# Inverse Laplace of the following s function: F(s) = \frac{1}{3} e^{-4s}

## Question:

Inverse Laplace of the following s function:

{eq}F(s) = \frac{1}{s^3} e^{-4s} {/eq}

## Answer and Explanation:

{eq}L^{-1}\left\{e^{-4s}\frac{1}{s^3}\right\} {/eq}

{eq}\mathrm{Apply\:inverse\:transform\:rule:\quad 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\}=H\left(t-a\right)f\left(t-a\right) \\ \mathrm{Where\:}\text{H}\left(t\right)\mathrm{\:is\:Heaviside\:step\:function} \\ \mathrm{For\:}\frac{1}{s^3}e^{-4s}:\quad F\left(s\right)=\frac{1}{s^3},\:\quad \:a=4 \\ L^{-1}\left\{\frac{1}{s^3}\right\} \\ =L^{-1}\left\{\frac{1}{2}\cdot \frac{2}{s^3}\right\} \\ \mathrm{Use\:the\:constant\:multiplication\:property\:of\:Inverse\:Laplace\:Transform:} \\ \mathrm{For\:function\:}f\left(t\right)\mathrm{\:and\:constant\:}a:\quad L^{-1}\left\{a\cdot f\left(t\right)\right\}=a\cdot L^{-1}\left\{f\left(t\right)\right\} \\ =\frac{1}{2}L^{-1}\left\{\frac{2}{s^3}\right\} \\ \mathrm{Use\:Inverse\:Laplace\:Transform\:table}:\quad \:L^{-1}\left\{\frac{\left(n\right)!}{s^{n+1}}\right\}=t^n \\ L^{-1}\left\{\frac{2}{s^3}\right\}=t^2 \\ =\frac{t^2}{2} {/eq}

So the final answer is

{eq}=\frac{\text{H}\left(t-4\right)\left(t-4\right)^2}{2} {/eq}