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; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Union

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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) $= \frac{s + 3}{(s + 3)^{2} + 4} + \frac{6}{(s + 3)^{2} + 4}$
$L^{- 1} [F (s)]$
$= e^{- 3 t} [L^{- 1} (\frac{s}{s^{2} + 4}) + L^{- 1} (\frac{6}{s^{2} + 4})]$
$= e^{- 3 t} cos 2 t + 3 e^{- 3 t} sin 2 t$

Suggest Corrections

Similar questions

Q. The inverse Laplace transform of

F (s) = \frac{s + 3}{s^{2} + 2 s + 1}

for

t \geq 0

Find the inverse Laplace transform of

G (s) = \frac{s}{s^{2} + 2 s + 2}; σ > - 1

Q. The inverse Laplace transform of the signal

X (s) = 1 n [1 + \frac{ω^{2}}{s^{2}}]

Q. Laplace transform of

cos (ω t) i s \frac{s}{s^{2} + ω^{2}}

. The Laplace transform of

e^{- 2 t} cos (4 t)

Q. The unilateral laplace transform of

ϕ (t) = t \int_{0}^{t} β e^{- β} d β i s \frac{1}{s^{2}} + \frac{P}{(s + 1)^{3}} - \frac{1}{(s + 1)^{2}} .

Then the value of P is________.

Join BYJU'S Learning Program

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