What is inverse oflaplace {1/(s(s-1))} ?

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

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inverse oflaplace {1/(s(s-1))}

inverse laplace

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

- H (t) + e^{t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s - 1 )} : - \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}

f (x) = cos (\frac{π x}{2})

\int \frac{x}{1 - x x} d x

y = (x + 3)^{2} - (x - 2)^{2}

(3, - 6), (5, - 5)

\frac{d y}{d x} = 9 x e^{y}

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