What is inverse oflaplace 5/(s(s-1)) ?

The answer to inverse oflaplace 5/(s(s-1)) is -5\H(t)+5e^t

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inverse oflaplace 5/(s(s-1))

inverse laplace

\frac{5}{s ( s - 1 )}

- 5 H (t) + 5 e^{t}

Solution steps

L^{- 1} {\frac{5}{s ( s - 1 )}}

Take the partial fraction of

\frac{5}{s ( s - 1 )} : - \frac{5}{s} + \frac{5}{s - 1}

= L^{- 1} {- \frac{5}{s} + \frac{5}{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{5}{s}} + L^{- 1} {\frac{5}{s - 1}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

L^{- 1} {\frac{5}{s - 1}} : 5 e^{t}

= - 5 H (t) + 5 e^{t}

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

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

f (x) = 12 x - 2 x

x \frac{d y}{d x} = 2 y

t^{2} cos (- 2 t)

What is inverse oflaplace 5/(s(s-1)) ?
The answer to inverse oflaplace 5/(s(s-1)) is -5\H(t)+5e^t