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)

f (t) = 2^{t^{3}}

sin (x), (\frac{π}{4})

y^{^{'''}} - 6 y^{^{''}} = 4 - cos (x)

x \to 2 + lim (\frac{x}{x ^{2} - 4})

\frac{d ^{2}}{d x ^{2}} (7 θ sin (θ))

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