What is inverse oflaplace 1/(s^2+6s+8) ?

The answer to inverse oflaplace 1/(s^2+6s+8) is 1/2 e^{-2t}-1/2 e^{-4t}

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inverse oflaplace 1/(s^2+6s+8)

inverse laplace

\frac{1}{s ^{2} + 6 s + 8}

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

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 6 s + 8}}

Take the partial fraction of

\frac{1}{s ^{2} + 6 s + 8} : \frac{1}{2 ( s + 2 )} - \frac{1}{2 ( s + 4 )}

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

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

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

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

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

- x^{2} + 4 x, x^{2} - 2 x

f (x) = 8 e^{x} cos (x)

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

f (x) = - x^{2} - 4 x + 1, at x = - 4

What is inverse oflaplace 1/(s^2+6s+8) ?
The answer to inverse oflaplace 1/(s^2+6s+8) is 1/2 e^{-2t}-1/2 e^{-4t}