# Partial fractions

1. Aug 16, 2008

### franky2727

got an exam coming up in a few days and half way through my question i ran into a partial fractions question instead of having the standard (1/(y+c)(y+d))= A/(y+c) + B(y+d) and multiplying out i had a double root so (1/(y+c)(y+c)) does this change the way i go about the question and are there any other similar situations that i am likely to run into?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 16, 2008

### Hootenanny

Staff Emeritus
Although it shouldn't affect the method you use, it will affect the form of your solution. In this case your solution will be of the form:

$$\frac{1}{\left(y+c\right)^2} = \frac{A}{y+c\right} + \frac{B}{\left(y+c\right)^2}$$

i.e. a linear denominator followed by a quadratic denominator. You should then follow your usual technique (e.g. equating coefficients) to solve for A and B.

3. Aug 16, 2008

### franky2727

are you sure this is right cos this gives me A=0 and B=1 giving me 1/(y+c)^2=1/(y+c)^2 which doesnt achive anything at all

4. Aug 16, 2008

### tiny-tim

Hi franky2727!

(since Hootenanny is offline …)

This is really for fractions like (P + Qy)/(y + c)2,

and it gives you partial fractions of the form A/(y + c) + B/(y + c)2, without any y on top.

In this case, you started without any y on top, so indeed you didn't achieve anything at all, because 1/(y + c)2 was already the simplest answer.

5. Aug 16, 2008

### franky2727

Ahh.. ok so i just end up with Ln|(y+c)^2|?

6. Aug 16, 2008

### tiny-tim

erm … if you started with ∫dy/y2 (you didn't say), then nooo …

ln(y + c)2 is just 2 ln(y + c), isn't it?

7. Aug 16, 2008

### franky2727

i feal so stupid :P log laws a bit rusty to say the least

8. Aug 17, 2008

### HallsofIvy

the log law is not the problem here.
$$\int \frac{1}{y^2}dy= \int y^{-2} dy$$
is NOT a logarithm.

9. Aug 17, 2008

### franky2727

no i started with dy/(v+1)2
so am i right to get ln |(v+1)2| and does that go to 2ln|V+1| i thought it wouldnt because the squared bit is inside the modulus not outside or does that not matter because squaring something gets rid of the need for a modulus anyway

10. Aug 17, 2008

### franky2727

so what does this end up as when i e-it-out (V+1)^2 or 2(V+1) or something different

11. Aug 17, 2008

### tiny-tim

Nooo …

Hint: what do you have to differentiate to get -1/x2?
(V+1)^2

12. Aug 17, 2008

### franky2727

is this even ment to be in logs ? 1/((V+1)^2 can be (V+1)^-2 which goes to (-1/V+1)?????

13. Aug 17, 2008

Bingo!