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

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

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

inverse laplace

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

- \frac{1}{8} e^{- 2 t} - \frac{e ^{- 2 t} t}{2} + \frac{1}{8} e^{2 t}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {- \frac{1}{8 ( s + 2 )} - \frac{1}{2 ( s + 2 ) ^{2}} + \frac{1}{8 ( 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}{8 ( s + 2 )}} - L^{- 1} {\frac{1}{2 ( s + 2 ) ^{2}}} + L^{- 1} {\frac{1}{8 ( s - 2 )}}

L^{- 1} {\frac{1}{8 ( s + 2 )}} : \frac{1}{8} e^{- 2 t}

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

L^{- 1} {\frac{1}{8 ( s - 2 )}} : \frac{1}{8} e^{2 t}

= - \frac{1}{8} e^{- 2 t} - \frac{e ^{- 2 t} t}{2} + \frac{1}{8} e^{2 t}

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

\frac{d}{d θ} (4 + 4 sin (θ))

\frac{d}{d x} (x - 3 x^{\frac{1}{3}})

\frac{d}{d x} (cos (3 x) \cdot 3)

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

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