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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

x \to + 6 - lim (\frac{1}{x - 6})

y^{^{'}} = 3 y^{2} sin (x)

\frac{d y}{d x} = 10 x + y - 10

\frac{\partial}{\partial y} (ln (x y^{2}))

\int \frac{1}{1 - u ^{2}} d u

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