What is inverse oflaplace 1/(s^3(s+3)) ?

The answer to inverse oflaplace 1/(s^3(s+3)) is 1/27 \H(t)-t/9+(t^2)/6-1/27 e^{-3t}

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inverse oflaplace 1/(s^3(s+3))

inverse laplace

\frac{1}{s ^{3} ( s + 3 )}

\frac{1}{27} H (t) - \frac{t}{9} + \frac{t ^{2}}{6} - \frac{1}{27} e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ^{3} ( s + 3 )} : \frac{1}{27 s} - \frac{1}{9 s ^{2}} + \frac{1}{3 s ^{3}} - \frac{1}{27 ( s + 3 )}

= L^{- 1} {\frac{1}{27 s} - \frac{1}{9 s ^{2}} + \frac{1}{3 s ^{3}} - \frac{1}{27 ( 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{1}{27 s}} - L^{- 1} {\frac{1}{9 s ^{2}}} + L^{- 1} {\frac{1}{3 s ^{3}}} - L^{- 1} {\frac{1}{27 ( s + 3 )}}

L^{- 1} {\frac{1}{27 s}} : \frac{1}{27} H (t)

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

L^{- 1} {\frac{1}{3 s ^{3}}} : \frac{t ^{2}}{6}

L^{- 1} {\frac{1}{27 ( s + 3 )}} : \frac{1}{27} e^{- 3 t}

= \frac{1}{27} H (t) - \frac{t}{9} + \frac{t ^{2}}{6} - \frac{1}{27} e^{- 3 t}

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\int_{0}^{2} \frac{x + 2}{x + 1} d x

\int 102 e^{0.06 t} d t

What is inverse oflaplace 1/(s^3(s+3)) ?
The answer to inverse oflaplace 1/(s^3(s+3)) is 1/27 \H(t)-t/9+(t^2)/6-1/27 e^{-3t}