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

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

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

inverse laplace

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

H (t - 1)

Solution steps

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

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 ^{1}}} (t - 1)

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

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

= H (t - 1)

f (x) = cos (3 x) ln (x^{4}), (1, 0)

\frac{d}{d x} (\frac{9 x ^{2}}{( x ^{2} + x ) ^{3}})

\int \frac{4 - x ^{2}}{x ^{4}} d x

ln (ln (8 x))

\frac{- x ^{3}}{y ^{3}}

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