What is inverse oflaplace (-1+3s^2)/(s*(s^2-1)) ?

The answer to inverse oflaplace (-1+3s^2)/(s*(s^2-1)) is \H(t)+e^{-t}+e^t

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inverse oflaplace (-1+3s^2)/(s*(s^2-1))

inverse laplace

\frac{- 1 + 3 s ^{2}}{s \cdot ( s ^{2} - 1 )}

H (t) + e^{- t} + e^{t}

Solution steps

L^{- 1} {\frac{- 1 + 3 s ^{2}}{s ( s ^{2} - 1 )}}

Take the partial fraction of

\frac{- 1 + 3 s ^{2}}{s ( s ^{2} - 1 )} : \frac{1}{s} + \frac{1}{s + 1} + \frac{1}{s - 1}

= L^{- 1} {\frac{1}{s} + \frac{1}{s + 1} + \frac{1}{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}{s}} + L^{- 1} {\frac{1}{s + 1}} + L^{- 1} {\frac{1}{s - 1}}

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

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

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

= H (t) + e^{- t} + e^{t}

\int \frac{x ^{3}}{( 1 + 5 x ^{2} ) ^{3}} d x

\int 4 x^{2} 3 x^{3} + 2 d x

\frac{d}{d x} (\frac{15 x ^{6}}{3 x})

n = 0 \sum \infty \frac{1}{n ^{0.5}}

\int e^{5 x} cos (5 x) d x

What is inverse oflaplace (-1+3s^2)/(s*(s^2-1)) ?
The answer to inverse oflaplace (-1+3s^2)/(s*(s^2-1)) is \H(t)+e^{-t}+e^t