Question

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

35.$f\left(t\right)=e^t\sinh t$

Short answer

The Laplace transform of given function is,

$L\left\{e^t\sinh t\right\}=\frac{1}{s\left(s-2\right)}$

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.

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

$L\left\{f\left(t^n\right)\right\}=\frac{n!}{s^{n+1}},s>0$

Solution

Consider the function$f\left(t\right)=e^t\sin ht$

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

We know that$\sin ht=\frac{e^t-e^{-t}}{2}$$\sin ht=\frac{e^t-e^{-t}}{2}$

$f\left(t\right)$Can be written as,$f\left(t\right)=e^t\sin ht=\frac{e^{2t}-1}{2}$

From the linearity property of the Laplace transform, we have

$$\begin{gathered} L\left\{f\left(t\right)\right\}=L\left\{e^t\sinh t\right\} \\ L\left\{f\left(t\right)\right\}=L\left\{\frac{e^{2t}-1}{2}\right\} \\ L\left\{f\left(t\right)\right\}=\frac{1}{2}L\left\{e^{2t}\right\}-\frac{1}{2}L\left\{1\right\} \\ L\left\{f\left(t\right)\right\}=\frac{1}{2}L\left\{e^{2t}\right\}-\frac{1}{2}L\left\{t^0\right\} \end{gathered}$$

From the theorem7.1.1,

$L\left\{e^{at}\right\}=\frac{1}{s-a},s>a$, $L\left\{f\left(t^n\right)\right\}=\frac{n!}{s^{n+1}},s>0$Where$n,a=0,1,2,3,............$

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

Therefore the required Laplace transform of function is,

$L\left\{e^t\sinh t\right\}=\frac{1}{s\left(s-2\right)}$