What is inverse oflaplace (e^{-5s})/(s^2) ?

The answer to inverse oflaplace (e^{-5s})/(s^2) is \H(t-5)(t-5)

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inverse oflaplace (e^{-5s})/(s^2)

inverse laplace

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

H (t - 5) (t - 5)

Solution steps

L^{- 1} {\frac{e ^{- 5 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 - 5) L^{- 1} {\frac{1}{s ^{2}}} (t - 5)

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

= H (t - 5) (t - 5)

x \to 3 + lim (\frac{( x ^{2} )}{x - 3})

\int_{0}^{1} 2 π (2 - y) (5 y^{2}) d y

\frac{\partial}{\partial y} (2 y e^{- x})

\frac{d}{d x} (\frac{3 x}{x ^{2} + 3} + 98.6)

(3 x^{2})^{2 x} \cdot (x^{2} + 2 x)^{2}

What is inverse oflaplace (e^{-5s})/(s^2) ?
The answer to inverse oflaplace (e^{-5s})/(s^2) is \H(t-5)(t-5)