What is inverse oflaplace 1/(s(s^2+7s+12)) ?

The answer to inverse oflaplace 1/(s(s^2+7s+12)) is 1/12 \H(t)-1/3 e^{-3t}+1/4 e^{-4t}

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inverse oflaplace 1/(s(s^2+7s+12))

inverse laplace

\frac{1}{s ( s ^{2} + 7 s + 12 )}

\frac{1}{12} H (t) - \frac{1}{3} e^{- 3 t} + \frac{1}{4} e^{- 4 t}

Solution steps

L^{- 1} {\frac{1}{s ( s ^{2} + 7 s + 12 )}}

Take the partial fraction of

\frac{1}{s ( s ^{2} + 7 s + 12 )} : \frac{1}{12 s} - \frac{1}{3 ( s + 3 )} + \frac{1}{4 ( s + 4 )}

= L^{- 1} {\frac{1}{12 s} - \frac{1}{3 ( s + 3 )} + \frac{1}{4 ( s + 4 )}}

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}{3 ( s + 3 )}} + L^{- 1} {\frac{1}{4 ( s + 4 )}}

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

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

L^{- 1} {\frac{1}{4 ( s + 4 )}} : \frac{1}{4} e^{- 4 t}

= \frac{1}{12} H (t) - \frac{1}{3} e^{- 3 t} + \frac{1}{4} e^{- 4 t}

\int sin^{2} (x) \cdot cos^{3} (x) d x

x^{2}

\frac{d}{d x} (\frac{5 x - 1}{( 6 x ^{2} - 2 ) ^{2}})

\frac{d}{d x} (ln (\frac{x}{4}) \cdot (x - 4))

\frac{d}{d x} (\frac{1}{2} (1 - cos (2 x)))

What is inverse oflaplace 1/(s(s^2+7s+12)) ?
The answer to inverse oflaplace 1/(s(s^2+7s+12)) is 1/12 \H(t)-1/3 e^{-3t}+1/4 e^{-4t}