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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

- \frac{8}{x ^{2}}

\int_{0}^{\infty} e^{x} d x

\int \frac{1}{1 + 49 x ^{2}} d x

\frac{\partial}{\partial y} (ln (x^{7} + y^{7}))

\frac{( x - 3 ) ^{2}}{3}, 9 - x

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