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}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

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

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

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

\int (5 + 4 x^{2})^{2} d x

\int_{0}^{\frac{π}{2}} sin^{5} (2 x) d x

\int tan^{3} (x) sec (x) d x

\int_{0}^{\frac{π}{8}} cos (2 x) d x

\frac{d y}{d x} = (x + y + 1)^{2}

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