What is inverse oflaplace 3/s (1-e^{-s}) ?

The answer to inverse oflaplace 3/s (1-e^{-s}) is 3\H(t)-3\H(t-1)

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inverse oflaplace 3/s (1-e^{-s})

inverse laplace

\frac{3}{s} (1 - e^{- s})

3 H (t) - 3 H (t - 1)

Solution steps

L^{- 1} {\frac{3}{s} (- e^{- s} + 1)}

Expand

\frac{3}{s} (1 - e^{- s}) : \frac{3}{s} - \frac{3 e ^{- s}}{s}

= L^{- 1} {\frac{3}{s} - \frac{3 e ^{- 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{3}{s}} - L^{- 1} {\frac{3 e ^{- s}}{s}}

L^{- 1} {\frac{3}{s}} : 3 H (t)

L^{- 1} {\frac{3 e ^{- s}}{s}} : 3 H (t - 1)

= 3 H (t) - 3 H (t - 1)

\int e^{4} d x

\int_{1}^{2} (3 - e^{x}) d x

y^{^{''}} + 4 y = e^{t} (cos (2 t) + sin (2 t))

f (x) = - 7 x^{2}

x \to 0 lim ((ln (\frac{1}{x}))^{x})

What is inverse oflaplace 3/s (1-e^{-s}) ?
The answer to inverse oflaplace 3/s (1-e^{-s}) is 3\H(t)-3\H(t-1)