What is inverse oflaplace 1-e^{-s} ?

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

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inverse oflaplace 1-e^{-s}

inverse laplace

1 - e^{- s}

δ (t) - δ (t - 1)

Solution steps

L^{- 1} {1 - e^{- 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} {1} - L^{- 1} {e^{- s}}

L^{- 1} {1} : δ (t)

L^{- 1} {e^{- s}} : δ (t - 1)

= δ (t) - δ (t - 1)

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

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

x \to 1 lim (\frac{7}{x ^{3} - 1})

4 t^{- \frac{3}{8}} + 5 ln (t) + ln (5)

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