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

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

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

inverse laplace

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

H (t - 1)

Solution steps

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

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

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

= H (t - 1)

(1 + e^{x}) y \frac{d y}{d x} = e^{y}

8 x \frac{d y}{d x} = 2 x e^{x} - 8 y + 6 x^{2}

\frac{sec ( 9 x )}{1 + sec ( 9 x )}

\frac{\partial}{\partial x} (\frac{- x - 9 y}{2 y - 7 x})

\frac{d}{d x} (3 sin (x) - 4 sin^{3} (x))

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