What is inverse oflaplace (s+4)/(s^2+6s+18) ?

The answer to inverse oflaplace (s+4)/(s^2+6s+18) is e^{-3t}cos(3t)+1/3 e^{-3t}sin(3t)

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inverse oflaplace (s+4)/(s^2+6s+18)

inverse laplace

\frac{s + 4}{s ^{2} + 6 s + 18}

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

Solution steps

L^{- 1} {\frac{s + 4}{s ^{2} + 6 s + 18}}

Expand

\frac{s + 4}{s ^{2} + 6 s + 18} : \frac{s + 3}{( s + 3 ) ^{2} + 9} + 1 \cdot \frac{1}{( s + 3 ) ^{2} + 9}

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

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

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

Refine

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

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

\int a x^{5} d x

y = x^{2}, y = 8 - x^{2}, y = 4 x + 11

x \to 3 lim ((x - 2)^{2})

\frac{d y}{d x} = \frac{k}{x ^{2}}

\int_{1}^{2} \frac{3 ln ( x )}{x ^{2}} d x

What is inverse oflaplace (s+4)/(s^2+6s+18) ?
The answer to inverse oflaplace (s+4)/(s^2+6s+18) is e^{-3t}cos(3t)+1/3 e^{-3t}sin(3t)