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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

\int z^{2} (\frac{1}{2 π} e^{- \frac{1}{2} z^{2}}) d z

\int 3 sin^{2} (x) d x

\frac{\partial}{\partial y} (x^{y} + ln (z))

\frac{d y}{d t} - 5 y = sin (2 t)

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