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

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

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inverse oflaplace 1/(2s+2)

inverse laplace

\frac{1}{2 s + 2}

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

Solution steps

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

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

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 + 1}}

Use Inverse Laplace Transform table:

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

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

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

\int \frac{x ^{4}}{1 + x ^{10}} d x

\int 3 e^{2 x + e^{2 x}} d x

\frac{6}{x}

\int (2 x^{4} - 1) (5 x^{3} + 6 x) d x

f (- 3) = - 2 x^{2} + 4 x - \frac{7}{x ^{2}}

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