What is inverse oflaplace (5e^{-s})/(s+1) ?

The answer to inverse oflaplace (5e^{-s})/(s+1) is \H(t-1)*5e^{-t+1}

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inverse oflaplace (5e^{-s})/(s+1)

inverse laplace

\frac{5 e ^{- s}}{s + 1}

H (t - 1) \cdot 5 e^{- t + 1}

Solution steps

L^{- 1} {\frac{5 e ^{- s}}{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{5}{s + 1}} (t - 1)

L^{- 1} {\frac{5}{s + 1}} : 5 e^{- t}

= H (t - 1) \cdot 5 e^{- (t - 1)}

= H (t - 1) \cdot 5 e^{- t + 1}

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What is inverse oflaplace (5e^{-s})/(s+1) ?
The answer to inverse oflaplace (5e^{-s})/(s+1) is \H(t-1)*5e^{-t+1}