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

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

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

inverse laplace

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

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

Solution steps

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

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

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

\int \frac{x ^{3}}{x ^{2} + 1} d x

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

e^{\frac{z}{( z - 3 )}}

9 y^{^{''}} + 18 y^{^{'}} + 11 y = 0

\int_{1}^{2} x 2^{x^{2}} d x

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