Question

laplace please

Answer 1

\[\mathcal L\]

Answer 2

@blessingkay please post your question

Answer 3

i think u mean inverse laplace of
\[\large \frac{3s+1}{s^2+6s+13}\]

Answer 4

just notice that
\[\frac{3s+1}{s^2+6s+13}=\frac{3(s+3)-8}{(s+3)^2+4}=3 \frac{(s+3)}{(s+3)^2+4}-8\frac{1}{(s+3)^2+4}\]

Using the Frequency shifting property of Laplace transform :

\[\mathscr{L}^{-1} [3 \frac{(s+3)}{(s+3)^2+4}-8\frac{1}{(s+3)^2+4}]= 3 e^{-3t} \cos 2t-8 e^{-3t} \sin 2t\]

Frequency shifting

\[\mathscr{L} [e^{at} f(t)]= F(s-a)\]

Answer 5

are u there?

Answer 6

@mukushia are u there

Answer 7

im here brother

Answer 8

i need final answer

Answer 9

final answer is up there 3e^(-3t) cos 2t ....

Answer 10

@mukushla are u there?

Answer 11

i need more explanation