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

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

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

inverse laplace

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

H (t - 2) - 2 H (t)

Solution steps

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

Expand

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

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

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

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

= H (t - 2) - 2 H (t)

\int x o t x^{2} d x

\frac{\partial}{\partial x} (1 + e^{x})

\frac{d}{d x} (\frac{1}{x \cdot ln ( x ^{9} )})

\frac{d}{d x} (\frac{2}{3 x ^{3}})

\int \frac{8}{x ^{7}} d x

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