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

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

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

inverse laplace

\frac{2}{s ( s + 1 ) ( s + 2 ) ( s + 3 )}

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

x \to \infty lim (- 2 x - 1)

4 x, x^{2}

\int (2 y + 1) d y

\int_{3}^{9} \frac{1}{x ln ( x )} d x

y = 2 x + 1, at x = \frac{1}{2}

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