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

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

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

inverse laplace

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

H (t) - H (t - 2)

Solution steps

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

Expand

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

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

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

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

x \to - \infty lim (1 - x^{3})

\int x ln (11 x) d x

x \to \frac{3 π}{2} lim (sin (x))

f (y) = ln (5 x^{2} - x + 1)

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