What is inverse oflaplace e^{-2s}*1/s ?

The answer to inverse oflaplace e^{-2s}*1/s is \H(t-2)

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inverse oflaplace e^{-2s}*1/s

inverse laplace

e^{- 2 s} \cdot \frac{1}{s}

H (t - 2)

Solution steps

L^{- 1} {e^{- 2 s} \frac{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 - 2) L^{- 1} {\frac{1}{s}} (t - 2)

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

= H (t - 2) H (t - 2)

= H (t - 2)

\int \frac{x ^{10} + 1 0 ^{x}}{10} d x

\frac{d}{d x} (\frac{5 x}{x ^{2} + 7})

x \to 0 lim (\frac{x}{2 x})

\int_{0}^{\frac{π}{6}} 1 - cos (6 x) d x

d er i v a t i v e y = 3 1 + 5 x

What is inverse oflaplace e^{-2s}*1/s ?
The answer to inverse oflaplace e^{-2s}*1/s is \H(t-2)