Question

In Problems 19–36 use theorem 7.1.1 to find$L\left\{f\left(t\right)\right\}.$

30.$f\left(t\right)={\left(e^t-e^{-t}\right)}^2$

Short answer

The Laplace transform of given function is,

$L\left\{{\left(e^t-e^{-t}\right)}^2\right\}=\frac{1}{s-2}+\frac{1}{s+2}-\frac{2}{s},s>a$

Step-by-step solution

Theorem

Here we determine the Laplace transform of the given function by comparing it to general form as provided in theorem 7.1.1.

$L\left\{e^{at}\right\}=\frac{1}{s-a},s>a$

Solution

Consider the function$f\left(t\right)={\left(e^t-e^{-t}\right)}^2$

The objective is to find $L\left\{f\left(t\right)\right\}$using the theorem.

From the linearity property of the Laplace transform, we have

$$\begin{gathered} L\left\{f\left(t\right)\right\}=L\left\{{\left(e^t-e^{-t}\right)}^2\right\} \\ L\left\{f\left(t\right)\right\}=L\left\{e^{2t}+e^{-2t}-2\right\} \\ L\left\{f\left(t\right)\right\}=L\left\{e^{2t}\right\}+L\left\{e^{-2t}\right\}-2L\left\{1\right\} \end{gathered}$$

From the theorem7.1.1,

$L\left\{e^{at}\right\}=\frac{1}{s-a},s>a$; Where$a=0,1,2,3,............$

$$\begin{gathered} L\left\{{\left(e^t-e^{-t}\right)}^2\right\}=\frac{1}{s-2}+\frac{1}{s+2}-2.\frac{0!}{s^{0+1}} \\ L\left\{{\left(e^t-e^{-t}\right)}^2\right\}=\frac{1}{s-2}+\frac{1}{s+2}-\frac{2}{s},s>a0 \end{gathered}$$

Therefore the required Laplace transform of function is,

$L\left\{{\left(e^t-e^{-t}\right)}^2\right\}=\frac{1}{s-2}+\frac{1}{s+2}-\frac{2}{s},s>a$