What is inverse oflaplace (e^{-3s})/(s^2) ?

The answer to inverse oflaplace (e^{-3s})/(s^2) is \H(t-3)(t-3)

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inverse oflaplace (e^{-3s})/(s^2)

inverse laplace

\frac{e ^{- 3 s}}{s ^{2}}

H (t - 3) (t - 3)

Solution steps

L^{- 1} {\frac{e ^{- 3 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 - 3) L^{- 1} {\frac{1}{s ^{2}}} (t - 3)

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

= H (t - 3) (t - 3)

(t + 1)^{2}

\int \frac{4 x + 6}{( x ^{2} + 2 ) \cdot ( x + 2 )} d x

f (x) = x^{\frac{5}{3}} - x^{\frac{2}{3}}

e^{- t} t

\frac{1}{x - 3}

What is inverse oflaplace (e^{-3s})/(s^2) ?
The answer to inverse oflaplace (e^{-3s})/(s^2) is \H(t-3)(t-3)