What is inverse oflaplace 1/(s(s+4)(s+2)) ?

The answer to inverse oflaplace 1/(s(s+4)(s+2)) is 1/8 \H(t)+1/8 e^{-4t}-1/4 e^{-2t}

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inverse oflaplace 1/(s(s+4)(s+2))

inverse laplace

\frac{1}{s ( s + 4 ) ( s + 2 )}

\frac{1}{8} H (t) + \frac{1}{8} e^{- 4 t} - \frac{1}{4} e^{- 2 t}

Solution steps

L^{- 1} {\frac{1}{s ( s + 4 ) ( s + 2 )}}

Take the partial fraction of

\frac{1}{s ( s + 4 ) ( s + 2 )} : \frac{1}{8 s} + \frac{1}{8 ( s + 4 )} - \frac{1}{4 ( s + 2 )}

= L^{- 1} {\frac{1}{8 s} + \frac{1}{8 ( s + 4 )} - \frac{1}{4 ( 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}{8 s}} + L^{- 1} {\frac{1}{8 ( s + 4 )}} - L^{- 1} {\frac{1}{4 ( s + 2 )}}

L^{- 1} {\frac{1}{8 s}} : \frac{1}{8} H (t)

L^{- 1} {\frac{1}{8 ( s + 4 )}} : \frac{1}{8} e^{- 4 t}

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

= \frac{1}{8} H (t) + \frac{1}{8} e^{- 4 t} - \frac{1}{4} e^{- 2 t}

\int \frac{ln ( 3 x + 1 )}{12 x + 4} d x

\frac{d ^{2}}{d x ^{2}} (x (x - 4)^{3})

e^{a x}

\int 5 x^{5} - x d x

\int sec^{9} (x) d x

What is inverse oflaplace 1/(s(s+4)(s+2)) ?
The answer to inverse oflaplace 1/(s(s+4)(s+2)) is 1/8 \H(t)+1/8 e^{-4t}-1/4 e^{-2t}