Question

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

36.$f\left(t\right)=e^{-t}\cosh t$

Short answer

The Laplace transform of given function is,

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

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.

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

Solution

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

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

$f\left(t\right)$We know that $\cos ht=\frac{e^t+e^{-t}}{2}$

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

From the linearity property of the Laplace transform, we have

role="math" localid="1663917715439" $$\begin{gathered} L\left\{f\left(t\right)\right\}=L\left\{e^{-t}\cosh t\right\} \\ L\left\{f\left(t\right)\right\}=L\left\{\frac{1+e^{-2t}}{2}\right\} \\ L\left\{f\left(t\right)\right\}=\frac{1}{2}L\left\{1\right\}+\frac{1}{2}L\left\{e^{-2t}\right\} \end{gathered}$$

From the theorem7.1.1,

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

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

Therefore the required Laplace transform of function is,

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