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

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

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

inverse laplace

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

- H (t) + e^{t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= - H (t) + e^{t}

x \to 3 lim ((1 - 3 x)^{\frac{1}{x}})

\int tan (2 x) d x

\frac{\partial}{\partial z} (6 y (z + 1)^{4})

\int (1 - 2 x) d x

x \to - \infty lim (1 x)

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