What is inverse oflaplace 2/(3s+1) ?

The answer to inverse oflaplace 2/(3s+1) is 2/3 e^{-t/3}

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inverse oflaplace 2/(3s+1)

inverse laplace

\frac{2}{3 s + 1}

\frac{2}{3} e^{- \frac{t}{3}}

Solution steps

L^{- 1} {\frac{2}{3 s + 1}}

= L^{- 1} {\frac{2}{3} \cdot \frac{1}{s + \frac{1}{3}}}

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{2}{3} L^{- 1} {\frac{1}{s + \frac{1}{3}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + \frac{1}{3}}} = e^{- \frac{t}{3}}

= \frac{2}{3} e^{- \frac{t}{3}}

y = \frac{4}{2 x + 8}, (0, \frac{1}{2})

\frac{d}{d x} (2 x^{2} + 4 x - 10)

\frac{d}{d x} (\frac{e ^{- x^{3}}}{x})

\frac{1}{s ^{2} + 2 s + 10}

\int \frac{5 x ^{2}}{( x - 2 ) ( x ^{2} + 4 )} d x

What is inverse oflaplace 2/(3s+1) ?
The answer to inverse oflaplace 2/(3s+1) is 2/3 e^{-t/3}