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

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

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

inverse laplace

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

H (t - a) (t - a)

Solution steps

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

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

= H (t - a) (t - a)

3 + 4 x^{2} - 5 x^{3}

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 e^{- 3 x}

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

n = 1 \sum \infty \frac{3 ^{n}}{4 ^{n + 1}}

x \to 1 lim (4 x - 1)

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