What is inverse oflaplace 5/(s^3-2s^2-3s) ?

The answer to inverse oflaplace 5/(s^3-2s^2-3s) is -5/3 \H(t)+5/4 e^{-t}+5/12 e^{3t}

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inverse oflaplace 5/(s^3-2s^2-3s)

inverse laplace

\frac{5}{s ^{3} - 2 s ^{2} - 3 s}

- \frac{5}{3} H (t) + \frac{5}{4} e^{- t} + \frac{5}{12} e^{3 t}

Solution steps

L^{- 1} {\frac{5}{s ^{3} - 2 s ^{2} - 3 s}}

Take the partial fraction of

\frac{5}{s ^{3} - 2 s ^{2} - 3 s} : - \frac{5}{3 s} + \frac{5}{4 ( s + 1 )} + \frac{5}{12 ( s - 3 )}

= L^{- 1} {- \frac{5}{3 s} + \frac{5}{4 ( s + 1 )} + \frac{5}{12 ( 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{5}{3 s}} + L^{- 1} {\frac{5}{4 ( s + 1 )}} + L^{- 1} {\frac{5}{12 ( s - 3 )}}

L^{- 1} {\frac{5}{3 s}} : \frac{5}{3} H (t)

L^{- 1} {\frac{5}{4 ( s + 1 )}} : \frac{5}{4} e^{- t}

L^{- 1} {\frac{5}{12 ( s - 3 )}} : \frac{5}{12} e^{3 t}

= - \frac{5}{3} H (t) + \frac{5}{4} e^{- t} + \frac{5}{12} e^{3 t}

(2005, 56), (2010, 59)

\int_{0}^{1} \frac{3}{x ^{2} - x - 2} d x

\frac{3}{x ^{2}}

n = 1 \sum \infty (- 1)^{n} e^{- n}

x \to 2 lim (3 x - 9)

What is inverse oflaplace 5/(s^3-2s^2-3s) ?
The answer to inverse oflaplace 5/(s^3-2s^2-3s) is -5/3 \H(t)+5/4 e^{-t}+5/12 e^{3t}