# Find the inverse laplace transform of the function Y(s)=frac{e^{-s}}{s(2s-1)}

Question
Laplace transform
Find the inverse laplace transform of the function
$$Y(s)=\frac{e^{-s}}{s(2s-1)}$$

2020-11-06
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,
$$A=-1$$
$$2A+B=0\dots2$$
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 , $$(e^{(\frac{t}{2}-\frac{1}{2})}-1)u(t-1)$$

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