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

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

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inverse oflaplace 1/(s^2-s)

inverse laplace

\frac{1}{s ^{2} - s}

- H (t) + e^{t}

Solution steps

L^{- 1} {\frac{1}{s ^{2} - s}}

Take the partial fraction of

\frac{1}{s ^{2} - s} : - \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 \frac{1}{5} lim (\frac{x}{10 x - 2})

\frac{d}{d x} ((cos (x))^{3})

x^{^{'}} + \frac{2}{1 + t} x = \frac{e ^{t}}{1 + t}

\int_{- 1}^{1} 3 x^{2} x^{3} + 5 d x

(x^{2} + 2 x)^{^{'}}

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