What is inverse oflaplace 1/(s(s+2)(s+6)) ?

The answer to inverse oflaplace 1/(s(s+2)(s+6)) is 1/12 \H(t)-1/8 e^{-2t}+1/24 e^{-6t}

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inverse oflaplace 1/(s(s+2)(s+6))

inverse laplace

\frac{1}{s ( s + 2 ) ( s + 6 )}

\frac{1}{12} H (t) - \frac{1}{8} e^{- 2 t} + \frac{1}{24} e^{- 6 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s + 2 ) ( s + 6 )} : \frac{1}{12 s} - \frac{1}{8 ( s + 2 )} + \frac{1}{24 ( s + 6 )}

= L^{- 1} {\frac{1}{12 s} - \frac{1}{8 ( s + 2 )} + \frac{1}{24 ( s + 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}{12 s}} - L^{- 1} {\frac{1}{8 ( s + 2 )}} + L^{- 1} {\frac{1}{24 ( s + 6 )}}

L^{- 1} {\frac{1}{12 s}} : \frac{1}{12} H (t)

L^{- 1} {\frac{1}{8 ( s + 2 )}} : \frac{1}{8} e^{- 2 t}

L^{- 1} {\frac{1}{24 ( s + 6 )}} : \frac{1}{24} e^{- 6 t}

= \frac{1}{12} H (t) - \frac{1}{8} e^{- 2 t} + \frac{1}{24} e^{- 6 t}

\int_{- 1}^{1} \frac{1}{x ^{2} - 1} d x

n = 0 \sum \infty 1800 (\frac{1}{3})^{n}

f (x) = 4 x e^{x}, (0, 0)

\int x (4 + x)^{10} d x

\frac{2 ( 1 - ln ( x ) )}{x ^{2}}

What is inverse oflaplace 1/(s(s+2)(s+6)) ?
The answer to inverse oflaplace 1/(s(s+2)(s+6)) is 1/12 \H(t)-1/8 e^{-2t}+1/24 e^{-6t}