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

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

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inverse oflaplace 1/(s^2)(1-e^{-s})

inverse laplace

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

t - H (t - 1) (t - 1)

Solution steps

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

Expand

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

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

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

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

= t - H (t - 1) (t - 1)

\frac{1}{( x ^{2} + 8 ) ^{5}}

\int \frac{x ^{3} e ^{x} - 2 x ^{2}}{x ^{3}} d x

\int p (p + 6)^{6} d p

f (x) = \frac{1}{x ^{2}}, (- 2, \frac{1}{4})

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