Question

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

$\mathbf{L}\left[{\mathbf{te}}^{\mathbf{-}\mathbf{3}\mathbf{t}}\mathbf{cos}\mathbf{3}\mathbf{t}\right]$

Short answer

The Laplace transform is $\mathbf{L}\left[{\mathbf{te}}^{\mathbf{-}\mathbf{3}\mathbf{t}}\mathbf{cos}\mathbf{3}\mathbf{t}\right]$is $\frac{{(s+3)}^2-9}{{\left({(s+3)}^2+9\right)}^2}$

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

Solve this tasks using the derivatives of transform theorem

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

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

Where,

$n=1$

$f\left(t\right)=e^{-3t}\cos 3t$

$F\left(s\right)=\frac{s+3}{{(s+3)}^2+9}$

We have,

$L\left[t{e}^{-3t}\cos 3t\right]=-\frac{d}{ds}\left(\frac{s+3}{{(s+3)}^2+9}\right)$

$=-\left(\frac{{(s+3)}^2+9-2{(s+3)}^2}{{({(s+3)}^2+9)}^2}\right)$

$=\frac{{(s+3)}^2-9}{{\left({(s+3)}^2+9\right)}^2}$

Therefore, the laplace transform of $L\left[t{e}^{-3t}\cos 3t\right]$is$\frac{{(s+3)}^2-9}{{\left({(s+3)}^2+9\right)}^2}$