What is inverse oflaplace (4se^{-s})/(2s^2+32) ?

The answer to inverse oflaplace (4se^{-s})/(2s^2+32) is \H(t-1)*2cos(4(t-1))

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inverse oflaplace (4se^{-s})/(2s^2+32)

inverse laplace

\frac{4 s e ^{- s}}{2 s ^{2} + 32}

H (t - 1) \cdot 2 cos (4 (t - 1))

Solution steps

L^{- 1} {\frac{4 e ^{- s} s}{2 s ^{2} + 32}}

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{4 s}{2 s ^{2} + 32}} (t - 1)

L^{- 1} {\frac{4 s}{2 s ^{2} + 32}} : 2 cos (4 t)

= H (t - 1) \cdot 2 cos (4 (t - 1))

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

\int \frac{x + 5}{x} d x

\frac{d}{d x} (\frac{1}{3} sin (x))

f (x) = (\frac{3}{5} x)^{9}

\frac{d}{d x} (\frac{21}{64 x ^{\frac{11}{4}}})

What is inverse oflaplace (4se^{-s})/(2s^2+32) ?
The answer to inverse oflaplace (4se^{-s})/(2s^2+32) is \H(t-1)*2cos(4(t-1))