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

The answer to inverse oflaplace 1/((s-1)(s^2-1)) is -1/4 e^t+(e^tt)/2+1/4 e^{-t}

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inverse oflaplace 1/((s-1)(s^2-1))

inverse laplace

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

- \frac{1}{4} e^{t} + \frac{e ^{t} t}{2} + \frac{1}{4} e^{- t}

Solution steps

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

Take the partial fraction of

\frac{1}{( s - 1 ) ( s ^{2} - 1 )} : - \frac{1}{4 ( s - 1 )} + \frac{1}{2 ( s - 1 ) ^{2}} + \frac{1}{4 ( s + 1 )}

= L^{- 1} {- \frac{1}{4 ( s - 1 )} + \frac{1}{2 ( s - 1 ) ^{2}} + \frac{1}{4 ( 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}{4 ( s - 1 )}} + L^{- 1} {\frac{1}{2 ( s - 1 ) ^{2}}} + L^{- 1} {\frac{1}{4 ( s + 1 )}}

L^{- 1} {\frac{1}{4 ( s - 1 )}} : \frac{1}{4} e^{t}

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

L^{- 1} {\frac{1}{4 ( s + 1 )}} : \frac{1}{4} e^{- t}

= - \frac{1}{4} e^{t} + \frac{e ^{t} t}{2} + \frac{1}{4} e^{- t}

\int x^{5} x^{3} + 5 d x

\int \frac{( x ^{3} + 64 )}{x ^{2} + 64} d x

\frac{1}{s ( s ^{2} + 34.5 s + 1000 )}

\int_{0}^{7} 1

x \to 5 lim (x^{2} - 2 x - 15)

What is inverse oflaplace 1/((s-1)(s^2-1)) ?
The answer to inverse oflaplace 1/((s-1)(s^2-1)) is -1/4 e^t+(e^tt)/2+1/4 e^{-t}