What is inverse oflaplace 1/(s^3+4s^2+3s) ?

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

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inverse oflaplace 1/(s^3+4s^2+3s)

inverse laplace

\frac{1}{s ^{3} + 4 s ^{2} + 3 s}

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

Solution steps

L^{- 1} {\frac{1}{s ^{3} + 4 s ^{2} + 3 s}}

Take the partial fraction of

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

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

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

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

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

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

\int (arctan (x)) d x

\frac{d}{d x} (cosh (ln (x)))

\int_{0}^{2} 6 d x

\int \frac{1}{x x - 1} d x

e^{- \frac{h}{7}}

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