Partical Fractions, Integration and Equating coefficients

[zodiacbrave]

Homework Statement

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

Homework Equations

1/(w^4-w^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

so,

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

 

[Hurkyl]

The book gives me a different answer..

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

I am pretty sure I messed up, probably at the start with factoring, can someone please help?

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...