What is inverse oflaplace 1/(2s^2+6s+4) ?

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

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inverse oflaplace 1/(2s^2+6s+4)

inverse laplace

\frac{1}{2 s ^{2} + 6 s + 4}

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

Solution steps

L^{- 1} {\frac{1}{2 s ^{2} + 6 s + 4}}

Take the partial fraction of

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

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

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

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

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

\frac{d}{d x} (\frac{2 ( x ^{2} - 9 )}{x ^{2} - 4})

n = 0 \sum \infty \frac{n !}{8 ^{n}}

(\frac{sin ( x )}{cos ( x )})^{^{'}}

y^{^{''}} + 6 y^{^{'}} + 11 y = 0

\int (2 - cos (x))^{2} d x

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