Question

In Problems 1-8use Theorem 7.4.1to evaluate the given Laplace transform.

$\mathbf{L}\left\{t^3{e}^t\right\}$

Short answer

$\mathrm{The}\mathrm{Laplace}\mathrm{transform}\mathrm{for}\mathrm{L}\left\{{\mathrm{t}}^3{\mathrm{e}}^{\mathrm{t}}\right\}\mathrm{is}\frac{6}{{(\mathrm{s}-1)}^4}$

Step-by-step solution

Determine the derivatives of transforms

The Laplace transform converts a linear differential equation to an algebraic equation, which may subsequently be solved using algebraic principles.

The inverse Laplace transform can then be used to solve the original differential equation.

_{$\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}{\mathbf{t}}^{\mathbf{\text{'}}}$}

Determine the laplace transform

The Laplace transform of $\mathrm{L}\left\{{\mathrm{t}}^3{\mathrm{e}}^{\mathrm{t}}\right\}$ is$\mathrm{L}\left\{{\mathrm{t}}^3{\mathrm{e}}^{\mathrm{t}}\right\}$

$\mathrm{F}\left(\mathrm{s}\right)=\frac{1}{\mathrm{s}-1}\to \frac{{\mathrm{d}}^3}{{\mathrm{ds}}^3}\left(\frac{1}{\mathrm{s}-1}\right)$

$=-\frac{6}{{(s-1)}^4}$

Use the derivatives of transform theorem

If $\mathrm{F}(\mathrm{s})=\mathcal{L}\{\mathrm{f}(\mathrm{t})\}$and $n=1,2,3,\dots$, then

$\mathcal{L}\left\{{\mathrm{t}}^{\mathrm{n}}\mathrm{f}\left(\mathrm{t}\right)\right\}={(-1)}^{\mathrm{n}}\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{ds}}^{\mathrm{n}}}\mathrm{F}\left(\mathrm{s}\right).$

${(-1)}^3\left(-\frac{6}{{(s-1)}^4}\right)=\frac{6}{{(s-1)}^4}$

Therefore, the laplace transform of $\mathrm{L}\left\{{\mathrm{t}}^3{\mathrm{e}}^{\mathrm{t}}\right\}$is $\frac{6}{{(s-1)}^4}$.