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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

0

y^{^{'}} = ry (s - y)

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

\int e^{- 6 x} sin (3 x) d x

x \to 3 lim (x^{2} + 4 x - 6)

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