What is inverse oflaplace (s^2+12)/(s^4-16) ?

The answer to inverse oflaplace (s^2+12)/(s^4-16) is -1/2 sin(2t)-1/2 e^{-2t}+1/2 e^{2t}

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inverse oflaplace (s^2+12)/(s^4-16)

inverse laplace

\frac{s ^{2} + 12}{s ^{4} - 16}

- \frac{1}{2} sin (2 t) - \frac{1}{2} e^{- 2 t} + \frac{1}{2} e^{2 t}

Solution steps

L^{- 1} {\frac{s ^{2} + 12}{s ^{4} - 16}}

Take the partial fraction of

\frac{s ^{2} + 12}{s ^{4} - 16} : - \frac{1}{s ^{2} + 4} - \frac{1}{2 ( s + 2 )} + \frac{1}{2 ( s - 2 )}

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

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

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

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

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

= - \frac{1}{2} sin (2 t) - \frac{1}{2} e^{- 2 t} + \frac{1}{2} e^{2 t}

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

x \to 0 lim (sin (x) ln (x))

\int \frac{29 x ^{2}}{x ^{2} - 4} d x

y = x^{2} + 2 x, y = x + 2

\int \frac{1}{8} t d t

What is inverse oflaplace (s^2+12)/(s^4-16) ?
The answer to inverse oflaplace (s^2+12)/(s^4-16) is -1/2 sin(2t)-1/2 e^{-2t}+1/2 e^{2t}