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

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

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inverse oflaplace 1/(s^3+s)

inverse laplace

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

H (t) - cos (t)

Solution steps

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

Take the partial fraction of

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

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

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}{s}} - L^{- 1} {\frac{s}{s ^{2} + 1}}

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

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

= H (t) - cos (t)

\int \frac{3 x ^{2} + 7 x + 3}{( x ^{2} + 1 ) ^{2}} d x

\frac{d}{d x} (8^{- x})

x \to 0 lim (\frac{tan ( 3 x )}{x})

\int cos (2 - \frac{x}{3}) d x

x \to - 2 lim (\frac{x - 2}{x ^{3} + 8})

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