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

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

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inverse oflaplace 2/(s^2-s)

inverse laplace

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

- 2 H (t) + 2 e^{t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= - 2 H (t) + 2 e^{t}

8 cos (4 x), 8 - 8 cos (4 x), 0, \frac{π}{4}

\frac{d}{d x} (\frac{2 x}{2 + x})

\frac{d}{d x} (cos (\frac{1 - e ^{4 x}}{1 + e ^{4 x}}))

\int x cos (3 π x^{2}) d x

ln (x) y^{^{'}} = \frac{y}{x}

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