
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!

Partical Fractions, Integration and Equating coefficients

  1. Feb 6, 2009 #1

    Split the function into partial fractions. 1/(w^4-w^3)

    2. Relevant equations


    3. The attempt at a solution

    I started by factoring the denominator to w^3(w-1) and re-writing the original function as

    (Aw^2+Bw+C)/w^3 + D/(w-1) and set it = 1/(w^3(w-1))

    I end up with 1 = (Aw^2+Bw+C)(w-1)+Dw^3

    if I set w = 0 then,

    -1 = C

    if i set w = 1 then,
    1 = D

    then I start organizing everything and I end up with,

    1 = [A + D]w^3 + [B-A]w^2 + [C-B]w - C


    0 = A + D
    0 = B - A
    0 = C - B

    since I know what D and C are,

    A = -1

    B = -1

    so my final answer is (-w^2 -w - 1) / (w^3) + 1/(w-1)

    The book gives me a different answer.. I am pretty sure I messed up, probably at the start with factoring, can someone please help?

    The book gives the answer to be 1/(w-1) -1/w - 1/w^2 - 1/w^3

    Thank you
    Last edited: Feb 6, 2009
  2. jcsd
  3. Feb 6, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What was the book's answer? Are you sure it's actually different?

    It's easy enough to check -- simply plug in a few values of w and see if your solution has the same value as the original fraction. Of course, this isn't a proof*, but if you made a mistake, it's extremely likely you'll catch it this way....

    *: Actually, it is a proof if you do it right and know the relevant theorems. But I digress....

Similar Discussions: Partical Fractions, Integration and Equating coefficients
  1. Partical Fractions (Replies: 7)