What is inverse oflaplace 1/(s^3-7s) ?

The answer to inverse oflaplace 1/(s^3-7s) is -1/7 \H(t)+1/14 e^{-sqrt(7)t}+1/14 e^{sqrt(7)t}

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inverse oflaplace 1/(s^3-7s)

inverse laplace

\frac{1}{s ^{3} - 7 s}

- \frac{1}{7} H (t) + \frac{1}{14} e^{- 7 t} + \frac{1}{14} e^{7 t}

Solution steps

L^{- 1} {\frac{1}{s ^{3} - 7 s}}

Take the partial fraction of

\frac{1}{s ^{3} - 7 s} : - \frac{1}{7 s} + \frac{1}{14 ( s + 7 )} + \frac{1}{14 ( s - 7 )}

= L^{- 1} {- \frac{1}{7 s} + \frac{1}{14 ( s + 7 )} + \frac{1}{14 ( s - 7 )}}

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}{7 s}} + L^{- 1} {\frac{1}{14 ( s + 7 )}} + L^{- 1} {\frac{1}{14 ( s - 7 )}}

L^{- 1} {\frac{1}{7 s}} : \frac{1}{7} H (t)

L^{- 1} {\frac{1}{14 ( s + 7 )}} : \frac{1}{14} e^{- 7 t}

L^{- 1} {\frac{1}{14 ( s - 7 )}} : \frac{1}{14} e^{7 t}

= - \frac{1}{7} H (t) + \frac{1}{14} e^{- 7 t} + \frac{1}{14} e^{7 t}

y^{^{'''}} + 64 y^{^{'}} = 0

f (x) = ln (100 sin^{2} (x))

\frac{d ^{2}}{d x ^{2}} (x^{\frac{8}{9}} + 6 x)

x \to - 1 lim (2 x + 4)

18 x

What is inverse oflaplace 1/(s^3-7s) ?
The answer to inverse oflaplace 1/(s^3-7s) is -1/7 \H(t)+1/14 e^{-sqrt(7)t}+1/14 e^{sqrt(7)t}