Question

In Problems1-8use Theorem 7.4.1to evaluate the given Laplace transform.

$\mathbf{}\mathbf{L}\left\{{\mathrm{te}}^{-10t}\right\}$

Short answer

The Laplace transform for$\mathrm{L}\left\{{\mathrm{te}}^{-10\mathrm{t}}\right\}\mathrm{is}\frac{1}{{(\mathrm{s}+10)}^2}$

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.

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

Determine the Laplace transform

The Laplace transform of$\mathcal{L}\left\{{\mathrm{te}}^{-10\mathrm{t}}\right\}$ is

$\mathrm{L}\left\{{\mathrm{te}}^{-10\mathrm{t}}\right\}$

$\mathrm{F}\left(\mathrm{s}\right)=\frac{1}{\mathrm{s}+10}\to \frac{\mathrm{d}}{\mathrm{ds}}\left(\frac{1}{\mathrm{s}+10}\right)$

$=-\frac{1}{{(\mathrm{s}+10)}^2}$

Use the derivatives of transform theorem

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

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

${(-1)}^1\left(-\frac{1}{{(\mathrm{s}+10)}^2}\right)=\frac{1}{{(\mathrm{s}+10)}^2}$

Therefore, the laplace transform of $\mathrm{L}\left\{{\mathrm{te}}^{-10\mathrm{t}}\right\}\mathrm{is}\frac{1}{{(\mathrm{s}+10)}^2}$