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

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

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inverse oflaplace s/(s^2-9)

inverse laplace

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

\frac{1}{2} e^{- 3 t} + \frac{1}{2} e^{3 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= \frac{1}{2} e^{- 3 t} + \frac{1}{2} e^{3 t}

\int_{0}^{1} - e^{- x} d x

\frac{d y}{d x} = 2 x + 3

x \to 0 - lim (\frac{2}{3 x ^{\frac{1}{3}}})

\int tan (4 θ) d θ

f (x) = 4 e^{x}

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