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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\int (3 e^{x} + 5 sec^{2} (x)) d x

\frac{\partial}{\partial y} (\frac{2 y}{x + z})

\frac{d}{d x} (8 (x^{3} - x)^{4})

f (x) = 1.58 e^{- 0.213 x}

f (x) = \frac{4 x}{e ^{x}}

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