What is inverse oflaplace (e^{-4s})/(s^2) ?

The answer to inverse oflaplace (e^{-4s})/(s^2) is \H(t-4)(t-4)

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inverse oflaplace (e^{-4s})/(s^2)

inverse laplace

\frac{e ^{- 4 s}}{s ^{2}}

H (t - 4) (t - 4)

Solution steps

L^{- 1} {\frac{e ^{- 4 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 - 4) L^{- 1} {\frac{1}{s ^{2}}} (t - 4)

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

= H (t - 4) (t - 4)

98.284 e^{- 0.28 x}

\int_{0}^{1} 1 0^{x} d x

x \to 2 lim (4 tan (3 x - 3))

\frac{d y}{d x}, x cos (5 y) + y^{5} e^{6 x} = e^{2 x}

n = 2 \sum \infty 1

What is inverse oflaplace (e^{-4s})/(s^2) ?
The answer to inverse oflaplace (e^{-4s})/(s^2) is \H(t-4)(t-4)