What is inverse oflaplace 1/(s^2-16) ?

The answer to inverse oflaplace 1/(s^2-16) is -1/8 e^{-4t}+1/8 e^{4t}

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inverse oflaplace 1/(s^2-16)

inverse laplace

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

- \frac{1}{8} e^{- 4 t} + \frac{1}{8} e^{4 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ^{2} - 16} : - \frac{1}{8 ( s + 4 )} + \frac{1}{8 ( s - 4 )}

= L^{- 1} {- \frac{1}{8 ( s + 4 )} + \frac{1}{8 ( 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}{8 ( s + 4 )}} + L^{- 1} {\frac{1}{8 ( s - 4 )}}

L^{- 1} {\frac{1}{8 ( s + 4 )}} : \frac{1}{8} e^{- 4 t}

L^{- 1} {\frac{1}{8 ( s - 4 )}} : \frac{1}{8} e^{4 t}

= - \frac{1}{8} e^{- 4 t} + \frac{1}{8} e^{4 t}

\int_{- \infty}^{0} e^{- x^{2}} d x

\int x \cdot ln^{4} (x) d x

x \to 2 - lim (- \frac{1}{x - 2})

in v erse l a pl a ce \frac{16}{s ^{2} + 9}

\frac{\partial}{\partial y} (\frac{4 x}{y})

What is inverse oflaplace 1/(s^2-16) ?
The answer to inverse oflaplace 1/(s^2-16) is -1/8 e^{-4t}+1/8 e^{4t}