Calculus Help Wanted:Antiderivative using Partial Fractions

sidhlyn · #1 $Old$ December 7th, 2008, 11:28 PM

I have spent hours on this problem. We were not instructed on how to solve when there are 3 parts.

Problem:
find the antideravitive for: 8dy/(y^3-4y)
which I think breaks down to: 8=A(y^2-4)+(By+C)y right?
So, A= -2 when y=0, but I cannot for the life of me figure out B & C.
?? C=4-2B & B=2-.5C ??

Thanks for any assistance anyone can offer.
I think I'll need step by step advice for isolating these variables.
My book does not offer any explanation what so ever.

Stumped again,
Sidhlyn

Jameson · #2 $Old$ December 8th, 2008, 01:59 AM

Your setup looks fine to me. You have two equations with two variables, B and C. Use whatever method of solving a system you like, but you should be able to solve this. Substitution? Matrix multiplication? Surely you remember how to solve a two variable system.

If not, no worries I'll help.

sidhlyn · #3 $Old$ December 8th, 2008, 07:09 AM

Re:
"Surely you remember how to solve a two variable system."

Hello, and thank you. However, no....I do not remember exactly how to do that. I am not a fresh out of high-school student. I tested into Pre-calculus when returning to college, then was able to fare very well in Pre-calculus and in Calculus 1.
However, it is situations like this that I become stumped over, because it requires a recall reaching back over 15 yrs.
I was able to find out how to solve with substitution via online tutorial, however....I don't understand what to do with this result: C=C or 0=0.
This still doesn't tell me what to use for C. I get the same result if I start by solving for B: B=B or 0=0. What do I do with this?
I know this might seem elementary, but please consider that I've been out of practice for around 15 years up until this summer when I started up with Pre-calculus. Something like this also drives me crazy, because I know it is probably very simple. :-(

Thank you,
Lynn.

Krizalid · #4 $Old$ December 8th, 2008, 08:15 AM

Let's talk about $\frac{1}{y^{3}-4y}.$

Since $y^3-4y=y(y^2-4),$ we have $\frac{1}{y^{3}-4y}=\frac{1}{y\left( y^{2}-4 \right)}.$ Now "let's use the numerator against denominator."

Observe that $\frac{1}{y\left( y^{2}-4 \right)}=\frac{1}{4}\cdot \frac{4}{y\left( y^{2}-4 \right)}=\frac{1}{4}\cdot \frac{y^{2}-\left( y^{2}-4 \right)}{y\left( y^{2}-4 \right)}=\frac{1}{4}\left( \frac{y}{y^{2}-4}-\frac{1}{y} \right).$

And this is easy to integrate.

sidhlyn · #5 $Old$ December 8th, 2008, 08:44 AM

This very frustrating.

I am having difficulty solving the 2 variable problem with:
8=A(y^2-4)+(By+C)y
I know how to integrate this once I figure out what B & C are.
I'm having an issue with the "elementary" aspect of this problem, not the calculus application.
Please....can someone please help me with what to do with:
C=C or 0=0 aka B=B or 0=0??

I have 8dy/(y^3-4y) to integrate which means:
8dy/(y^3-4y)= A/y + (By+C)/(y^2-4) correct?
So, I have 8=A(y^2-4)+(By+C)y.
I can solve for A by assigning "0" to y. Then A= -4.

My problem is solving for B and C when y=2 (so that A drops out).

Am I going about solving this antiderivative wrong?

Sorry...I don't know how to make my keyboard write out problems in the way they should actually appear.

Thank you to whomever can assist.

sidhlyn · #6 $Old$ December 8th, 2008, 11:59 AM

This very frustrating.
I'm moving this post over from the Calculus section, because it is still unresolved & this is still driving me bananas while I'm sitting at work.

1.)I am having difficulty solving the 2 variable problem with:
8=A(y^2-4)+(By+C)y
The original problem is to find the antiderivative for: 8dy/(y^3-4y) using partial fractions.
I have come up with:
8dy/(y^3-4y)= A/y + (By+C)/(y^2-4) correct?
So, then I have 8=A(y^2-4)+(By+C)y.
I can solve for A by assigning "0" to y. Then A= -4.

My problem is solving for B and C when y=2 (so that A drops out).

I know how to integrate this once I figure out what B & C are.
I'm having an issue with the "elementary" aspect of this problem, not the calculus application. I am out of practice with more of the basic applicatioins like this than more complicated ones, because I just started back with Pre-calculus after a 15 year hiatus away from any mathematical academic exposure.
Please....can someone please help me with what to do with:
C=C or 0=0 aka B=B or 0=0??

Am I going about solving this antiderivative wrong?
The answer is supposed to be: ln[(y^2-4)/y^2] + C

2.)
The other one I'm getting an answer of: 2*ln|5-x| + 2*ln|5+x| + C for the antiderivative problem: 20/(25-x^2)*dx.
The instructor gave the answer as: -2*ln|5-x| + 2*ln|5+x| + C.
What am I doing wrong yeilding a positive instead of the negative he has listed?
My work:
Partial fractions into: A/(5+x) + B/(5-x)
Gives: 20=A(5+x)+B(5-x) to solve for.
With x= -5, 20=A(0)+B(10) yields B=2,
then with x=5, 20=A(10)+B(0) yields A=2 also.
Then I work the antiderivative for 2dx/(5-x) + 2dx/(5+x) pulling the 2 out front of each. So, where is the negative 2 coming from?

