What is inverse oflaplace 1/(s^4(s^2-4)) ?

The answer to inverse oflaplace 1/(s^4(s^2-4)) is -t/(16)-(t^3)/(24)-1/64 e^{-2t}+1/64 e^{2t}

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inverse oflaplace 1/(s^4(s^2-4))

inverse laplace

\frac{1}{s ^{4} ( s ^{2} - 4 )}

- \frac{t}{16} - \frac{t ^{3}}{24} - \frac{1}{64} e^{- 2 t} + \frac{1}{64} e^{2 t}

Solution steps

L^{- 1} {\frac{1}{s ^{4} ( s ^{2} - 4 )}}

Take the partial fraction of

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

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

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

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

L^{- 1} {\frac{1}{4 s ^{4}}} : \frac{t ^{3}}{24}

L^{- 1} {\frac{1}{64 ( s + 2 )}} : \frac{1}{64} e^{- 2 t}

L^{- 1} {\frac{1}{64 ( s - 2 )}} : \frac{1}{64} e^{2 t}

= - \frac{t}{16} - \frac{t ^{3}}{24} - \frac{1}{64} e^{- 2 t} + \frac{1}{64} e^{2 t}

y^{2} - 4 x = 4, 4 x - y = 16

x \to 1 lim (7 - 2 x - x^{2})

\frac{d}{d x} ((\frac{- sin ( x )}{1 - cos ( x )})^{2})

\int x (x - 1) d x

\frac{d}{d x} (\frac{( 1 - x ^{2} )}{1 + x ^{2}})

What is inverse oflaplace 1/(s^4(s^2-4)) ?
The answer to inverse oflaplace 1/(s^4(s^2-4)) is -t/(16)-(t^3)/(24)-1/64 e^{-2t}+1/64 e^{2t}