What is inverse oflaplace 1/((s(s^2+6s-27))) ?

The answer to inverse oflaplace 1/((s(s^2+6s-27))) is -1/27 \H(t)+1/36 e^{3t}+1/108 e^{-9t}

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inverse oflaplace 1/((s(s^2+6s-27)))

inverse laplace

\frac{1}{( s ( s ^{2} + 6 s - 27 ) )}

- \frac{1}{27} H (t) + \frac{1}{36} e^{3 t} + \frac{1}{108} e^{- 9 t}

Solution steps

L^{- 1} {\frac{1}{( s ( s ^{2} + 6 s - 27 ) )}}

Take the partial fraction of

\frac{1}{s ( s ^{2} + 6 s - 27 )} : - \frac{1}{27 s} + \frac{1}{36 ( s - 3 )} + \frac{1}{108 ( s + 9 )}

= L^{- 1} {- \frac{1}{27 s} + \frac{1}{36 ( s - 3 )} + \frac{1}{108 ( 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}{27 s}} + L^{- 1} {\frac{1}{36 ( s - 3 )}} + L^{- 1} {\frac{1}{108 ( s + 9 )}}

L^{- 1} {\frac{1}{27 s}} : \frac{1}{27} H (t)

L^{- 1} {\frac{1}{36 ( s - 3 )}} : \frac{1}{36} e^{3 t}

L^{- 1} {\frac{1}{108 ( s + 9 )}} : \frac{1}{108} e^{- 9 t}

= - \frac{1}{27} H (t) + \frac{1}{36} e^{3 t} + \frac{1}{108} e^{- 9 t}

\frac{3 v + v ^{2}}{v + 1} = x \frac{d v}{d x} + v

\frac{d y}{d x} = \frac{y + 1}{x}

20 + 53 e^{k t}

\frac{3 x - 1}{( x - 1 ) ^{3}}

\int 6 x + 2 d x

What is inverse oflaplace 1/((s(s^2+6s-27))) ?
The answer to inverse oflaplace 1/((s(s^2+6s-27))) is -1/27 \H(t)+1/36 e^{3t}+1/108 e^{-9t}