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

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

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inverse oflaplace 1/(s+1/a)

inverse laplace

\frac{1}{s + \frac{1}{a}}

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

Solution steps

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}}

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

x, \frac{1}{x ^{2}}, 2

\frac{d}{d a} (sin^{4} (a) - cos^{4} (a))

\frac{1}{4} y^{^{''}} + 4 y = 2 tan (4 t) - \frac{1}{3} e^{3 t}

\int (1 + y^{2}) d y

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