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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{s ( 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^{- t}

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

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

f (x) = \frac{1}{x + 2}, at x = 7

f (x) = \frac{x ^{2}}{x + 5}

\frac{d ^{4} y}{d x ^{4}} + 2 \frac{d ^{2} y}{d x ^{2}} + y = 0

\frac{d}{d x} (2 e^{x} + 4 x e^{x} + x^{2} e^{x})

\int_{4}^{9} \frac{1}{9 - x} d x

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