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

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

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inverse oflaplace 1/(s^2-1)

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{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}

\int_{0}^{π} cos (x) sin (x) d x

\int_{0}^{\frac{π}{6}} 64 sin^{2} (x) d x

\int \frac{cos ( \frac{π}{x ^{31}} )}{x ^{32}} d x

y = \frac{e ^{x}}{1 + e ^{x}}

\frac{d}{d x} (\frac{x ^{2} + 2 x - 23}{x - 4})

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