What is inverse oflaplace 1/(s^3+8s^2+12s) ?

The answer to inverse oflaplace 1/(s^3+8s^2+12s) is 1/12 \H(t)-1/8 e^{-2t}+1/24 e^{-6t}

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inverse oflaplace 1/(s^3+8s^2+12s)

inverse laplace

\frac{1}{s ^{3} + 8 s ^{2} + 12 s}

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

Solution steps

L^{- 1} {\frac{1}{s ^{3} + 8 s ^{2} + 12 s}}

Take the partial fraction of

\frac{1}{s ^{3} + 8 s ^{2} + 12 s} : \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}

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

f (x) = 4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1

n = 0 \sum \infty 2 \cdot 0. 2^{n}

x \to 0 lim (\frac{5 ^{x + 1} - 5}{x})

\int_{0}^{\infty} 8 sin^{2} (x) d x

What is inverse oflaplace 1/(s^3+8s^2+12s) ?
The answer to inverse oflaplace 1/(s^3+8s^2+12s) is 1/12 \H(t)-1/8 e^{-2t}+1/24 e^{-6t}