Question

In Problems 55-58 use the given Laplace transform and the result in Problem 54 to find the indicated Laplace transform. Assume that a and k are positive constants.

57.$\mathcal{L}\mathbf{\{}\mathbf{1}\mathbf{-}\mathbf{cost}\mathbf{\}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}\left(s^2+1\right)}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathcal{L}\mathbf{\{}\mathbf{1}\mathbf{-}\mathbf{coskt}\mathbf{\}}$

Short answer

$\mathcal{L}\{1-\mathrm{coskt}\}=\frac{{\mathrm{k}}^2}{\mathrm{s}\left({\mathrm{s}}^2+{\mathrm{k}}^2\right)}$

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for $\mathbf{t}\mathbf{\ge }\mathbf{0}$. Then the integral

$\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathbf{e}}^{\mathbf{-}\mathbf{xt}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

is said to be the Laplace transform of f, provided that the integral converges.

Use Laplace Transform

$\mathrm{F}\left(\mathrm{s}\right)=\mathcal{L}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}=\mathcal{L}\{1-\mathrm{cost}\}=\frac{1}{\mathrm{s}\left({\mathrm{s}}^2+1\right)}$

By definition of the change of scale theorem.

$\mathcal{L}\left\{\mathrm{f}\right(\mathrm{at}\left)\right\}=\frac{1}{\mathrm{a}}\mathrm{F}\left(\frac{\mathrm{s}}{\mathrm{a}}\right)$

Thus,$\mathcal{L}\{1-\mathrm{coskt}\}\stackrel{\left(2\right)}{=}\frac{1}{\mathrm{k}}\mathrm{F}\left(\frac{\mathrm{s}}{\mathrm{k}}\right)$

$\stackrel{\left(1\right)}{=}\frac{1}{k}\left(\frac{1}{\frac{s}{k}\left({\left(\frac{s}{k}\right)}^2+1\right)}\right)$

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

$=\frac{1}{\mathrm{k}}\left(\frac{{\mathrm{k}}^3}{\mathrm{s}\left({\mathrm{s}}^2+{\mathrm{k}}^2\right)}\right)$

$=\frac{{\mathrm{k}}^2}{\mathrm{s}\left({\mathrm{s}}^2+{\mathrm{k}}^2\right)}$

Therefore,$\mathcal{L}\{1-\mathrm{coskt}\}=\frac{{\mathrm{k}}^2}{\mathrm{s}\left({\mathrm{s}}^2+{\mathrm{k}}^2\right)}$