What is inverse oflaplace 4/((s^2-9)) ?

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

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inverse oflaplace 4/((s^2-9))

inverse laplace

\frac{4}{( s ^{2} - 9 )}

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\int_{1}^{2} \frac{1}{x} - \frac{1}{x ^{2}} d x

\int \frac{y}{y ^{2} + 2 y + 1} d y

x \to 0 lim (6 x - 1)

\int \frac{- 3}{x ^{2} x ^{2} - 9} d x

\int cos (ln (7 x)) d x

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