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{( x ^{3} )}{3} + x

\int_{8}^{11} 2 arccot (x) d x

t^{2} e^{- 2 t} sin (2 t)

\frac{\partial}{\partial x} (2 x + y)

7 y^{^{'}} - 14 y = 0

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