Partial Fractions Exam: Double Root Question?

[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?

 

[Hootenanny]

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?

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.

 

[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 doesn't achive anything at all

 

[tiny-tim]

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 doesn't achive anything at all

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.

 

[franky2727]

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

 

[tiny-tim]

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

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?

 

[franky2727]

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

 

[HallsofIvy]

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

 

[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 wouldn't 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

 

[franky2727]

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

 

[tiny-tim]

no i started with dy/(v+1)²
so am i right to get ln |(v+1)²|

Nooo …

Hint: what do you have to differentiate to get -1/x²?

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

(V+1)^2

 

[franky2727]

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

 

[tiny-tim]

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

Bingo!