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

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

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

inverse laplace

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

2 e^{- 3 t} - e^{3 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= 2 e^{- 3 t} - e^{3 t}

\frac{d}{d x} (tan (4 x))

\frac{d y}{d x} + \frac{y}{x - 2} = 5 (x - 2) y^{\frac{1}{2}}

f (x) = \frac{6 x ^{2} - 1}{2 x}

\frac{d}{d x} (\frac{1}{2 x ^{2} + 1})

\int (x^{2} - 7 x + 3) d x

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