What is inverse oflaplace 1/(s(s+8)) ?

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

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inverse oflaplace 1/(s(s+8))

inverse laplace

\frac{1}{s ( s + 8 )}

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

Solution steps

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

Take the partial fraction of

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

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

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 + 8 )}}

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

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

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

\int x^{x} (1 + ln (x)) d x

\frac{x ^{4} + 9}{x ^{2}}

\frac{d x}{d t} = 2 (2 x t)

\int_{- 1}^{8} 2 - 3 x d x

x \to - 1 + lim (\frac{1}{( x + 1 ) ^{2}})

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