What is inverse oflaplace 2/(s^3(s^2-9)) ?

The answer to inverse oflaplace 2/(s^3(s^2-9)) is -2/81 \H(t)-(t^2)/9+1/81 e^{-3t}+1/81 e^{3t}

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inverse oflaplace 2/(s^3(s^2-9))

inverse laplace

\frac{2}{s ^{3} ( s ^{2} - 9 )}

- \frac{2}{81} H (t) - \frac{t ^{2}}{9} + \frac{1}{81} e^{- 3 t} + \frac{1}{81} e^{3 t}

Solution steps

L^{- 1} {\frac{2}{s ^{3} ( s ^{2} - 9 )}}

Take the partial fraction of

\frac{2}{s ^{3} ( s ^{2} - 9 )} : - \frac{2}{81 s} - \frac{2}{9 s ^{3}} + \frac{1}{81 ( s + 3 )} + \frac{1}{81 ( s - 3 )}

= L^{- 1} {- \frac{2}{81 s} - \frac{2}{9 s ^{3}} + \frac{1}{81 ( s + 3 )} + \frac{1}{81 ( s - 3 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{2}{81 s}} - L^{- 1} {\frac{2}{9 s ^{3}}} + L^{- 1} {\frac{1}{81 ( s + 3 )}} + L^{- 1} {\frac{1}{81 ( s - 3 )}}

L^{- 1} {\frac{2}{81 s}} : \frac{2}{81} H (t)

L^{- 1} {\frac{2}{9 s ^{3}}} : \frac{t ^{2}}{9}

L^{- 1} {\frac{1}{81 ( s + 3 )}} : \frac{1}{81} e^{- 3 t}

L^{- 1} {\frac{1}{81 ( s - 3 )}} : \frac{1}{81} e^{3 t}

= - \frac{2}{81} H (t) - \frac{t ^{2}}{9} + \frac{1}{81} e^{- 3 t} + \frac{1}{81} e^{3 t}

\frac{d ^{2}}{d x ^{2}} (3 x - \frac{8}{x})

x \to 4 + lim (\frac{x - 4}{∣ x - 4 ∣})

x \to 0 lim (\frac{sin ^{2} ( x )}{x})

\frac{d}{d x} (ln (\frac{∣ x - 1 ∣}{∣ x - 3 ∣}))

\int 2 x sec^{2} (3 x) x d x

What is inverse oflaplace 2/(s^3(s^2-9)) ?
The answer to inverse oflaplace 2/(s^3(s^2-9)) is -2/81 \H(t)-(t^2)/9+1/81 e^{-3t}+1/81 e^{3t}