What is inverse oflaplace s/(s^2-16) ?

The answer to inverse oflaplace s/(s^2-16) is 1/2 e^{-4t}+1/2 e^{4t}

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inverse oflaplace s/(s^2-16)

inverse laplace

\frac{s}{s ^{2} - 16}

\frac{1}{2} e^{- 4 t} + \frac{1}{2} e^{4 t}

Solution steps

L^{- 1} {\frac{s}{s ^{2} - 16}}

Take the partial fraction of

\frac{s}{s ^{2} - 16} : \frac{1}{2 ( s + 4 )} + \frac{1}{2 ( s - 4 )}

= L^{- 1} {\frac{1}{2 ( s + 4 )} + \frac{1}{2 ( s - 4 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{2 ( s + 4 )}} + L^{- 1} {\frac{1}{2 ( s - 4 )}}

L^{- 1} {\frac{1}{2 ( s + 4 )}} : \frac{1}{2} e^{- 4 t}

L^{- 1} {\frac{1}{2 ( s - 4 )}} : \frac{1}{2} e^{4 t}

= \frac{1}{2} e^{- 4 t} + \frac{1}{2} e^{4 t}

y^{^{'}} = x^{2} - y - 2, y (0) = 1

\int \frac{x ^{2}}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\frac{d}{d x} (ln (7 x^{3}))

(x^{2} + y + y^{2}) d x = x d y

What is inverse oflaplace s/(s^2-16) ?
The answer to inverse oflaplace s/(s^2-16) is 1/2 e^{-4t}+1/2 e^{4t}