Question

In Problems 1-8use Theorem 7.4to evaluate the given Laplace transform.

$\mathbf{L}\left\{t^2\mathrm{cost}\right\}$

Short answer

The Laplace transform for$\mathrm{L}\left\{{\mathrm{t}}^2\mathrm{cost}\right\}$is$\frac{2s^3-6s}{{\left(s^2+1\right)}^3}$

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.

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

Determine the laplace transform

The Laplace transform of $\mathrm{L}\left\{{\mathrm{t}}^2\mathrm{cost}\right\}$is$\mathrm{L}\left\{{\mathrm{t}}^2\mathrm{cost}\right\}$

$$\begin{gathered} F\left(s\right)=\frac{s}{s^2+1}\to \frac{d^2}{d{s}^2}\left(\frac{s}{s^2+1}\right) \\ =\frac{8s^3-6s^3-6s}{{\left(s^2+1\right)}^3} \end{gathered}$$

Use the derivatives of transform theorem

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

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

${(-1)}^2\left(\frac{2s^3-6s}{{(s^2+1)}^3}\right)=\frac{2s^3-6s}{{(s^2+1)}^3}$

Therefore, the laplace transform of $\mathrm{L}\left\{{\mathrm{t}}^2\mathrm{cost}\right\}$is $\frac{2s^3-6s}{{\left(s^2+1\right)}^3}$