# Prove {F}{left({s}right)}=frac{1}{{{s}-{2}}} text{ then } f{{left({t}right)}} is a) {e}^{{{2}{t}}}{u}{left({t}right)} b) u(t+2) c) u(t-2) d) {e}^{{-{2}{t}}}{u}{left({t}right)}

Prove is
a) ${e}^{2t}u\left(t\right)$
b) $u\left(t+2\right)$
c) $u\left(t-2\right)$
d) ${e}^{-2t}u\left(t\right)$
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Derrick
Step 1
Given $F\left(s\right)=\frac{1}{s-2}$
We have to find the f(t)
Use definition of inverse Laplace transform, which is given below
$f\left(t\right)={L}^{-1}\left\{F\left(s\right)\right\}\dots \left(1\right)$
Step 2
Take inverse Laplace transform of $F\left(s\right)=\frac{1}{s-2}$
Hence, ${L}^{-1}\left\{F\right\}\left(s\right)={L}^{-1}\left\{\frac{1}{s-2}\right\}\dots \left(2\right)$
From equation (1) and equation (2)
$f\left(t\right)={L}^{-1}\left\{\frac{1}{s-2}\right\}$
$={e}^{2t}u\left(t\right)$
Therefore, $f\left(t\right)={e}^{2t}u\left(t\right)$
Hence, option (a) is correct.