What is inverse oflaplace 1/(s^3+6s^2+9s) ?

The answer to inverse oflaplace 1/(s^3+6s^2+9s) is 1/9 \H(t)-1/9 e^{-3t}-(e^{-3t}t)/3

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inverse oflaplace 1/(s^3+6s^2+9s)

inverse laplace

\frac{1}{s ^{3} + 6 s ^{2} + 9 s}

\frac{1}{9} H (t) - \frac{1}{9} e^{- 3 t} - \frac{e ^{- 3 t} t}{3}

Solution steps

L^{- 1} {\frac{1}{s ^{3} + 6 s ^{2} + 9 s}}

Take the partial fraction of

\frac{1}{s ^{3} + 6 s ^{2} + 9 s} : \frac{1}{9 s} - \frac{1}{9 ( s + 3 )} - \frac{1}{3 ( s + 3 ) ^{2}}

= L^{- 1} {\frac{1}{9 s} - \frac{1}{9 ( s + 3 )} - \frac{1}{3 ( s + 3 ) ^{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}{9 s}} - L^{- 1} {\frac{1}{9 ( s + 3 )}} - L^{- 1} {\frac{1}{3 ( s + 3 ) ^{2}}}

L^{- 1} {\frac{1}{9 s}} : \frac{1}{9} H (t)

L^{- 1} {\frac{1}{9 ( s + 3 )}} : \frac{1}{9} e^{- 3 t}

L^{- 1} {\frac{1}{3 ( s + 3 ) ^{2}}} : \frac{e ^{- 3 t} t}{3}

= \frac{1}{9} H (t) - \frac{1}{9} e^{- 3 t} - \frac{e ^{- 3 t} t}{3}

\int \frac{3}{10 - t} d t

\int 35 arctan (x) d x

7 (x^{2} + y^{2}) d x + 8 x y d y = 0

x \to 0 - lim (\frac{cos ( x )}{x})

y = \frac{x + 1}{e ^{x}}

What is inverse oflaplace 1/(s^3+6s^2+9s) ?
The answer to inverse oflaplace 1/(s^3+6s^2+9s) is 1/9 \H(t)-1/9 e^{-3t}-(e^{-3t}t)/3