Question

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

$\mathbf{33}\mathbf{.}{\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{\left(s^2+a^2\right)\left(s^2+b^2\right)}\right\}\mathbf{;}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{asinbt}\mathbf{-}\mathbf{bsinat}$

Short answer

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

Step-by-step solution

To Find use the formula for laplace tranform

Take Laplace transform of both side.

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

role="math" localid="1664031160015" $=\mathrm{a}\mathcal{L}\left\{\mathrm{sinbt}\right\}-\mathrm{b}\mathcal{L}\left\{\mathrm{sinat}\right\}\left(\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}=\mathcal{L}\left\{\mathrm{sinkt}\right\}\right)$

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

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

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

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

Divide each side of the equation

Hence,

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

Take inverse Laplace transform of both side.

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

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

Divide each side of the equation ${\mathrm{by}}_{\mathrm{\iota }\text{'}\text{'}}\mathrm{ab}\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)$.

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

Final proof

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