Question

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

27.$f\left(t\right)=1+e^{4t}$

Short answer

The Laplace transform of given function is,

$L\left\{1+e^{4t}\right\}=\frac{1}{s}+\frac{1}{s-4},s>0$

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$f\left(t\right)=\left(1+e^{4t}\right)$

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

$$\begin{gathered} L\left\{f\left(t\right)\right\}=L\left\{1+e^{4t}\right\} \\ L\left\{f\left(t\right)\right\}=L\left\{1\right\}+L\left\{e^{4t}\right\} \end{gathered}$$

From the theorem7.1.1,

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

$L\left\{1+e^{4t}\right\}=\frac{1}{s}+\frac{1}{s-4},s>0$

Therefore the required Laplace transform of function is,

$L\left\{1+e^{4t}\right\}=\frac{1}{s}+\frac{1}{s-4},s>0$