Question

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

$\mathbf{L}\left\{{\mathbf{te}}^{\mathbf{2}\mathbf{t}}\mathbf{sin}\mathbf{6}\mathbf{t}\right\}$

Short answer

The Laplace transform for$\mathrm{L}\left\{{\mathrm{te}}^{2\mathrm{t}}\sin 6\mathrm{t}\right\}$is$-\frac{24-12s}{{\left({(s-2)}^2+36\right)}^2}$

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.

$\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}{\int }_0^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$

Determine the Laplace transform

The Laplace transform of $\mathrm{L}\left\{{\mathrm{te}}^{2\mathrm{t}}\sin 6\mathrm{t}\right\}$ is

$L\left\{t{e}^{2t}\sin 6t\right\}$

$\text{F(s)=L}{\text{{e}}^{2t}sin6t\}$

$=\frac{6}{{(s-2)}^2+36}\to$$\frac{d}{ds}\left(\frac{6}{{(s-2)}^2+36}\right)$

$=\frac{24-12s}{{({(s-2)}^2+36)}^2}$

Use the derivatives of transform theorem

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

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

$(-1)\left(\frac{24-12s}{{({(s-2)}^2+36)}^2}\right)=-\frac{24-12s}{{({(s-2)}^2+36)}^2}$

The Laplace transform for$\mathrm{L}\left\{{\mathrm{te}}^{2\mathrm{t}}\sin 6\mathrm{t}\right\}$is$-\frac{24-12s}{{\left({(s-2)}^2+36\right)}^2}$