The inverse Laplace transform of s-9s2-6s-13 is

F(s)=s+9s^2+6s+13 =s+3+6(s+3)^2+4 F(s)=s+3(s+3)^2+4+6(s+3)^2+4 L^ 1[F(s)] =e^ 3t [ L^ 1 ( ss^2+4 )+L^ 1 ( 6s^2+4 ) ] =e^ 3tcos 2t+3e^ 3tsin 2t nbsp; ...

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The inverse L...

Question

The inverse Laplace transform of $\frac{s + 9}{s^{2} + 6 s + 13}$ is

cos 2 t + 9 sin 2 t

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e^{- 3 t} cos 2 t - 3 e^{- 3 t} sin 2 t

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

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

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Solution

The correct option is D $e^{- 3 t} cos 2 t + 3 e^{- 3 t} sin 2 t$
F(s) $= \frac{s + 9}{s^{2} + 6 s + 13}$
$= \frac{s + 3 + 6}{(s + 3)^{2} + 4}$
F(s) $= \frac{s + 3}{(s + 3)^{2} + 4} + \frac{6}{(s + 3)^{2} + 4}$
$L^{- 1} [F (s)]$
$= e^{- 3 t} [L^{- 1} (\frac{s}{s^{2} + 4}) + L^{- 1} (\frac{6}{s^{2} + 4})]$
$= e^{- 3 t} cos 2 t + 3 e^{- 3 t} sin 2 t$

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The inverse Laplace transform of s+9s2+6s+13 is

The correct option is D e−3tcos2t+3e−3tsin2tF(s)=s+9s2+6s+13 =s+3+6(s+3)2+4 F(s)=s+3(s+3)2+4+6(s+3)2+4 L−1[F(s)] =e−3t[L−1(ss2+4)+L−1(6s2+4)] =e−3tcos2t+3e−3tsin2t

The inverse Laplace transform of $\frac{s + 9}{s^{2} + 6 s + 13}$ is