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

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

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

inverse laplace

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

f (x) = 3 ln (x) + 4 e^{x} - 5 x^{2}

\frac{d y}{d x} = y^{2} (e^{x} cos (3 x))

\int \frac{17 x ^{3}}{3 + x ^{2}} d x

\int (x - \frac{5}{x}) d x

- 4 x^{2} + 200 x + 375

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