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

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

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

inverse laplace

\frac{( 18 )}{s ( s ^{2} + 6 s + 18 )}

H (t) - e^{- 3 t} cos (3 t) - e^{- 3 t} sin (3 t)

Solution steps

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

Take the partial fraction of

\frac{18}{s ( s ^{2} + 6 s + 18 )} : \frac{1}{s} + \frac{- s - 6}{s ^{2} + 6 s + 18}

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

Expand

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

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

L^{- 1} {\frac{1}{s}} : H (t)

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)

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

Refine

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

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

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