Question

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

31.$f\left(t\right)=4t^2-5\sin 3t$

Short answer

The Laplace transform of given function is,

$L\left\{4t^2-5\sin 3t\right\}=\frac{8}{s^3}-\frac{15}{s^2+9},s>0$

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.

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

$L\left\{\sin at\right\}=\frac{a}{s^2+a^2},s>0$

Solution

Consider the function$f\left(t\right)=4t^2-5\sin 3t$

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

$$\begin{gathered} L\left\{f\left(t\right)\right\}=L\left\{4t^2-5\sin 3t\right\} \\ L\left\{f\left(t\right)\right\}=4L\left\{t^2\right\}-5L\left\{\sin 3t\right\} \end{gathered}$$

From the theorem7.1.1,

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

$$\begin{gathered} L\left\{4t^2-5\sin 3t\right\}=4.\left(\frac{2!}{s^{2+1}}\right)-5.\frac{3}{s^2+3^2} \\ L\left\{4t^2-5\sin 3t\right\}=\frac{8}{s^3}-\frac{15}{s^2+9},s>0 \end{gathered}$$

Therefore the required Laplace transform of function is,

data-custom-editor="chemistry" $L\left\{4t^2-5\sin 3t\right\}=\frac{8}{s^3}-\frac{15}{s^2+9},s>0$