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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\int \frac{1}{3 - 5 x ^{2}} d x

\frac{\partial}{\partial x} (x^{3} - 12 x y + 8 y^{3})

(x^{2} - 5 x + 1)^{4}

\int_{1}^{3} (2 x^{2} - 12 x + 13) d x

\int (\frac{ln ( x )}{x})^{2} d x

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