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

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

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

inverse laplace

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

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

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

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

L^{- 1} {\frac{1}{1 - s}} : - e^{t}

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

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

\frac{1}{x - 5}

\frac{d}{d x} ((A x^{2} + B x) \cdot sin (x))

n = 0 \sum \infty \frac{n ^{n}}{n !}

f (x) = \frac{- 2}{2 x - 14}

x \to 1 - lim (\frac{1}{ln ( x )})

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