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

The answer to inverse oflaplace 1/((s-1)(s)) is e^t-\H(t)

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

inverse laplace

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

e^{t} - H (t)

Solution steps

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

Take the partial fraction of

\frac{1}{( s - 1 ) s} : \frac{1}{s - 1} - \frac{1}{s}

= L^{- 1} {\frac{1}{s - 1} - \frac{1}{s}}

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}{s - 1}} - L^{- 1} {\frac{1}{s}}

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

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

= e^{t} - H (t)

\frac{d}{d x} (\frac{1 - 2 x}{x ^{2}})

2 x^{2} - 12 x + 11

\frac{\partial}{\partial x} (2 + y - x^{2} + 9)

\int_{0}^{l n (x)} sin (e^{x}) d x

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

What is inverse oflaplace 1/((s-1)(s)) ?
The answer to inverse oflaplace 1/((s-1)(s)) is e^t-\H(t)