What is inverse oflaplace (e^{-s})/((s^2+8)) ?

The answer to inverse oflaplace (e^{-s})/((s^2+8)) is \H(t-1) 1/(2sqrt(2))sin(2sqrt(2)(t-1))

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inverse oflaplace (e^{-s})/((s^2+8))

inverse laplace

\frac{e ^{- s}}{( s ^{2} + 8 )}

H (t - 1) \frac{1}{2 2} sin (22 (t - 1))

Solution steps

L^{- 1} {\frac{e ^{- s}}{( s ^{2} + 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 ^{2} + 8}} (t - 1)

L^{- 1} {\frac{1}{s ^{2} + 8}} : \frac{1}{2 2} sin (22 t)

= H (t - 1) \frac{1}{2 2} sin (22 (t - 1))

\frac{d}{d x} (\frac{3 x - 1}{2 x + 1})

y^{^{'}} + 4 x^{- 1} y = x^{2}, y (1) = - 2

n = 1 \sum \infty \frac{5 ^{n}}{n ^{5}}

\frac{1}{x + 2}, 0

k = 2 \sum \infty \frac{1}{k} - \frac{1}{4 ^{k}}

What is inverse oflaplace (e^{-s})/((s^2+8)) ?
The answer to inverse oflaplace (e^{-s})/((s^2+8)) is \H(t-1) 1/(2sqrt(2))sin(2sqrt(2)(t-1))