What is inverse oflaplace (e^{-4s})/(s-2) ?

The answer to inverse oflaplace (e^{-4s})/(s-2) is \H(t-4)e^{2(t-4)}

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inverse oflaplace (e^{-4s})/(s-2)

inverse laplace

\frac{e ^{- 4 s}}{s - 2}

H (t - 4) e^{2 (t - 4)}

Solution steps

L^{- 1} {\frac{e ^{- 4 s}}{s - 2}}

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 - 4) L^{- 1} {\frac{1}{s - 2}} (t - 4)

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

= H (t - 4) e^{2 (t - 4)}

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What is inverse oflaplace (e^{-4s})/(s-2) ?
The answer to inverse oflaplace (e^{-4s})/(s-2) is \H(t-4)e^{2(t-4)}