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

The answer to inverse oflaplace-1/(s(s-1)(2s+1)(s+2)) is 1/2 \H(t)-1/9 e^t-4/9 e^{-t/2}+1/18 e^{-2t}

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

inverse laplace

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

\frac{1}{2} H (t) - \frac{1}{9} e^{t} - \frac{4}{9} e^{- \frac{t}{2}} + \frac{1}{18} e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s - 1 ) ( 2 s + 1 ) ( s + 2 )} : - \frac{1}{2 s} + \frac{1}{9 ( s - 1 )} + \frac{8}{9 ( 2 s + 1 )} - \frac{1}{18 ( s + 2 )}

= L^{- 1} {\frac{1}{2 s} - \frac{1}{9 ( s - 1 )} - \frac{8}{9 ( 2 s + 1 )} + \frac{1}{18 ( 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{1}{2 s}} - L^{- 1} {\frac{1}{9 ( s - 1 )}} - L^{- 1} {\frac{8}{9 ( 2 s + 1 )}} + L^{- 1} {\frac{1}{18 ( s + 2 )}}

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

L^{- 1} {\frac{1}{9 ( s - 1 )}} : \frac{1}{9} e^{t}

L^{- 1} {\frac{8}{9 ( 2 s + 1 )}} : \frac{4}{9} e^{- \frac{t}{2}}

L^{- 1} {\frac{1}{18 ( s + 2 )}} : \frac{1}{18} e^{- 2 t}

= \frac{1}{2} H (t) - \frac{1}{9} e^{t} - \frac{4}{9} e^{- \frac{t}{2}} + \frac{1}{18} e^{- 2 t}

\int_{0}^{7} 2 π x (7 x - x^{2}) d x

3 \frac{d y}{d x} + 5 y = 1

2 y^{^{'}} = e^{\frac{x}{2}} + y

\int 3^{2 - x} d x

What is inverse oflaplace-1/(s(s-1)(2s+1)(s+2)) ?
The answer to inverse oflaplace-1/(s(s-1)(2s+1)(s+2)) is 1/2 \H(t)-1/9 e^t-4/9 e^{-t/2}+1/18 e^{-2t}