What is inverse oflaplace 1/(6s+1) ?

The answer to inverse oflaplace 1/(6s+1) is 1/6 e^{-t/6}

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inverse oflaplace 1/(6s+1)

inverse laplace

\frac{1}{6 s + 1}

\frac{1}{6} e^{- \frac{t}{6}}

Solution steps

L^{- 1} {\frac{1}{6 s + 1}}

= L^{- 1} {\frac{1}{6} \cdot \frac{1}{s + \frac{1}{6}}}

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}{6} L^{- 1} {\frac{1}{s + \frac{1}{6}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + \frac{1}{6}}} = e^{- \frac{t}{6}}

= \frac{1}{6} e^{- \frac{t}{6}}

f (x) = \frac{1}{x}, g (x) = - \frac{1}{3} x + 3, [0, 12]

\int_{0}^{1} \frac{4 x}{e ^{x^{2}}} d x

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{0.5})

\int \frac{4 - x}{x ^{4} - x ^{2}} d x

x \to 2 lim (2 x + c)

What is inverse oflaplace 1/(6s+1) ?
The answer to inverse oflaplace 1/(6s+1) is 1/6 e^{-t/6}