What is inverse oflaplace 1/(s(s^2+5s+6)) ?

The answer to inverse oflaplace 1/(s(s^2+5s+6)) is 1/6 \H(t)-1/2 e^{-2t}+1/3 e^{-3t}

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inverse oflaplace 1/(s(s^2+5s+6))

inverse laplace

\frac{1}{s ( s ^{2} + 5 s + 6 )}

\frac{1}{6} H (t) - \frac{1}{2} e^{- 2 t} + \frac{1}{3} e^{- 3 t}

Solution steps

L^{- 1} {\frac{1}{s ( s ^{2} + 5 s + 6 )}}

Take the partial fraction of

\frac{1}{s ( s ^{2} + 5 s + 6 )} : \frac{1}{6 s} - \frac{1}{2 ( s + 2 )} + \frac{1}{3 ( s + 3 )}

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

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

L^{- 1} {\frac{1}{6 s}} : \frac{1}{6} H (t)

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

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

= \frac{1}{6} H (t) - \frac{1}{2} e^{- 2 t} + \frac{1}{3} e^{- 3 t}

x \to 0 lim (x sin (\frac{1}{x ^{2}}))

f (x) = \frac{x}{x - 1}

f (x) = 6^{x}

\int \frac{1}{x ^{2} 25 + x ^{2}} d x

\frac{d}{d x} (sin^{2} (x) - sin (x^{2}))

What is inverse oflaplace 1/(s(s^2+5s+6)) ?
The answer to inverse oflaplace 1/(s(s^2+5s+6)) is 1/6 \H(t)-1/2 e^{-2t}+1/3 e^{-3t}