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

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

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

inverse laplace

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

\frac{1}{3} H (t) - \frac{1}{3} cos (3 t)

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{3 s}} : \frac{1}{3} H (t)

L^{- 1} {\frac{s}{3 ( s ^{2} + 3 )}} : \frac{1}{3} cos (3 t)

= \frac{1}{3} H (t) - \frac{1}{3} cos (3 t)

x \to 0 lim (\frac{3}{x} - \frac{3}{e ^{x} - 1})

x \to 0 lim (\frac{1}{x ^{2}} + 2)

\int \frac{cos ( x )}{x} d x

\int_{0}^{6} 9 sin (x + 1) d x

- 4 (8 x^{2} + 8)^{- 6}

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