What is inverse oflaplace (e^{-2s})/(s(s+1.1)) ?

The answer to inverse oflaplace (e^{-2s})/(s(s+1.1)) is \H(t-2)(10/11-0.90909…e^{-1.1(t-2)})

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inverse oflaplace (e^{-2s})/(s(s+1.1))

inverse laplace

\frac{e ^{- 2 s}}{s ( s + 1.1 )}

H (t - 2) (\frac{10}{11} - 0.90909 \dots e^{- 1.1 (t - 2)})

Solution steps

L^{- 1} {\frac{e ^{- 2 s}}{s ( s + 1.1 )}}

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 - 2) L^{- 1} {\frac{1}{s ( s + 1.1 )}} (t - 2)

L^{- 1} {\frac{1}{s ( s + 1.1 )}} : \frac{10}{11} H (t) - 0.90909 \dots e^{- 1.1 t}

= H (t - 2) (\frac{10}{11} H (t - 2) - 0.90909 \dots e^{- 1.1 (t - 2)})

H (t - 2) = {01 t < 2 t \geq 2

= {0 \frac{10}{11} - 0.90909 \dots e^{- 1.1 (t - 2)} t < 2 t \geq 2

= H (t - 2) (\frac{10}{11} - 0.90909 \dots e^{- 1.1 (t - 2)})

\int_{0}^{1} x sin (π x) d x

y = \frac{x}{( e ^{x} )}

(e^{t^{2}})^{^{'}}

y = \frac{2}{x - 8}, (10, 1)

\int \frac{2 x}{5 - x ^{4}} d x

What is inverse oflaplace (e^{-2s})/(s(s+1.1)) ?
The answer to inverse oflaplace (e^{-2s})/(s(s+1.1)) is \H(t-2)(10/11-0.90909…e^{-1.1(t-2)})