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} (sin^{4} (2 x co) s^{4} 3 x)

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

\frac{\partial}{\partial y} (5 x + 4 y)

\frac{d}{d x} (\frac{7}{x + 1})

\int_{1}^{5} w^{2} ln (w) d w

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