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)

n = 0 \sum \infty \frac{1}{n e ^{n}}

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

n = 0 \sum \infty \frac{2}{n ^{2} + 1}

x \to 3 lim (\frac{4 x - 4}{5 x - 1})

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

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