What is inverse oflaplace 1/(s(4s^2-1)) ?

The answer to inverse oflaplace 1/(s(4s^2-1)) is -\H(t)+1/2 e^{-t/2}+1/2 e^{t/2}

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inverse oflaplace 1/(s(4s^2-1))

inverse laplace

\frac{1}{s ( 4 s ^{2} - 1 )}

- H (t) + \frac{1}{2} e^{- \frac{t}{2}} + \frac{1}{2} e^{\frac{t}{2}}

Solution steps

L^{- 1} {\frac{1}{s ( 4 s ^{2} - 1 )}}

Take the partial fraction of

\frac{1}{s ( 4 s ^{2} - 1 )} : - \frac{1}{s} + \frac{1}{2 s + 1} + \frac{1}{2 s - 1}

= L^{- 1} {- \frac{1}{s} + \frac{1}{2 s + 1} + \frac{1}{2 s - 1}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{1}{s}} + L^{- 1} {\frac{1}{2 s + 1}} + L^{- 1} {\frac{1}{2 s - 1}}

L^{- 1} {\frac{1}{s}} : H (t)

L^{- 1} {\frac{1}{2 s + 1}} : \frac{1}{2} e^{- \frac{t}{2}}

L^{- 1} {\frac{1}{2 s - 1}} : \frac{1}{2} e^{\frac{t}{2}}

= - H (t) + \frac{1}{2} e^{- \frac{t}{2}} + \frac{1}{2} e^{\frac{t}{2}}

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What is inverse oflaplace 1/(s(4s^2-1)) ?
The answer to inverse oflaplace 1/(s(4s^2-1)) is -\H(t)+1/2 e^{-t/2}+1/2 e^{t/2}