What is inverse oflaplace 1/s e^{-s} ?

The answer to inverse oflaplace 1/s e^{-s} is \H(t-1)

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inverse oflaplace 1/s e^{-s}

inverse laplace

\frac{1}{s} e^{- s}

H (t - 1)

Solution steps

L^{- 1} {e^{- s} \frac{1}{s}}

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

L^{- 1} {\frac{1}{s}} : H (t)

= H (t - 1) H (t - 1)

= H (t - 1)

\int 5 x (2 x + 5) (x - 3) d x

t^{2} y^{^{'}} + t y = 7, y (1) = 4

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

x \to \infty lim (\frac{0}{1})

\frac{\partial}{\partial x} (\frac{( 9 y + 7 z )}{x + u})

What is inverse oflaplace 1/s e^{-s} ?
The answer to inverse oflaplace 1/s e^{-s} is \H(t-1)