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

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

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inverse oflaplace s/(s^2-1)

inverse laplace

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

\frac{1}{2} e^{- t} + \frac{1}{2} e^{t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\frac{d}{d x} (- ln (1 - \frac{1}{1 + e ^{- x}}))

\frac{d ^{2}}{d x ^{2}} (e^{8 x^{2}})

n = 1 \sum \infty 4^{- 2 n + 1}

\frac{d}{d x} (\frac{1}{4 - x ^{2}})

x \to 1 lim ((1 - x^{2})^{1 - x})

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