Question

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

23.$f\left(t\right)=t^2+6t-3$

Short answer

The Laplace transform of given function is,

$L\left\{t^2+6t-3\right\}=\frac{2}{s^3}+\frac{6}{s^2}-\frac{3}{s},s>0$

Step-by-step solution

Definition

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

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

Solution

Consider the function$f\left(t\right)=t^2+6t-3$

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

$$\begin{gathered} L\left\{f\left(t\right)\right\}=L\left\{t^2\right\}+6L\left\{t\right\}-3L\left\{1\right\} \\ L\left\{f\left(t\right)\right\}=L\left\{t^2\right\}+6L\left\{t\right\}-3L\left\{t^0\right\} \end{gathered}$$

From the theorem7.1.1,

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

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

$$\begin{gathered} L\left\{t^2+6t-3\right\}=\left\{\frac{2!}{s^{2+1}}\right\}+6\left\{\frac{1!}{s^{1+1}}\right\}-3L\left\{\frac{0!}{s^{0+1}}\right\} \\ L\left\{t^2+6t-3\right\}=\frac{2}{s^3}+\frac{6}{s^2}-\frac{3}{s} \end{gathered}$$

Therefore the required Laplace transform of function

$L\left\{t^2+6t-3\right\}=\frac{2}{s^3}+\frac{6}{s^2}-\frac{3}{s},s>0$