What is inverse oflaplace 1/(2s^3+10s^2+12s) ?

The answer to inverse oflaplace 1/(2s^3+10s^2+12s) is 1/12 \H(t)-1/4 e^{-2t}+1/6 e^{-3t}

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inverse oflaplace 1/(2s^3+10s^2+12s)

inverse laplace

\frac{1}{2 s ^{3} + 10 s ^{2} + 12 s}

\frac{1}{12} H (t) - \frac{1}{4} e^{- 2 t} + \frac{1}{6} e^{- 3 t}

Solution steps

L^{- 1} {\frac{1}{2 s ^{3} + 10 s ^{2} + 12 s}}

Take the partial fraction of

\frac{1}{2 s ^{3} + 10 s ^{2} + 12 s} : \frac{1}{12 s} - \frac{1}{4 ( s + 2 )} + \frac{1}{6 ( s + 3 )}

= L^{- 1} {\frac{1}{12 s} - \frac{1}{4 ( s + 2 )} + \frac{1}{6 ( 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}{12 s}} - L^{- 1} {\frac{1}{4 ( s + 2 )}} + L^{- 1} {\frac{1}{6 ( s + 3 )}}

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

L^{- 1} {\frac{1}{4 ( s + 2 )}} : \frac{1}{4} e^{- 2 t}

L^{- 1} {\frac{1}{6 ( s + 3 )}} : \frac{1}{6} e^{- 3 t}

= \frac{1}{12} H (t) - \frac{1}{4} e^{- 2 t} + \frac{1}{6} e^{- 3 t}

y = x^{4} cos (4 x^{2})

y = 17 x, y = x^{2}

\int sin^{2} (t) cos (t) d t

x \to 3 lim (5 x + 4)

\int 0.0025 x - 0.075 d x

What is inverse oflaplace 1/(2s^3+10s^2+12s) ?
The answer to inverse oflaplace 1/(2s^3+10s^2+12s) is 1/12 \H(t)-1/4 e^{-2t}+1/6 e^{-3t}