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

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

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inverse oflaplace 1/(as+1)

inverse laplace

\frac{1}{a s + 1}

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

Solution steps

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

= 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}

x \to 1 lim (\frac{I n x}{x - 1})

\int ln (k x) d x

\frac{d}{d x} (\frac{3 x ^{2}}{2 y ^{2}})

\frac{\partial}{\partial x} (cos (z))

\frac{d}{d x} (e^{0.5} + 5 x - 5)

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