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

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

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

inverse laplace

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

4 e^{- 2 t} - 4 e^{- 4 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= 4 e^{- 2 t} - 4 e^{- 4 t}

\frac{d}{d x} (lo g_{10} ((arcsin (x))^{2}))

\int_{- 1}^{1} (x^{2} - 2 x + 3) d x

\frac{\partial}{\partial x} (y^{4} - 6 x y)

y^{^{'}} + 4 y = e^{3 x}

x \to 0 lim (\frac{1}{1})

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