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)

f (x) = - 8 x^{3} - 52 x^{2} + 3

\frac{d y}{d x} = 2 y + x

\int cos^{3} (\frac{x}{17}) d x

\frac{d y}{d x}, sin (x y^{3}) = y^{3}

\int \frac{1 - x ^{2}}{4} d x

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