Question

Partial Fractions Decomposition\[\frac{s-1}{[(s-1)^2+1](s^2-s)}\]

Answer 1

\[\frac{As+B}{(s-1)^2+1}+\frac{C}{s}+\frac{D}{s-1}=\frac{s-1}{[(s-1)^2+1]s(s-1)}\]

Answer 2

Is that a good start?

Answer 3

\[(As+B)s(s-1)+C[(s-1)^2+1](s-1)+Ds[(s-1)^2+1]=s-1\]

Answer 4

For s = 1:\[D=0.\]For s = 0:\[C=\frac{1}{2}.\]For s = -1:\[-2A+2B=3.\]For s = 2:\[2A+B=0.\]Therefore,\[B=1,\]\[A=-\frac{1}{2},\]and\[\frac{-1/2s+1}{(s-1)^2+1}+\frac{1/2}{s}=\frac{s-1}{[(s-1)^2+1]s(s-1)}.\]Is this correct?

Answer 5

Rewriting:\[-\frac{1}{2}\frac{s}{(s-1)^2+1}+\frac{1}{(s-1)^2+1}+\frac{1}{2}\frac{1}{s}=\frac{s-1}{[(s-1)^2+1]s(s-1)}.\]

Answer 6

thats exactly what i got but i went ahead in canceled the s-1's at the beginning

Answer 7

it looks good

Answer 8

\[\frac{1}{s(s^2-2s+2)}=\frac{A}{s}+\frac{Bs+D}{s^2-2s+2}=\frac{As^2-2As+2A+Bs^2+Ds}{s(s^2-2s+2)}\]
=>\[1=s^2(A+B)+s(-2A+D)+(2A)\]
=>
\[A+B=0; -2A+D=0; 2A=1 \]

Answer 9

A=1/2
B=-1/2
D=1

Answer 10

\[\frac{s-1}{((s-1)^2+1)(s^2-s)}\]
\[\frac{s-1}{((s-1)^2+1)\ s(s-1)}\]
\[\frac{1}{((s-1)^2+1)\ s}\]
\[\frac{1}{s(s^2-2s+1)+s}\]
\[\frac{1}{s^3-2s^2+s+s}\]
\[\frac{1}{s^3-2s^2+2s}\]
\[\frac{1}{s(s^2-2s+2)}\]
\[s=\frac{2\pm\sqrt{-4}}{2}\]so its irreducible there across the reals
\[\frac{1}{s(s^2-2s+2)}=\frac{C_1}{s}+\frac{C_2s+C_3}{s^2-2s+2}\]

Answer 11

so awesome!

Answer 12

how would you partial it across the complex and have it work out?

Answer 13

why would you want to partial it across the complex?

Answer 14

so that all you uners are linears

Answer 15

denoms ... unders

Answer 16

\[\frac{1}{s(s^2-2s+2)}=\frac{C_1}{s}+\frac{C_2}{s+(1+i)}+\frac{C_3}{s+(1-i)}\] or some such

Answer 17

i spose those should be subtractions

Answer 18

i tend to get lost in complexes, i just dont work with them enough