Question

Use theorem 7.1.1 to find$\mathbf{L}\left\{\mathbf{f}\left(\mathbf{t}\right)\right\}\mathbf{,}$if$\mathbf{f}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}{\mathbf{t}}^{\mathbf{4}}\mathbf{?}$$\mathbf{f}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}{\mathbf{t}}^{\mathbf{4}}\mathbf{?}$

Short answer

The Laplace transform of given function is$L\left\{2t^4\right\}=\left\{\frac{48}{s^5}\right\}$.

Step-by-step solution

Determine the formulas:

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

Consider the expression for the general form as:

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

Determine the comparison to obtain the Laplace transform

Consider the function$f\left(t\right)=2t^4$.

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

From the theorem7.1.1,

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

Since f is defined as in Theorem 7.1.1 can be formally justified for n a positive integer using integration by parts to first show that,

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

Here,$n=4$

$L\left\{2t^4\right\}=2\left\{\frac{4!}{s^{4+1}}\right\}$

Determine the Laplace transform:

Simplification,

$L\left\{2t^4\right\}=2\left\{\frac{4\times 3\times 2\times 1}{s^5}\right\}$

$L\left\{2t^4\right\}=\left\{\frac{48}{s^5}\right\}$

Therefore the required Laplace transform of function is,

$L\left\{2t^4\right\}=\left\{\frac{48}{s^5}\right\}$