Question

Use Theorem to find ℒ{f(t)}.
f(t) = (et − e−t)2

Answer 1

`(3s^2-4)/(s(s^2-4))`

Explanation:

Here, the question is to find the Laplace transformation of
`f(t)`

where,
`f(t)=(e^t-e^(-t))^2`

We know that for
`t>0`,
`\mathcal{L}\{f(t)\} =\int_0^(\infty) e^(-st)f(t)dt`

`\mathcal{L}\{e^(at)\}=\int_0^(\infty) e^(-st)e^(at)dt`

`\Rightarrow \mathcal{L}\{e^(at)\}=\int_0^(\infty) e^((a-s)t)dt`

`\Rightarrow \mathcal{L}\{e^(at)\}=\left[\frac { e^((a-s))}{a-s}\right]_0^(\infty)`

`\Rightarrow \mathcal{L}\{e^(at)\}=\frac { 1}{s-a}\cdots(i)`

Here, we have
`f(t)=(e^t-e^(-t))^2`

`\Rightarrow f(t)=(e^t)^2+(e^(-t))^2-2(e^t)(e^(-t))`

`\Rightarrow f(t)=e^(2t)+e^(-2t)-2e^(0t)`

So, the laplace transformation of
`f(t)` is

`\mathcal{L}\{f(t)\}=\mathcal{L}\{e^(2t)+e^(-2t)-2e^(0t)\}`

`\Rightarrow \mathcal{L}\{f(t)\}=\mathcal{L}\{e^(2t)\}+\mathcal{L}\{e^(-2t)\}+\mathcal{L}\{e^(0)\}`

Now, using equation
`(i)`

`\mathcal{L}\{f(t)\}=(1)/(s-2)+(1)/(s-(-2))+(1)/(s-0)`

`\Rightarrow \mathcal{L}\{f(t)\}=(1)/(s-2)+(1)/(s+2)+(1)/(s)`

`\Rightarrow \mathcal{L}\{f(t)\}=(3s^2-4)/(s(s^2-4))`

Hence, the Lapelace transformation of
`f(t)=(e^t-e^(-t))^2` is

`(3s^2-4)/(s(s^2-4))`.