What is inverse oflaplace 1/(s^2+6s+34) ?

The answer to inverse oflaplace 1/(s^2+6s+34) is e^{-3t} 1/5 sin(5t)

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inverse oflaplace 1/(s^2+6s+34)

inverse laplace

\frac{1}{s ^{2} + 6 s + 34}

e^{- 3 t} \frac{1}{5} sin (5 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 6 s + 34}}

\frac{1}{s ^{2} + 6 s + 34} = \frac{1}{( s + 3 ) ^{2} + 25}

= L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 25}}

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} + 25} : a = - 3, F (s) = \frac{1}{s ^{2} + 25}

= e^{- 3 t} L^{- 1} {\frac{1}{s ^{2} + 25}}

L^{- 1} {\frac{1}{s ^{2} + 25}} : \frac{1}{5} sin (5 t)

= e^{- 3 t} \frac{1}{5} sin (5 t)

1 - e^{- t}

cos (x), sin (x), [0, \frac{π}{4}]

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What is inverse oflaplace 1/(s^2+6s+34) ?
The answer to inverse oflaplace 1/(s^2+6s+34) is e^{-3t} 1/5 sin(5t)