What is inverse oflaplace 1/(s^2-6s+18) ?

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

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inverse oflaplace 1/(s^2-6s+18)

inverse laplace

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

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

Solution steps

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

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

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s - 3 ) ^{2} + 9} : a = 3, F (s) = \frac{1}{s ^{2} + 9}

= e^{3 t} L^{- 1} {\frac{1}{s ^{2} + 9}}

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

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

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