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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

L^{- 1} {\frac{1}{4 ( s - 1 ) ^{2}}} : \frac{e ^{t} t}{4}

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

\frac{d}{d x} (- ln^{4} (2 sin^{2} (x)^{3} - 1))

x y^{^{'}} = \frac{1}{x y}

\int (\frac{x ^{3}}{1 + x ^{4}}) d x

\frac{\partial}{\partial x} (e^{4 x y} ln (7 y))

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