What is inverse oflaplace (3(s+3))/(s^2+6s+8) ?

The answer to inverse oflaplace (3(s+3))/(s^2+6s+8) is 3/2 e^{-2t}+3/2 e^{-4t}

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inverse oflaplace (3(s+3))/(s^2+6s+8)

inverse laplace

\frac{3 ( s + 3 )}{s ^{2} + 6 s + 8}

\frac{3}{2} e^{- 2 t} + \frac{3}{2} e^{- 4 t}

Solution steps

L^{- 1} {\frac{3 ( s + 3 )}{s ^{2} + 6 s + 8}}

Take the partial fraction of

\frac{3 ( s + 3 )}{s ^{2} + 6 s + 8} : \frac{3}{2 ( s + 2 )} + \frac{3}{2 ( s + 4 )}

= L^{- 1} {\frac{3}{2 ( s + 2 )} + \frac{3}{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{3}{2 ( s + 2 )}} + L^{- 1} {\frac{3}{2 ( s + 4 )}}

L^{- 1} {\frac{3}{2 ( s + 2 )}} : \frac{3}{2} e^{- 2 t}

L^{- 1} {\frac{3}{2 ( s + 4 )}} : \frac{3}{2} e^{- 4 t}

= \frac{3}{2} e^{- 2 t} + \frac{3}{2} e^{- 4 t}

x \to \infty lim (3^{x} - x)

n = 1 \sum \infty \frac{n}{3 n + 2}

\frac{\partial}{\partial x} (y^{5} cos (6 x) \cdot 6)

\frac{d}{d x} (\frac{1 + ln ( x )}{1 - ln ( x )})

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

What is inverse oflaplace (3(s+3))/(s^2+6s+8) ?
The answer to inverse oflaplace (3(s+3))/(s^2+6s+8) is 3/2 e^{-2t}+3/2 e^{-4t}