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)

n = 1 \sum \infty \frac{3}{n ^{2} + n}

\int \frac{4 x ^{2} + 6}{x ^{3} + 3 x} d x

f (x) = 7 x - 1

y^{^{''}} + 4 y^{^{'}} + 6 y = 0

\int_{- 1}^{1} (x^{2} + 1) d x

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