What is inverse oflaplace 1/((s^3+4s)) ?

The answer to inverse oflaplace 1/((s^3+4s)) is 1/4 \H(t)-1/4 cos(2t)

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inverse oflaplace 1/((s^3+4s))

inverse laplace

\frac{1}{( s ^{3} + 4 s )}

\frac{1}{4} H (t) - \frac{1}{4} cos (2 t)

Solution steps

L^{- 1} {\frac{1}{( s ^{3} + 4 s )}}

Take the partial fraction of

\frac{1}{s ^{3} + 4 s} : \frac{1}{4 s} - \frac{s}{4 ( s ^{2} + 4 )}

= L^{- 1} {\frac{1}{4 s} - \frac{s}{4 ( s ^{2} + 4 )}}

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}{4 s}} - L^{- 1} {\frac{s}{4 ( s ^{2} + 4 )}}

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

L^{- 1} {\frac{s}{4 ( s ^{2} + 4 )}} : \frac{1}{4} cos (2 t)

= \frac{1}{4} H (t) - \frac{1}{4} cos (2 t)

10 (t - 1)

(2 x - 5 y + 2) d x + (1 - 6 y - 5 x) d y = 0

\frac{\partial}{\partial x} ((2 x + 3 y)^{7})

\frac{d}{d x} (- 16 x^{2} + 64 x)

\int \frac{x ^{- \frac{1}{3}}}{1 + x ^{\frac{1}{6}}} d x

What is inverse oflaplace 1/((s^3+4s)) ?
The answer to inverse oflaplace 1/((s^3+4s)) is 1/4 \H(t)-1/4 cos(2t)