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

The answer to inverse oflaplace-1/(s^2) is -t

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inverse oflaplace-1/(s^2)

inverse laplace

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

- t

Solution steps

L^{- 1} {- \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)}

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

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

= - t

\int e^{- c x} d x

\frac{\partial}{\partial x} (y + \frac{1}{x + y})

\frac{d}{d x} (lo g_{3} (sec (2 x)))

x \to 0 lim (tan (x) - sec (x))

(t e)^{3} e^{- 2 t} e^{- 2}

What is inverse oflaplace-1/(s^2) ?
The answer to inverse oflaplace-1/(s^2) is -t