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

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

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inverse oflaplace (e^{1-s})/(s^2-1)

inverse laplace

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

e H (t - 1) sinh (t - 1)

Solution steps

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

= e^{1} H (t - 1) sinh (t - 1)

Refine

e^{1} H (t - 1) sinh (t - 1) : e H (t - 1) sinh (t - 1)

= e H (t - 1) sinh (t - 1)

\frac{e ^{x^{2}} - 1}{x ^{2}}

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

y = \frac{2}{x}, (\frac{1}{16}, 8)

\frac{3 x ^{6}}{cos ( x ) e ^{x}}

x^{2}, \frac{1}{2} x^{2}, 3 x

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