What is inverse oflaplace ((1-e^{-5s}))/(s^2) ?

The answer to inverse oflaplace ((1-e^{-5s}))/(s^2) is t-\H(t-5)(t-5)

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inverse oflaplace ((1-e^{-5s}))/(s^2)

inverse laplace

\frac{( 1 - e ^{- 5 s} )}{s ^{2}}

t - H (t - 5) (t - 5)

Solution steps

L^{- 1} {\frac{( 1 - e ^{- 5 s} )}{s ^{2}}}

Expand

= L^{- 1} {\frac{1}{s ^{2}} - \frac{e ^{- 5 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 ^{- 5 s}}{s ^{2}}}

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

L^{- 1} {\frac{e ^{- 5 s}}{s ^{2}}} : H (t - 5) (t - 5)

= t - H (t - 5) (t - 5)

\int \frac{1}{x ln ( x ) ln ^{2} ( x ) - 4} d x

\int (\frac{x ^{2}}{( 7 + x ^{3} ) ^{2}}) d x

\frac{\partial}{\partial x} (\frac{x \cdot y}{x - y})

f (x) = - \frac{6 x}{x ^{2} + 1}, at x = 0

x \cdot cos (x)

What is inverse oflaplace ((1-e^{-5s}))/(s^2) ?
The answer to inverse oflaplace ((1-e^{-5s}))/(s^2) is t-\H(t-5)(t-5)