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

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

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inverse oflaplace a/(s+a)

inverse laplace

\frac{a}{s + a}

a e^{- a t}

Solution steps

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= a 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}

= a e^{- a t}

\int 4 x^{3} d x

\frac{d}{d x} (\frac{1}{9 x})

\int x a^{x} d x

\int \frac{x ^{2}}{( 196 + x ^{2} ) ^{2}} d x

\frac{\partial}{\partial x} (x y - 2 x - 2 y - x^{2} - y^{2})

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