What is inverse oflaplace 3/((8z+16)) ?

The answer to inverse oflaplace 3/((8z+16)) is 3/8 e^{-2t}

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inverse oflaplace 3/((8z+16))

inverse laplace

\frac{3}{( 8 z + 16 )}

\frac{3}{8} e^{- 2 t}

Solution steps

L^{- 1} {\frac{3}{( 8 z + 16 )}}

= L^{- 1} {\frac{3}{8} \cdot \frac{1}{z + 2}}

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{3}{8} L^{- 1} {\frac{1}{z + 2}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{z + 2}} = e^{- 2 t}

= \frac{3}{8} e^{- 2 t}

\frac{d}{d x} (\frac{2}{( x + 1 ) ( - x + 1 )})

t \to 1 lim (cos (t π))

f (x) = \frac{2}{x ^{3}}

y = \frac{12}{1 + x ^{4}}, y = 6 x^{2}

\int x sin (5 x^{2}) d x

What is inverse oflaplace 3/((8z+16)) ?
The answer to inverse oflaplace 3/((8z+16)) is 3/8 e^{-2t}