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{\partial}{\partial x} (7 + x ln (x y - 5))

\frac{d}{d y} (2 y)

\int_{0}^{x} y lo g_{10} (y) d y

n = 3 \sum \infty \frac{4 ^{n}}{7 ^{n}}

\frac{d y}{d x} = 4 x^{3} y^{2}

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