What is inverse oflaplace 1/(1+sa) ?

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

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inverse oflaplace 1/(1+sa)

inverse laplace

\frac{1}{1 + s a}

\frac{e ^{- \frac{t}{a}}}{a}

Solution steps

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

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

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

Use Inverse Laplace Transform table:

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

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

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

= \frac{e ^{- \frac{t}{a}}}{a}

\frac{d}{d x} f (x) = 8^{(x^{2} + 4 x)}

\frac{d y}{d x} = e^{2 x + 3 y}, y (0) = 0

\int_{6}^{\infty} x e^{- 2 x} d x

\int - 5 cos (x) d x

y^{^{''}} - 8 y^{^{'}} + 10 y = 0

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