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}

\frac{d}{d x} (\frac{2}{x} + x)

\frac{\partial}{\partial v} (\frac{1}{u ^{2} + v ^{2}})

\int \frac{7 + 6 x}{1 + x ^{2}} d x

x \to 3 lim (2 x - 5)

x \to \frac{π}{x} lim (x cos (x))

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