What is inverse oflaplace {1/(s^3+5s)} ?

The answer to inverse oflaplace {1/(s^3+5s)} is 1/5 \H(t)-1/5 cos(sqrt(5)t)

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inverse oflaplace {1/(s^3+5s)}

inverse laplace

{\frac{1}{s ^{3} + 5 s}}

\frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

Solution steps

L^{- 1} {\frac{1}{s ^{3} + 5 s}}

Take the partial fraction of

\frac{1}{s ^{3} + 5 s} : \frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 5 )}

= L^{- 1} {\frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 5 )}}

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}{5 s}} - L^{- 1} {\frac{s}{5 ( s ^{2} + 5 )}}

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

L^{- 1} {\frac{s}{5 ( s ^{2} + 5 )}} : \frac{1}{5} cos (5 t)

= \frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

What is inverse oflaplace {1/(s^3+5s)} ?
The answer to inverse oflaplace {1/(s^3+5s)} is 1/5 \H(t)-1/5 cos(sqrt(5)t)