What is inverse oflaplace (2s-6)/(s^2-9) ?

The answer to inverse oflaplace (2s-6)/(s^2-9) is 2cosh(3t)-2sinh(3t)

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inverse oflaplace (2s-6)/(s^2-9)

inverse laplace

\frac{2 s - 6}{s ^{2} - 9}

2 cosh (3 t) - 2 sinh (3 t)

Solution steps

L^{- 1} {\frac{2 s - 6}{s ^{2} - 9}}

Expand

= L^{- 1} {\frac{2 s}{s ^{2} - 9} - \frac{6}{s ^{2} - 9}}

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)}

= 2 L^{- 1} {\frac{s}{s ^{2} - 9}} - 6 L^{- 1} {\frac{1}{s ^{2} - 9}}

L^{- 1} {\frac{s}{s ^{2} - 9}} : cosh (3 t)

L^{- 1} {\frac{1}{s ^{2} - 9}} : \frac{1}{3} sinh (3 t)

= 2 cosh (3 t) - 6 \cdot \frac{1}{3} sinh (3 t)

Refine

2 cosh (3 t) - 6 \frac{1}{3} sinh (3 t) : 2 cosh (3 t) - 2 sinh (3 t)

= 2 cosh (3 t) - 2 sinh (3 t)

x \to 0 lim (\frac{( sin ( 8 x ) )}{3 x})

\frac{d}{d x} (cos (3 x + 3))

x \to - 2 + lim (\frac{2 x + 1}{2 + x})

cos (x^{2}) \cdot 2 x

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What is inverse oflaplace (2s-6)/(s^2-9) ?
The answer to inverse oflaplace (2s-6)/(s^2-9) is 2cosh(3t)-2sinh(3t)