What is inverse oflaplace 1/(s(s^2+5)) ?

The answer to inverse oflaplace 1/(s(s^2+5)) is 1/5 \H(t)-1/5 cos(sqrt(5)t)

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inverse oflaplace 1/(s(s^2+5))

inverse laplace

\frac{1}{s ( s ^{2} + 5 )}

\frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

Solution steps

L^{- 1} {\frac{1}{s ( s ^{2} + 5 )}}

Take the partial fraction of

\frac{1}{s ( s ^{2} + 5 )} : \frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 5 )}

= L^{- 1} {\frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 5 )}}

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}{5 s}} - L^{- 1} {\frac{s}{5 ( s ^{2} + 5 )}}

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

L^{- 1} {\frac{s}{5 ( s ^{2} + 5 )}} : \frac{1}{5} cos (5 t)

= \frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

x \to 0 lim (sin (3 x))

j \to \infty lim (\frac{2 e ^{j} + e ^{- j} + 2 e ^{4} + 1}{e ^{j} + e ^{4}})

y = sec^{2} (x)

n \to \infty lim (i = 1 \sum n \frac{2 i}{n} \cdot \frac{2}{n})

\frac{\partial}{\partial x} (\frac{1}{x + 2})

What is inverse oflaplace 1/(s(s^2+5)) ?
The answer to inverse oflaplace 1/(s(s^2+5)) is 1/5 \H(t)-1/5 cos(sqrt(5)t)