What is inverse oflaplace s/(s+a) ?

The answer to inverse oflaplace s/(s+a) is -ae^{-at}+1

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inverse oflaplace s/(s+a)

inverse laplace

\frac{s}{s + a}

- a e^{- a t} + 1

Solution steps

L^{- 1} {\frac{s}{s + a}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{s}{s + a} : F (s) = \frac{1}{s + a}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s + a}}) + L^{- 1} {\frac{1}{s + a}} (0)

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

\frac{d}{d t} (e^{- a t}) = - a e^{- a t}

Evaluate

e^{- a t}

at point

0 = 1

= - a e^{- a t} + 1

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What is inverse oflaplace s/(s+a) ?
The answer to inverse oflaplace s/(s+a) is -ae^{-at}+1