What is inverse oflaplace 5/(s(s+1)(s+2)) ?

The answer to inverse oflaplace 5/(s(s+1)(s+2)) is 5/2 \H(t)-5e^{-t}+5/2 e^{-2t}

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inverse oflaplace 5/(s(s+1)(s+2))

inverse laplace

\frac{5}{s ( s + 1 ) ( s + 2 )}

\frac{5}{2} H (t) - 5 e^{- t} + \frac{5}{2} e^{- 2 t}

Solution steps

L^{- 1} {\frac{5}{s ( s + 1 ) ( s + 2 )}}

Take the partial fraction of

\frac{5}{s ( s + 1 ) ( s + 2 )} : \frac{5}{2 s} - \frac{5}{s + 1} + \frac{5}{2 ( s + 2 )}

= L^{- 1} {\frac{5}{2 s} - \frac{5}{s + 1} + \frac{5}{2 ( s + 2 )}}

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}{2 s}} - L^{- 1} {\frac{5}{s + 1}} + L^{- 1} {\frac{5}{2 ( s + 2 )}}

L^{- 1} {\frac{5}{2 s}} : \frac{5}{2} H (t)

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

L^{- 1} {\frac{5}{2 ( s + 2 )}} : \frac{5}{2} e^{- 2 t}

= \frac{5}{2} H (t) - 5 e^{- t} + \frac{5}{2} e^{- 2 t}

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What is inverse oflaplace 5/(s(s+1)(s+2)) ?
The answer to inverse oflaplace 5/(s(s+1)(s+2)) is 5/2 \H(t)-5e^{-t}+5/2 e^{-2t}