What is inverse oflaplace (2s+12)/(s^2+6s+18) ?

The answer to inverse oflaplace (2s+12)/(s^2+6s+18) is 2e^{-3t}cos(3t)+2e^{-3t}sin(3t)

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inverse oflaplace (2s+12)/(s^2+6s+18)

inverse laplace

\frac{2 s + 12}{s ^{2} + 6 s + 18}

2 e^{- 3 t} cos (3 t) + 2 e^{- 3 t} sin (3 t)

Solution steps

L^{- 1} {\frac{2 s + 12}{s ^{2} + 6 s + 18}}

Expand

\frac{2 s + 12}{s ^{2} + 6 s + 18} : 2 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 9} + 6 \cdot \frac{1}{( s + 3 ) ^{2} + 9}

= L^{- 1} {2 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 9} + 6 \cdot \frac{1}{( s + 3 ) ^{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 + 3}{( s + 3 ) ^{2} + 9}} + 6 L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 9}}

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

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

= 2 e^{- 3 t} cos (3 t) + 6 e^{- 3 t} \frac{1}{3} sin (3 t)

Refine

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

= 2 e^{- 3 t} cos (3 t) + 2 e^{- 3 t} sin (3 t)

\int \frac{5 x ^{3} - 4 x}{1 - x ^{4}} d x

\int_{0}^{1} \frac{x - 7}{x ^{2} - 5 x + 6} d x

y^{^{'}} + 6 y = e^{4 t}

\frac{d}{d x} (sec^{5} (x^{2} - 1))

3 x^{2} + 1

What is inverse oflaplace (2s+12)/(s^2+6s+18) ?
The answer to inverse oflaplace (2s+12)/(s^2+6s+18) is 2e^{-3t}cos(3t)+2e^{-3t}sin(3t)