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

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

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

inverse laplace

\frac{3}{s ( s ^{2} + 6 s )}

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

Solution steps

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

Take the partial fraction of

\frac{3}{s ( s ^{2} + 6 s )} : - \frac{1}{12 s} + \frac{1}{2 s ^{2}} + \frac{1}{12 ( s + 6 )}

= L^{- 1} {- \frac{1}{12 s} + \frac{1}{2 s ^{2}} + \frac{1}{12 ( 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}{2 s ^{2}}} + L^{- 1} {\frac{1}{12 ( s + 6 )}}

L^{- 1} {\frac{1}{12 s}} : \frac{1}{12} H (t)

L^{- 1} {\frac{1}{2 s ^{2}}} : \frac{t}{2}

L^{- 1} {\frac{1}{12 ( s + 6 )}} : \frac{1}{12} e^{- 6 t}

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

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