What is inverse oflaplace 1/(s^3+6s) ?

The answer to inverse oflaplace 1/(s^3+6s) is 1/6 \H(t)-1/6 cos(sqrt(6)t)

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inverse oflaplace 1/(s^3+6s)

inverse laplace

\frac{1}{s ^{3} + 6 s}

\frac{1}{6} H (t) - \frac{1}{6} cos (6 t)

Solution steps

L^{- 1} {\frac{1}{s ^{3} + 6 s}}

Take the partial fraction of

\frac{1}{s ^{3} + 6 s} : \frac{1}{6 s} - \frac{s}{6 ( s ^{2} + 6 )}

= L^{- 1} {\frac{1}{6 s} - \frac{s}{6 ( s ^{2} + 6 )}}

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}{6 s}} - L^{- 1} {\frac{s}{6 ( s ^{2} + 6 )}}

L^{- 1} {\frac{1}{6 s}} : \frac{1}{6} H (t)

L^{- 1} {\frac{s}{6 ( s ^{2} + 6 )}} : \frac{1}{6} cos (6 t)

= \frac{1}{6} H (t) - \frac{1}{6} cos (6 t)

\frac{d y}{d x} = \frac{x y + x ^{2} e ^{\frac{y}{x}}}{x ^{2}}

\int - 20

2 y^{^{'}} + y = \frac{e ^{2 x}}{2}

y = 9 x, y = \frac{8}{3} x, y = 52 - x^{2}

x^{2} y^{^{'}} = y - x y, y (- 1) = - 1

What is inverse oflaplace 1/(s^3+6s) ?
The answer to inverse oflaplace 1/(s^3+6s) is 1/6 \H(t)-1/6 cos(sqrt(6)t)