What is inverse oflaplace (8e^{(-3s)})/(s^3) ?

The answer to inverse oflaplace (8e^{(-3s)})/(s^3) is \H(t-3)*4(t-3)^2

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inverse oflaplace (8e^{(-3s)})/(s^3)

inverse laplace

\frac{8 e ^{(- 3 s)}}{s ^{3}}

H (t - 3) \cdot 4 (t - 3)^{2}

Solution steps

L^{- 1} {\frac{8 e ^{(- 3 s)}}{s ^{3}}}

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{8}{s ^{3}}} (t - 3)

L^{- 1} {\frac{8}{s ^{3}}} : 4 t^{2}

= H (t - 3) \cdot 4 (t - 3)^{2}

\frac{1}{1 - y}

\frac{d}{d x} (3 x^{2} cos (x) cot (x))

x = 6 y^{2}, x = 16 + 2 y^{2}

x \to - \infty lim (\frac{5 x ^{2} - sin ( 3 x )}{x ^{2} + 10})

\frac{d}{d x} (3^{x^{2} - 3})

What is inverse oflaplace (8e^{(-3s)})/(s^3) ?
The answer to inverse oflaplace (8e^{(-3s)})/(s^3) is \H(t-3)*4(t-3)^2