What is inverse oflaplace (e^{-6s})/(s^2+2s-8) ?

The answer to inverse oflaplace (e^{-6s})/(s^2+2s-8) is \H(t-6)(1/6 e^{2(t-6)}-1/6 e^{-4(t-6)})

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inverse oflaplace (e^{-6s})/(s^2+2s-8)

inverse laplace

\frac{e ^{- 6 s}}{s ^{2} + 2 s - 8}

H (t - 6) (\frac{1}{6} e^{2 (t - 6)} - \frac{1}{6} e^{- 4 (t - 6)})

Solution steps

L^{- 1} {\frac{e ^{- 6 s}}{s ^{2} + 2 s - 8}}

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 - 6) L^{- 1} {\frac{1}{s ^{2} + 2 s - 8}} (t - 6)

L^{- 1} {\frac{1}{s ^{2} + 2 s - 8}} : \frac{1}{6} e^{2 t} - \frac{1}{6} e^{- 4 t}

= H (t - 6) (\frac{1}{6} e^{2 (t - 6)} - \frac{1}{6} e^{- 4 (t - 6)})

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What is inverse oflaplace (e^{-6s})/(s^2+2s-8) ?
The answer to inverse oflaplace (e^{-6s})/(s^2+2s-8) is \H(t-6)(1/6 e^{2(t-6)}-1/6 e^{-4(t-6)})