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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

y = 2 C x + C (sin (x) + x cos (x))

\frac{\partial}{\partial x} (\frac{x y}{y - x})

n = 1 \sum \infty (\frac{7}{3})^{n}

\int_{6}^{8} \frac{66}{( x - 6 ) ^{3}} d x

x \to 4 lim (x^{\frac{1}{5}})

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