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

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

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

inverse laplace

\frac{e ^{- s}}{s - 8}

H (t - 1) e^{8 (t - 1)}

Solution steps

L^{- 1} {\frac{e ^{- s}}{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 - 1) L^{- 1} {\frac{1}{s - 8}} (t - 1)

L^{- 1} {\frac{1}{s - 8}} : e^{8 t}

= H (t - 1) e^{8 (t - 1)}

\int 4 - 2 x d x

2 x - y \frac{d y}{d x} = 3

\int \frac{x ^{2} + 1}{x ^{4}} d x

500 - 500 t^{- 1}

2 + x, - 1

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