Question

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

28.$f\left(t\right)=t^2-e^{-9t}+5$

Short answer

The Laplace transform of given function is,

$L\left\{t^2-e^{-9t}+5\right\}=\frac{2}{s^2}+\frac{1}{s+9}+\frac{5}{s},s>0$

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$f\left(t\right)=t^2-e^{-9t}+5$

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

$$\begin{gathered} L\left\{f\left(t\right)\right\}=L\left\{t^2-e^{-9t}+5\right\} \\ L\left\{f\left(t\right)\right\}=L\left\{t^2\right\}-L\left\{e^{-9t}\right\}+5L\left\{1\right\} \end{gathered}$$

From the theorem7.1.1,

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

$$\begin{gathered} L\left\{t^2-e^{-9t}+5\right\}=\frac{2!}{s^{2+1}}-\frac{1}{s+9}+5.\frac{01}{s^{0+1}} \\ L\left\{t^2-e^{-9t}+5\right\}=\frac{2}{s^2}+\frac{1}{s+9}+\frac{5}{s},s>0 \end{gathered}$$

Therefore the required Laplace transform of function is,

$L\left\{t^2-e^{-9t}+5\right\}=\frac{2}{s^2}+\frac{1}{s+9}+\frac{5}{s},s>0$