Question

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

$\mathbf{32}\mathbf{.}{\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{s^2\left(s^2+a^2\right)}\right\}\mathbf{;}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{at}\mathbf{-}\mathbf{sinat}$

Short answer

${\mathcal{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}=\frac{1}{{\mathrm{a}}^3}(\mathrm{at}-\mathrm{sinat})$

Step-by-step solution

To Find use the formula for laplace tranform

Take Laplace transform of both side.

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

$=\mathrm{a}\mathcal{L}\left\{\mathrm{t}\right\}-\mathcal{L}\left\{\mathrm{sinat}\right\}\left(\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}=\mathcal{L}\left\{\mathrm{sinkt}\right\},\frac{\mathrm{n}!}{{\mathrm{s}}^{\mathrm{n}+1}}=\mathcal{L}\left\{{\mathrm{t}}^{\mathrm{n}}\right\}\right)$

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

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

$=\frac{{\mathrm{as}}^2+{\mathrm{a}}^3-{\mathrm{as}}^2}{{\mathrm{s}}^2\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}$

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

Divide each side of the equation

Hence,

$\mathcal{L}\{\mathrm{at}-\mathrm{sinat}\}=\frac{{\mathrm{a}}^3}{{\mathrm{s}}^2\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}$

Take inverse Laplace transform of both side.

$\mathrm{at}-\mathrm{sinat}={\mathcal{L}}^{-1}\left\{\frac{{\mathrm{a}}^3}{{\mathrm{s}}^2\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}$

$\mathrm{at}-\mathrm{sinat}={\mathrm{a}}^3{\mathcal{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}$

Divide each side of the equation by${}_{\text{'}\text{'}\text{'}}{\mathrm{a}}^3\$.$

${\mathcal{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}=\frac{1}{{\mathrm{a}}^3}(\mathrm{at}-\mathrm{sinat})$

Final proof

${\mathcal{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2\left({\mathrm{s}}^2+{\mathrm{a}}^2\right)}\right\}=\frac{1}{{\mathrm{a}}^3}(\mathrm{at}-\mathrm{sinat})$