What is inverse oflaplace (e^{-s})/(s(s+1)) ?

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

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inverse oflaplace (e^{-s})/(s(s+1))

inverse laplace

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

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

Solution steps

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

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 + 1 )}} (t - 1)

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

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

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

= {0 1 - e^{- t + 1} t < 1 t \geq 1

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

93.25 v^{^{'}} + v = 0.001 t, v (0) = 0

\frac{\partial}{\partial y} (x^{2} + 6 x y + 2 y^{2})

x = y, x = 2 - y

\int y^{3} e^{2 x} d y

\frac{x}{1 - ln ( x - 2 )}

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