- #1
zodiacbrave
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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
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