Question

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

$\mathbf{L}\mathbf{\left\{}\mathbf{tsinh}\mathbf{3}\mathbf{t}\mathbf{\right\}}$

Short answer

The Laplace transform for$\mathrm{L}\{\mathrm{tsinh}3\mathrm{t}\}$is$\frac{6s}{{\left(s^2-9\right)}^2}$

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.

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

Determine the laplace transform

The Laplace transform of $ℒ\left\{t\sinh 3t\right\}$is

role="math" localid="1668583917336" $ℒ\left\{t\sinh 3t\right\}$

$F\left(s\right)=\frac{3}{s^2-9}\to \frac{d}{ds}\left(\frac{3}{s^2-9}\right)$

$=-\frac{6s}{{(s^2-9)}^2}$

Use the derivatives of transform theorem

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

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

$(-1)\left(-\frac{6\mathrm{s}}{{({\mathrm{s}}^2-9)}^2}\right)=\frac{6\mathrm{s}}{{({\mathrm{s}}^2-9)}^2}$

Therefore, the laplace transform of$ℒ\left\{t\sinh 3t\right\}$ is$\frac{6s}{{(s^2-9)}^2}$ .