Question

Use theorem 7.1.1 to find$L\left\{f\left(t\right)\right\},$if$f\left(t\right)=\sinh kt?$

Short answer

The Laplace transform of given function is,

$L\left\{\sinh kt\right\}=\frac{k}{s^2-k^2}$

Step-by-step solution

Step 1:

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

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

Step 2: 

Consider the function$f\left(t\right)=\sin hkt$

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

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

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

From the linearity property of the Laplace transform, we have

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

From the theorem7.1.1,

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

$$\begin{gathered} L\left\{\sinh kt\right\}=\frac{1}{2}\frac{1}{s-k}-\frac{1}{2}\frac{1}{s+k} \\ L\left\{\sinh kt\right\}=\frac{k}{s^2-k^2} \end{gathered}$$

Therefore the required Laplace transform of function is,

$L\left\{\sinh kt\right\}=\frac{k}{s^2-k^2}$