What is inverse oflaplace (e^{-2s})/(s-9) ?

The answer to inverse oflaplace (e^{-2s})/(s-9) is \H(t-2)e^{9(t-2)}

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inverse oflaplace (e^{-2s})/(s-9)

inverse laplace

\frac{e ^{- 2 s}}{s - 9}

H (t - 2) e^{9 (t - 2)}

Solution steps

L^{- 1} {\frac{e ^{- 2 s}}{s - 9}}

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 - 9}} (t - 2)

L^{- 1} {\frac{1}{s - 9}} : e^{9 t}

= H (t - 2) e^{9 (t - 2)}

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What is inverse oflaplace (e^{-2s})/(s-9) ?
The answer to inverse oflaplace (e^{-2s})/(s-9) is \H(t-2)e^{9(t-2)}