What is inverse oflaplace 1/(s^3(s+4)) ?

The answer to inverse oflaplace 1/(s^3(s+4)) is 1/64 \H(t)-t/(16)+(t^2)/8-1/64 e^{-4t}

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inverse oflaplace 1/(s^3(s+4))

inverse laplace

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

\frac{1}{64} H (t) - \frac{t}{16} + \frac{t ^{2}}{8} - \frac{1}{64} e^{- 4 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ^{3} ( s + 4 )} : \frac{1}{64 s} - \frac{1}{16 s ^{2}} + \frac{1}{4 s ^{3}} - \frac{1}{64 ( s + 4 )}

= L^{- 1} {\frac{1}{64 s} - \frac{1}{16 s ^{2}} + \frac{1}{4 s ^{3}} - \frac{1}{64 ( s + 4 )}}

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}{64 s}} - L^{- 1} {\frac{1}{16 s ^{2}}} + L^{- 1} {\frac{1}{4 s ^{3}}} - L^{- 1} {\frac{1}{64 ( s + 4 )}}

L^{- 1} {\frac{1}{64 s}} : \frac{1}{64} H (t)

L^{- 1} {\frac{1}{16 s ^{2}}} : \frac{t}{16}

L^{- 1} {\frac{1}{4 s ^{3}}} : \frac{t ^{2}}{8}

L^{- 1} {\frac{1}{64 ( s + 4 )}} : \frac{1}{64} e^{- 4 t}

= \frac{1}{64} H (t) - \frac{t}{16} + \frac{t ^{2}}{8} - \frac{1}{64} e^{- 4 t}

\frac{d}{d x} (10 x^{10} e^{x})

\int \frac{1}{- t ^{2} + 8 t - 15} d t

x + e^{x} - \frac{1}{x}

\int \frac{( 4 x - 5 )}{( x - 2 ) ( x + 3 )} d x

\int \frac{y}{x ( x + y )} d x

What is inverse oflaplace 1/(s^3(s+4)) ?
The answer to inverse oflaplace 1/(s^3(s+4)) is 1/64 \H(t)-t/(16)+(t^2)/8-1/64 e^{-4t}