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

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

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

inverse laplace

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

H (t - 2) (t - 2)

Solution steps

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

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

= H (t - 2) (t - 2)

x \to - 2 lim (8)

f (x) = \frac{168}{( x + 4 ) ^{5}}

f (x) = x e^{x}

\int ln (7 x^{5}) d x

\frac{\partial}{\partial x} (e^{x} sin (y))

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