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

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

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inverse oflaplace 1/(3s+1)

inverse laplace

\frac{1}{3 s + 1}

\frac{1}{3} e^{- \frac{t}{3}}

Solution steps

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

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

y^{^{''}} + 14 y^{^{'}} + 49 y = t^{- 2} e^{- 7 t}

\int (z + 2)^{2} d z

\frac{d}{d x} (cos (2^{x} \cdot x^{5}))

\frac{d}{d x} (9 + 9 - x)

\int (15 x + 2) d x

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