What is inverse oflaplace 1/(s(s^2+16)) ?

The answer to inverse oflaplace 1/(s(s^2+16)) is 1/16 \H(t)-1/16 cos(4t)

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inverse oflaplace 1/(s(s^2+16))

inverse laplace

\frac{1}{s ( s ^{2} + 16 )}

\frac{1}{16} H (t) - \frac{1}{16} cos (4 t)

Solution steps

L^{- 1} {\frac{1}{s ( s ^{2} + 16 )}}

Take the partial fraction of

\frac{1}{s ( s ^{2} + 16 )} : \frac{1}{16 s} - \frac{s}{16 ( s ^{2} + 16 )}

= L^{- 1} {\frac{1}{16 s} - \frac{s}{16 ( s ^{2} + 16 )}}

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}{16 s}} - L^{- 1} {\frac{s}{16 ( s ^{2} + 16 )}}

L^{- 1} {\frac{1}{16 s}} : \frac{1}{16} H (t)

L^{- 1} {\frac{s}{16 ( s ^{2} + 16 )}} : \frac{1}{16} cos (4 t)

= \frac{1}{16} H (t) - \frac{1}{16} cos (4 t)

\frac{d}{d s} (\frac{1}{s ^{2} + 1})

4 x^{3} e^{- x}

\int cos^{3} (\frac{x}{5}) d x

f (x) = (x^{3} - 16 x)^{14}, (4, 0)

x \to 8 lim (5)

What is inverse oflaplace 1/(s(s^2+16)) ?
The answer to inverse oflaplace 1/(s(s^2+16)) is 1/16 \H(t)-1/16 cos(4t)