What is inverse oflaplace 7/(s^2-9) ?

The answer to inverse oflaplace 7/(s^2-9) is -7/6 e^{-3t}+7/6 e^{3t}

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inverse oflaplace 7/(s^2-9)

inverse laplace

\frac{7}{s ^{2} - 9}

- \frac{7}{6} e^{- 3 t} + \frac{7}{6} e^{3 t}

Solution steps

L^{- 1} {\frac{7}{s ^{2} - 9}}

Take the partial fraction of

\frac{7}{s ^{2} - 9} : - \frac{7}{6 ( s + 3 )} + \frac{7}{6 ( s - 3 )}

= L^{- 1} {- \frac{7}{6 ( s + 3 )} + \frac{7}{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{7}{6 ( s + 3 )}} + L^{- 1} {\frac{7}{6 ( s - 3 )}}

L^{- 1} {\frac{7}{6 ( s + 3 )}} : \frac{7}{6} e^{- 3 t}

L^{- 1} {\frac{7}{6 ( s - 3 )}} : \frac{7}{6} e^{3 t}

= - \frac{7}{6} e^{- 3 t} + \frac{7}{6} e^{3 t}

\int_{- 1}^{1} f (x) d x

\int (2 x - x^{2}) d x

x \to 0 lim ([x^{2}] - [x^{3}])

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

\int x^{2} e^{x + 4} d x

What is inverse oflaplace 7/(s^2-9) ?
The answer to inverse oflaplace 7/(s^2-9) is -7/6 e^{-3t}+7/6 e^{3t}