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

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

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inverse oflaplace e^{1-s}

inverse laplace

e^{1 - s}

e^{t} δ (t - 1)

Solution steps

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

Apply inverse transform rule:

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

then

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

For

e^{1 - s} : a = 1, F (s) = e^{- 1 s}

= e^{t} L^{- 1} {e^{- 1 \cdot s}}

L^{- 1} {e^{- 1 s}} : δ (t - 1)

= e^{t} δ (t - 1)

ln (arctan (x + 1))

\frac{\partial}{\partial t} (\frac{rs}{s ^{2} + t ^{2}})

\frac{d}{d x} ((8 x^{2} + 3) (3 - x^{2} + 7 x))

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

\frac{d}{d x} (\frac{1}{x ^{2} ln ( x )})

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