Find the inverse Laplace transforms of the functions given. Accurately sketch the time functions. a) F(s)=frac{3e^{-2s}}{s(s+3)} b) F(s)=frac{e^{-2s}}{s(s+1)} c) F(s)=frac{e^{-2s}-e^{-3s}}{2}

Step 1
given that a) $F(s)=\frac{3e^{-2s}}{s(s+3)}$
b) $F(s)=\frac{e^{-2s}}{s(s+1)}$
find the inverse laplace transform of the above functions
Step 2
a)first lets find the inverse laplace transform of $\frac{3}{s(s+3)}$ let it write interms of partial fractions
$\frac{3}{s(s+3)}=\frac{A}{s}+\frac{B}{(s+3)}$
$3=A(s+3)+Bs$
$\text{put }s=0\text{ then }3=3A,A=1$
$\text{put }s=-3\text{ then }3=-3B,B=-1$
lets substitute A and B values
$\frac{3}{s(s+3)}=\frac{1}{s}-\frac{1}{(s+3)}$
$L^{-1}\left[\frac{3}{s(s+3)}\right]=L^{-1}[\frac{1}{s}-\frac{1}{(s+3)}]$
$=1-e^{-3t}$
$L^{-1}\left[\frac{3e^{-2s}}{s(s+3)}\right]=(1-e^{-3(t-2)})u(t-2)$
Step 3
b) $F(s)=\frac{e^{-2s}}{s(s+1)}$
First lets find the inverse laplace transform of $\frac{1}{s(s+1)}$ for this lets write interms of partial fractions
$\frac{1}{s(s+1)}=\frac{A}{s}+\frac{B}{(s+1)}$
$1=A(s+1)+Bs$
$\text{put }s=0\text{ then }1=A,A=1$
$\text{put }s=-1\text{ then }1=-B,B=-1$
$\frac{1}{s(s+1)}=\frac{1}{s}+\frac{1}{(s+1)}$
$L^{-1}\left[\frac{1}{s(s+1)}\right]=L^{-1}[\frac{1}{s}+\frac{1}{(s+1)}]$
$=L^{-1}\left[\frac{1}{s}\right]-L^{-1}\left[\frac{1}{(s+1)}\right]$
$=1-e^{-t}$
next inverse laplace transform of $\frac{e^{-2s}}{s(s+1)}$ is $(1-e^{-(t-2)})u(t-2)$