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)

\frac{\partial}{\partial y} (3 x + x^{2} y)

x \to 1 lim ((a x^{2} - b x)^{2})

\frac{d}{d x} (2 c e^{2 x})

\frac{d y}{d x} = \frac{( 2 x ( y - 2 ) )}{x ^{2} + 1}, y (0) = 4

t \to 0 lim (e^{- t})

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