What is inverse oflaplace 1/(s^3(s-1)) ?

The answer to inverse oflaplace 1/(s^3(s-1)) is -\H(t)-t-(t^2)/2+e^t

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inverse oflaplace 1/(s^3(s-1))

inverse laplace

\frac{1}{s ^{3} ( s - 1 )}

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

Solution steps

L^{- 1} {\frac{1}{s ^{3} ( s - 1 )}}

Take the partial fraction of

\frac{1}{s ^{3} ( s - 1 )} : - \frac{1}{s} - \frac{1}{s ^{2}} - \frac{1}{s ^{3}} + \frac{1}{s - 1}

= L^{- 1} {- \frac{1}{s} - \frac{1}{s ^{2}} - \frac{1}{s ^{3}} + \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 ^{3}}} + 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 ^{3}}} : \frac{t ^{2}}{2}

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

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

\frac{\partial}{\partial x} (cos (2 y - 2 x))

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{4} d x

x \to 0 lim (\frac{1}{x} - \frac{1}{∣ x ∣})

x \to 3 lim (x^{2} + 7 x - 5)

\frac{d}{d x} (x^{2} cos (y))

What is inverse oflaplace 1/(s^3(s-1)) ?
The answer to inverse oflaplace 1/(s^3(s-1)) is -\H(t)-t-(t^2)/2+e^t