Question

how to turn \[\frac{1}{s^3(s^2+1)}\] into partial fractions? im having difficulties :/

Answer 1

\[\frac1{s^3(s^2+1)}=\frac As+\frac B{s^2}+\frac C{s^3}+\frac{Dx+E}{s^2+1}\]so far okay @lgbasallote ?

Answer 2

wait...in partial fractions dont i do \[\frac As + \frac{B}{s^2} + \frac{C}{s^3}?\]

Answer 3

yes @TuringTest that's what i did

Answer 4

so you agree with my line lgb\[\frac1{s^3(s^2+1)}=\frac As+\frac B{s^2}+\frac C{s^3}+\frac{Dx+E}{s^2+1}\]now multiply both sides by \(s^3(s^2+1)\) and see what you get

Answer 5

i got As^2 (s^2 + 1)( + Bs(s^2 + 1) + C(s^2 + 1) + Ds (s^3) + Es^3

Answer 6

=1

Answer 7

yep, now collect like terms

Answer 8

by root sub i got when s = 0; C = 1

after that i got nothing

Answer 9

i tried gaussian-ing...but no juice

Answer 10

it's one or the other, and since you have most variables attatched to an s you really do have to do the gaussian way I think

Answer 11

\[As^4+ As^2+ Bs^3 + Bs+ Cs^2 + C + Ds^4 + Es^3=1\]let me know if you see a mistake...

Answer 12

yup same as mine...

Answer 13

i now got D = 0

Answer 14

\[(A+D)s^4+ (B+E)s^3+ (A+C)s^2 + Bs + C =1\]still the same?

Answer 15

wait...how did you get ss^4 in a o.O

Answer 16

uhh darn messy solution..i see it now

Answer 17

okay now i have A = -1 and D = 1

im stuck with B and E

Answer 18

let me try to reason it out
C=1
A+C=0 -> A=-1
A+D=0 -> D=1
B=0 (since the coefficient of s is 0)
so
B+E=0 -> E=0

Answer 19

.....oh

Answer 20

of course...

Answer 21

you didn't realize that you knew B
I've made that mistake wayyyyyy to many times to do it again !
(that's what I tell myself every time at least)

Answer 22

http://www.wolframalpha.com/input/?i=laplace+transform+-2%2Bt^2%2B2+cos%28t%29