Question

Please helpme wid inverse Laplace of

(6s^3+4s^2+16)/(s^5+4s^3)

thanx

Answer 1

factor s^3 out of the bottom and then use partial fractions

answer: 2t^2+3sin2t

Answer 2

How factor s^3 out of the bottom?

Answer 3

How can you not, there's a common factor of s^3

Answer 4

OKKK! 2(3s^3+2s^2+8)/(s^2+4)s^3

Answer 5

then use partials and proceed as normal

Answer 6

2(3s^3+2s^2+8)=A(S^2+4)+B(S^3) is correct?

Answer 7

I'm assuming that's your set up of the partial and you haven't solved the partial fraction yet. If so, that set up is incorrect because s^2+4 is a irreducible quadratic and s^3 is repeated 3 times.. Though, when you actually solve the partial fraction you will find the other constants are actually 0 so it is indeed left in the form you put it as, but I believe you do need to solve for those 0 constants in order to find your remaining 2 missing coefficients anyways

Answer 8

(6s^3+4s^2+16)=A(S^2+4)+B(S^3); A=4, B=6

(6s^3+4s^2+16)/(s^5+4s^3)= 4/S^3 + 6/(S^2+4)

Answer 9

yeah that's right not take inverse laplace

Answer 10

simplifying and inv I have to get your answer =2t^2+3sin2t, right????????? thankuuuuuu

Answer 11

yeah