What is inverse oflaplace 1/(2s+6) ?

The answer to inverse oflaplace 1/(2s+6) is 1/2 e^{-3t}

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inverse oflaplace 1/(2s+6)

inverse laplace

\frac{1}{2 s + 6}

\frac{1}{2} e^{- 3 t}

Solution steps

L^{- 1} {\frac{1}{2 s + 6}}

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

Use Inverse Laplace Transform table:

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

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

= \frac{1}{2} e^{- 3 t}

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\int \frac{e ^{5 x} - e ^{2 x} + 1}{e ^{x}} d x

\int 8 x (ln (x^{5}))^{2} d x

\frac{d y}{d x}, x y = 0

\int_{0}^{π} cos^{2} (x) d x

What is inverse oflaplace 1/(2s+6) ?
The answer to inverse oflaplace 1/(2s+6) is 1/2 e^{-3t}