What is inverse oflaplace (e^{-10s})/(s^2) ?

The answer to inverse oflaplace (e^{-10s})/(s^2) is \H(t-10)(t-10)

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inverse oflaplace (e^{-10s})/(s^2)

inverse laplace

\frac{e ^{- 10 s}}{s ^{2}}

H (t - 10) (t - 10)

Solution steps

L^{- 1} {\frac{e ^{- 10 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 - 10) L^{- 1} {\frac{1}{s ^{2}}} (t - 10)

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

= H (t - 10) (t - 10)

x = - 3, x = 3, y = 2 e^{2 x}, y = e^{2 x} + 1

\int_{0}^{2 π} π cos^{2} (x) d x

\frac{d}{d x} (ab)

\int \frac{8 x + 10}{2 x ^{2} + 5 x} d x

\int \frac{1}{x 256 x ^{2} - 256} d x

What is inverse oflaplace (e^{-10s})/(s^2) ?
The answer to inverse oflaplace (e^{-10s})/(s^2) is \H(t-10)(t-10)