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

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

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inverse oflaplace 1/(s-1)*e^{-s}

inverse laplace

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

H (t - 1) e^{t - 1}

Solution steps

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

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 - 1}} (t - 1)

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

= H (t - 1) e^{t - 1}

\frac{d}{d x} (\frac{350}{x ^{2} + y ^{2}})

(x^{3} e^{- 3 x})^{^{'}}

y = x^{2}

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

\int_{0}^{2} 1 + \frac{5}{4 + x ^{2}} d x

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