2t)​

Answer 1

Answer:

Using trigonometric identities and the properties of trigonometric functions, the given expression involving the hyperbolic cosine can be written as

cosh2(3t)=12(cosh(6t)+1)(1)

Consulting or reviewing a table of Laplace transforms, it is found that for the hyperbolic cosine function we have the following solution:

Lt[cosh(at)](s)=ss2−a2

The Laplace transform of result (1) is the sum of the following Laplace transforms:

Lt[12cosh(6t)](s)=s2(s2−36)

Lt[12](s)=12s

Thus the Laplace transform of the expression in this question is given by:

Lt[cosh2(3t)](s)=Lt[12(cosh(6t)+1)](s)=12(ss2−36+1s)

And the Laplace transform is equal to

s2−18s(s2−36)

Below is a graphical representation of the solution obtained (made with Mathematica):

As a related result, for an arbitrary number h, the Laplace transform is given by (verified with Mathematica):

Lt[cosh2(ht)](s)=s2−2h2s3−4h2s

Answer 2

Given: The function$cosh^2(2t)$.

We have to find the Laplace of$cosh^2(2t)$.

For which we are using the concept of Laplace transform.

The Laplace of the given function is calculated by, $F(s)=L(f(t))=\int_{0}^{\infty} e^{-s t} f(t) d t$

We have, $cosh^2(2t)$

$=>\mathcal{L}\left(\cosh ^{2}(2 t)\right)\\=>\mathcal{L}\left(\cosh ^{2}(2 t)\right)=\frac{s^{2}-8}{s\left(s^{2}-16\right)}$

Hence, the Laplace of the given function is$\mathcal{L}\left(\cosh ^{2}(2 t)\right)=\frac{s^{2}-8}{s\left(s^{2}-16\right)}$.