What is inverse oflaplace 1/(s^2-9) ?

The answer to inverse oflaplace 1/(s^2-9) is -1/6 e^{-3t}+1/6 e^{3t}

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inverse oflaplace 1/(s^2-9)

inverse laplace

\frac{1}{s ^{2} - 9}

- \frac{1}{6} e^{- 3 t} + \frac{1}{6} e^{3 t}

Solution steps

L^{- 1} {\frac{1}{s ^{2} - 9}}

Take the partial fraction of

\frac{1}{s ^{2} - 9} : - \frac{1}{6 ( s + 3 )} + \frac{1}{6 ( s - 3 )}

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

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

L^{- 1} {\frac{1}{6 ( s + 3 )}} : \frac{1}{6} e^{- 3 t}

L^{- 1} {\frac{1}{6 ( s - 3 )}} : \frac{1}{6} e^{3 t}

= - \frac{1}{6} e^{- 3 t} + \frac{1}{6} e^{3 t}

(2 x - 4)^{4} (x^{2} + x + 1)^{5}

\int \frac{6 x ^{2}}{x ^{3} - 1} d x

3 s

y = 7 x + 5

9 (x^{3} - x)^{4}

What is inverse oflaplace 1/(s^2-9) ?
The answer to inverse oflaplace 1/(s^2-9) is -1/6 e^{-3t}+1/6 e^{3t}