Question

In Problems 3 - 24 fill in the blanks or answer true or false.

18.${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{e^{-5s}}{s^2}\right\}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Short answer

$L^{-1}\left\{\frac{e^{-5s}}{s^2}\right\}=\mathrm{U}\left(t-5\right)\left(t-5\right)$

Step-by-step solution

Consider the function

The function is ,

$L^{-1}\left\{\frac{e^{-5s}}{s^2}\right\}$

$\mathrm{If} L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right) \mathrm{then} L^{-1}\left\{e^{-as}F\left(s\right)\right\}=U\left(t-a\right)f\left(t-a\right)$65;.

Compute the Laplace inverse

Consider the function.

$L^{-1}\left\{\frac{e^{-5s}}{s^2}\right\}$

For$\frac{e^{-5s}}{s^2}:F\left(s\right)=\frac{1}{s^2}, a=5$

$L^{-1}\left\{\frac{e^{-5s}}{s^2}\right\}=\mathrm{U}\left(t-5\right)L^{-1}\left\{\frac{1}{s^2}\right\}\left(t-5\right)$

$L^{-1}\left\{\frac{1}{s^2}\right\}=t$
Therefore, the Laplace inverse of $L^{-1}\left\{\frac{e^{-5s}}{s^2}\right\}$

$L^{-1}\left\{\frac{e^{-5s}}{s^2}\right\}=\mathrm{U}\left(t-5\right)\left(t-5\right)$