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

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

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

inverse laplace

\frac{e ^{- s}}{s ( s + 2 )}

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

Solution steps

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

L^{- 1} {\frac{1}{s ( s + 2 )}} : \frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

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

H (t - 1) = {01 t < 1 t \geq 1

= {0 \frac{1}{2} - \frac{1}{2} e^{- 2 (t - 1)} t < 1 t \geq 1

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

x \to 0 lim (x^{2} cos (\frac{9}{x}))

\int 14 - 20 - 3 16 - 2 x d x

\int \frac{2 x}{9 x ^{2} - 12 x + 8} d x

y = x + x, (9, 12)

\frac{d}{d x} (\frac{x ^{3}}{12} + \frac{2}{3} x^{- 3})

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