What is inverse oflaplace (3s+3)/(s^2-9) ?

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

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inverse oflaplace (3s+3)/(s^2-9)

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

16 y^{^{''}} - 24 y^{^{'}} + 9 y = 0

\frac{d ^{2}}{d x ^{2}} (x^{\frac{1}{7}})

f (x) = 2 x^{3} - 4 x - 8 x + 6

\frac{\partial}{\partial y} (e^{3 x + 4 y} cos (5 z))

y^{^{'}} - a y (t) + b = 0

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