What is inverse oflaplace 9/(s(s+1)(s+9)) ?

The answer to inverse oflaplace 9/(s(s+1)(s+9)) is \H(t)-9/8 e^{-t}+1/8 e^{-9t}

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inverse oflaplace 9/(s(s+1)(s+9))

inverse laplace

\frac{9}{s ( s + 1 ) ( s + 9 )}

H (t) - \frac{9}{8} e^{- t} + \frac{1}{8} e^{- 9 t}

Solution steps

L^{- 1} {\frac{9}{s ( s + 1 ) ( s + 9 )}}

Take the partial fraction of

\frac{9}{s ( s + 1 ) ( s + 9 )} : \frac{1}{s} - \frac{9}{8 ( s + 1 )} + \frac{1}{8 ( s + 9 )}

= L^{- 1} {\frac{1}{s} - \frac{9}{8 ( s + 1 )} + \frac{1}{8 ( s + 9 )}}

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}{s}} - L^{- 1} {\frac{9}{8 ( s + 1 )}} + L^{- 1} {\frac{1}{8 ( s + 9 )}}

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

L^{- 1} {\frac{9}{8 ( s + 1 )}} : \frac{9}{8} e^{- t}

L^{- 1} {\frac{1}{8 ( s + 9 )}} : \frac{1}{8} e^{- 9 t}

= H (t) - \frac{9}{8} e^{- t} + \frac{1}{8} e^{- 9 t}

\int \frac{x ^{2} - 1}{x ^{2} + 1} d x

y^{^{''}} - 16 y^{^{'}} + 100 y = 0

\frac{\partial}{\partial y} (4 x y^{3} (z^{2} + 1)^{3})

f (x) = \frac{e ^{3 x}}{x ^{15}}

x \to + 2 - lim (x - 2)

What is inverse oflaplace 9/(s(s+1)(s+9)) ?
The answer to inverse oflaplace 9/(s(s+1)(s+9)) is \H(t)-9/8 e^{-t}+1/8 e^{-9t}