Question

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

Answer 1

Answer:

$\frac{3s^2-4}{s(s^2-4)}$

Step-by-step 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$(i)$

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

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

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

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

$\frac{3s^2-4}{s(s^2-4)}$.