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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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