What is inverse oflaplace 5/(s^2) ?

The answer to inverse oflaplace 5/(s^2) is 5t

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inverse oflaplace 5/(s^2)

inverse laplace

\frac{5}{s ^{2}}

5 t

Solution steps

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

Use Inverse Laplace Transform table:

L^{- 1} {\frac{( n ) !}{s ^{n + 1}}} = t^{n}

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

= 5 t

\int e^{2 t} cos (e^{t}) d t

x \to + 3 + lim (\frac{x + 9}{x - 3})

\frac{d}{d x} (3 tan^{2} (x))

\frac{r}{r ^{2} + 9}

x \to 0 lim (\frac{8}{x ^{2} + x} - \frac{8}{x})

What is inverse oflaplace 5/(s^2) ?
The answer to inverse oflaplace 5/(s^2) is 5t