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 \frac{1}{x ( x - 9 ) ^{2}} d x

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

f (x) = ln (6 x^{2} + 1)

e^{5 t} t

\frac{d}{d x} (x^{2} - 6 x + 9)

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}