What is inverse oflaplace 1/((s+1)(s-1)) ?

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

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inverse oflaplace 1/((s+1)(s-1))

inverse laplace

\frac{1}{( s + 1 ) ( s - 1 )}

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

Solution steps

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

Take the partial fraction of

\frac{1}{( s + 1 ) ( s - 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} (a r (x) cot (\frac{x}{y}))

x \to 0 lim (x^{\frac{1}{1 - x}})

n = 1 \sum \infty \frac{1}{2 + 5 ^{n}}

x \to - 2 lim (x^{4} + 5)

\frac{\partial}{\partial y} (sin (3 x - 6 y))

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