What is inverse oflaplace (e^{-3s})/(s^3+4s) ?

The answer to inverse oflaplace (e^{-3s})/(s^3+4s) is \H(t-3)(1/4-1/4 cos(2(t-3)))

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inverse oflaplace (e^{-3s})/(s^3+4s)

inverse laplace

\frac{e ^{- 3 s}}{s ^{3} + 4 s}

H (t - 3) (\frac{1}{4} - \frac{1}{4} cos (2 (t - 3)))

Solution steps

L^{- 1} {\frac{e ^{- 3 s}}{s ^{3} + 4 s}}

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 - 3) L^{- 1} {\frac{1}{s ^{3} + 4 s}} (t - 3)

L^{- 1} {\frac{1}{s ^{3} + 4 s}} : \frac{1}{4} H (t) - \frac{1}{4} cos (2 t)

= H (t - 3) (\frac{1}{4} H (t - 3) - \frac{1}{4} cos (2 (t - 3)))

H (t - 3) = {01 t < 3 t \geq 3

= {0 \frac{1}{4} - \frac{1}{4} cos (2 (t - 3)) t < 3 t \geq 3

= H (t - 3) (\frac{1}{4} - \frac{1}{4} cos (2 (t - 3)))

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

\int \frac{- 3 x ^{3}}{x ^{2} + 9} d x

\frac{d}{d x} (\frac{x y}{z})

x \to 7 lim (2 cos (x))

\frac{d}{d x} ((1 + x)^{3 x})

What is inverse oflaplace (e^{-3s})/(s^3+4s) ?
The answer to inverse oflaplace (e^{-3s})/(s^3+4s) is \H(t-3)(1/4-1/4 cos(2(t-3)))