Question

In Problems 31–34 find the given inverse Laplace transform by fnding the Laplace transform of the indicated function f.

$\mathbf{34}\mathbf{.}{\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{s}{\left(s^2+a^2\right)\left(s^2+b^2\right)}\right\}\mathbf{;}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cosbt}\mathbf{-}\mathbf{cosat}$

Short answer

${\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{\left({\mathrm{s}}^2+{\mathrm{b}}^2\right)\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}=\frac{1}{{\mathrm{a}}^2-{\mathrm{b}}^2}(\mathrm{cosbt}-\mathrm{cosat})$

Step-by-step solution

To Find use the formula for laplace tranform

Take Laplace transform of both side.

$\mathcal{L}\left\{f\right(t\left)\right\}$$=\mathcal{L}\{\mathrm{cosbt}-\mathrm{cosat}\}(\mathcal{L}\{\mathrm{\alpha f}\left(\mathrm{t}\right)+\mathrm{\beta g}\left(\mathrm{t}\right)\}=\mathrm{\alpha }\mathcal{L}\{\mathrm{f}\left(\mathrm{t}\right)\}+\mathrm{\beta }\mathcal{L}\{\mathrm{g}\left(\mathrm{t}\right)\left\}\right)$

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

$=\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{b}}^2}-\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{a}}^2}$

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

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

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

Divide each side of the equation

Hence,

$\mathcal{L}\{\mathrm{cosbt}-\mathrm{cosat}\}=\frac{\mathrm{s}\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}{\left({\mathrm{s}}^2+{\mathrm{b}}^2\right)\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}$

Take inverse Laplace transform of both sides.

$\mathrm{cosbt}-\mathrm{cosat}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}{\left({\mathrm{s}}^2+{\mathrm{b}}^2\right)\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}$

$\mathrm{cosbt}-\mathrm{cosat}=\left({\mathrm{a}}^2-{\mathrm{b}}^2\right){\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{\left({\mathrm{s}}^2+{\mathrm{b}}^2\right)\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}$

$\mathrm{cosbt}-\mathrm{cosat}=\left({\mathrm{a}}^2-{\mathrm{b}}^2\right){\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{\left({\mathrm{s}}^2+{\mathrm{b}}^2\right)\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}$

Divide each side of the equation by ${\mathrm{b}}_{\mathrm{m}}{\mathrm{a}}^2-{\mathrm{b}}^2$.

${\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{\left({\mathrm{s}}^2+{\mathrm{b}}^2\right)\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}=\frac{1}{{\mathrm{a}}^2-{\mathrm{b}}^2}(\mathrm{cosbt}-\mathrm{cosat})$

Final proof

${\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{\left({\mathrm{s}}^2+{\mathrm{b}}^2\right)\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}=\frac{1}{{\mathrm{a}}^2-{\mathrm{b}}^2}(\mathrm{cosbt}-\mathrm{cosat})$