Question

In Problems 1-8use Theorem7.4.1to evaluate the given Laplace transform.

$\mathbf{L}\mathbf{\left\{}\mathbf{tcos}\mathbf{2}\mathbf{t}\mathbf{\right\}}$

Short answer

The Laplace transform for$\mathrm{L}\left\{\mathrm{tcos}2\mathrm{t}\right\}$is$\frac{s^2-4}{{\left(s^2+4\right)}^2}$

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.

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

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,

$n=1$

$f\left(t\right)=\cos 2t$

$F\left(s\right)=\frac{s}{s^2+4}$

We have,

$L\left[t\cos 2t\right]=-\frac{d}{ds}\frac{s}{s^2+4}$

Therefore, the laplace transform of $L\left\{t\cos 2t\right\}$ is $\frac{s^2-4}{{(s^2+4)}^2}$.