What is inverse oflaplace (4s)/(s^2-1) ?

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

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inverse oflaplace (4s)/(s^2-1)

inverse laplace

\frac{4 s}{s ^{2} - 1}

2 e^{- t} + 2 e^{t}

Solution steps

L^{- 1} {\frac{4 s}{s ^{2} - 1}}

Take the partial fraction of

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

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

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

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

= 2 e^{- t} + 2 e^{t}

\int 10 x 5 x^{2} - 4 d x

\int_{0}^{x^{3}} sin (t) d t

x \to - 5 + lim (ln (x + 5))

\frac{s + 5}{( s + 3 ) ( s - 1 )}

x \to 9 lim (\frac{- 1}{x - 9})

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