What is inverse oflaplace 2/(3(s+1)^2-27) ?

The answer to inverse oflaplace 2/(3(s+1)^2-27) is 1/9 e^{2t}-1/9 e^{-4t}

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inverse oflaplace 2/(3(s+1)^2-27)

inverse laplace

\frac{2}{3 ( s + 1 ) ^{2} - 27}

\frac{1}{9} e^{2 t} - \frac{1}{9} e^{- 4 t}

Solution steps

L^{- 1} {\frac{2}{3 ( s + 1 ) ^{2} - 27}}

Take the partial fraction of

\frac{2}{3 ( s + 1 ) ^{2} - 27} : \frac{1}{9 ( s - 2 )} - \frac{1}{9 ( s + 4 )}

= L^{- 1} {\frac{1}{9 ( s - 2 )} - \frac{1}{9 ( 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}{9 ( s - 2 )}} - L^{- 1} {\frac{1}{9 ( s + 4 )}}

L^{- 1} {\frac{1}{9 ( s - 2 )}} : \frac{1}{9} e^{2 t}

L^{- 1} {\frac{1}{9 ( s + 4 )}} : \frac{1}{9} e^{- 4 t}

= \frac{1}{9} e^{2 t} - \frac{1}{9} e^{- 4 t}

\int \frac{x + 5}{x ^{2} + 9} d x

\int \frac{x - 1}{( x - 2 ) ( x - 3 )} d x

y^{^{'}} + e^{x} y = 7 e^{x}

(7 x + 1) (x^{2} - 6)

\frac{d}{d y} (a^{2} - y^{2})

What is inverse oflaplace 2/(3(s+1)^2-27) ?
The answer to inverse oflaplace 2/(3(s+1)^2-27) is 1/9 e^{2t}-1/9 e^{-4t}