Find the inverse laplace transform of the function Y(s)=e−ss(2s−1)

Step 1
The Laplace transform is given $Y(s)=\frac{e^{-s}}{s(2s-1)}$
Convert the laplace transform into partial derivative ,
$\frac{1}{s(2s-1)}=\frac{A}{s}+\frac{B}{2s-1}$
$1=A(2s-1)+B(s)$
$1=2As-A+Bs$
$1=s(2A+B)-A$
Compare the coefficient of terms,
$A=-1$
$2A+B=0\dots 2$
Step 2 Substitute value of A in equation 2,
$2(-1)+B=0$
$B=2$
The value of A is -1 and B is 2.
The expression is written as,
$\frac{1}{s(2s-1)}=\frac{-1}{s}+\frac{2}{2s-1}$
$=\frac{-1}{s}+\frac{1}{(s-\frac{1}{2}}$
For finding the inverse Laplace transform , use theorem below.
$L^{-1}[e^{-sT}F(s)]=f(t-T)u(t-T)$
Step 3
Applying the theorem , the given Laplace transform is written as,
$L^{-1}[e^{-s}(\frac{-1}{s}+\frac{1}{(s-\frac{1}{2}}]=(e^{(\frac{t}{2}-\frac{1}{2})}-1)u(t-1)$
The inverse Laplace transform of the function $Y(s)=\frac{e^{-s}}{s(2s-1)}$ is , $(e^{(\frac{t}{2}-\frac{1}{2})}-1)u(t-1)$