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

The answer to inverse oflaplace (2s-6)/(s^2+9) is 2cos(3t)-2sin(3t)

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

inverse laplace

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

2 cos (3 t) - 2 sin (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}} : cos (3 t)

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

= 2 cos (3 t) - 6 \cdot \frac{1}{3} sin (3 t)

Refine

2 cos (3 t) - 6 \frac{1}{3} sin (3 t) : 2 cos (3 t) - 2 sin (3 t)

= 2 cos (3 t) - 2 sin (3 t)

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

\frac{d}{d x} (\frac{2 tan ( x )}{x})

n = 0 \sum \infty \frac{1}{5 n + 1}

2 x^{3} - 4 x + 6

\frac{\partial}{\partial x} (5 - 3 x + 6 y - 2 x^{2} y)

What is inverse oflaplace (2s-6)/(s^2+9) ?
The answer to inverse oflaplace (2s-6)/(s^2+9) is 2cos(3t)-2sin(3t)