What is inverse oflaplace 1/(s+a) ?

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

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inverse oflaplace 1/(s+a)

inverse laplace

\frac{1}{s + a}

e^{- a t}

Solution steps

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

Use Inverse Laplace Transform table:

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

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

= e^{- a t}

x^{3} - 3 x^{2} + x - 5

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\int 3 x 1 - x^{2} d x

4 x y^{^{'}} = 2 x e^{x} - 4 y + 6 x^{2}

\int x sin^{3} (x^{2}) cos (x^{2}) d x

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