Sorry...I don't know how to make my keyboard write out problems in the way they should actually appear.

Thank you oodles to whomever can assist!

P.S. I had 5 pages of homework with different antiderivitive applications and only had issues with 3 of the problems.
Believe me...I have gone over and over these + one other word problem literally for HOURS & cannot get to the right answers!
Anyone who loves puzzles must understand how extremely frustrating this is for me. I work full time, so I can't just drop by the school for assistance with this.

running-gag · #7 $Old$ December 8th, 2008, 12:54 PM

Hi

If I understood correctly your problem is to find a primitive of 8dy/(y^3-4y)

Here is something that can help you

$\frac{8}{y^3 - 4y} = -\frac{2}{y} + \frac{1}{y+2}+ \frac{1}{y-2}$

sidhlyn · #8 $Old$ December 8th, 2008, 12:58 PM

Yes...find the antiderivative for: 8dy/(y^3-4y)

But- how did you get B=1 & C=1??

Thanks for your attention.

running-gag · #9 $Old$ December 8th, 2008, 01:22 PM

y²-4 = (y-2)(y+2)

Therefore

$\frac{8}{y^3 - 4y} = \frac{a}{y} + \frac{b}{y+2}+ \frac{c}{y-2}$

To find a, b and c you just have to use the same denominator

$\frac{8}{y^3 - 4y} = \frac{a(y^2-4) + by(y-2) + cy(y+2)}{y^3 - 4y}$

$\frac{8}{y^3 - 4y} = \frac{(a+b+c)y^2 + 2(c-b)y - 4a}{y^3 - 4y}$

This leads to the system
a+b+c=0
c-b=0
-4a=8

sidhlyn · #10 $Old$ December 8th, 2008, 02:09 PM

Gosh....thank you so much....
But....
"This leads to the system
a+b+c=0
c-b=0
-4a=8 "
is the very part I don't get.
I don't know how you got these figures. ?
I don't understand how to get to (A+B+C)y^2 + 2(C-B)y-4A?
Where did the extra A, B, & C come from.
I'm sorry....our instructer did not show us this way, but the textbook has it set up like how you've explained.
The textbook does not break down HOW to do that either.
They just print it as you did & then gives the "systems." How do you get to the systems?

I did get the antiderivative finally, but I really WANT to know HOW to do those necessary middle steps that I'm missing.
LOL- I think I was going about it the "long" way and was able to come up with A= -4 & C=1, but got stumped on solving for B because there's no way to zero out the C with "y^2= -2y" from: 8=A(y^2-4)=B(y^2-2y)+C(y^2+2y) with plugging in "y" values as we were shown by our instructor.

When you get the chance, can I please have the breakdown on how to get those "systems?" <Anyone out there who has the patience to list that out- please?> I appreciate your help & can use the info provided on my homework, but I have just got to know how to do that. I'm sure it's probably something simple I'm not seeing.

Thanks a million!!

CaptainBlack · #11 $Old$ December 8th, 2008, 02:52 PM

Quote:

Originally Posted by sidhlyn $View Post$

I have spent hours on this problem. We were not instructed on how to solve when there are 3 parts.

Problem:
find the antideravitive for: 8dy/(y^3-4y)
which I think breaks down to: 8=A(y^2-4)+(By+C)y right?
So, A= -2 when y=0, but I cannot for the life of me figure out B & C.
?? C=4-2B & B=2-.5C ??

Thanks for any assistance anyone can offer.
I think I'll need step by step advice for isolating these variables.
My book does not offer any explanation what so ever.

Stumped again,
Sidhlyn

Try finding $A,B$ and $C$ such that:

$\frac{8}{y^3-4y}=\frac{A}{y}+\frac{B}{y-2}+\frac{C}{y+2}$

CB

mr fantastic · #12 $Old$ December 8th, 2008, 04:39 PM

Quote:

Originally Posted by sidhlyn $View Post$

Gosh....thank you so much....
But....
"This leads to the system
a+b+c=0
c-b=0
-4a=8 "
is the very part I don't get.
I don't know how you got these figures. ?
I don't understand how to get to (A+B+C)y^2 + 2(C-B)y-4A?
Where did the extra A, B, & C come from.
I'm sorry....our instructer did not show us this way, but the textbook has it set up like how you've explained.
The textbook does not break down HOW to do that either.
They just print it as you did & then gives the "systems." How do you get to the systems?

I did get the antiderivative finally, but I really WANT to know HOW to do those necessary middle steps that I'm missing.
LOL- I think I was going about it the "long" way and was able to come up with A= -4 & C=1, but got stumped on solving for B because there's no way to zero out the C with "y^2= -2y" from: 8=A(y^2-4)=B(y^2-2y)+C(y^2+2y) with plugging in "y" values as we were shown by our instructor.

When you get the chance, can I please have the breakdown on how to get those "systems?" <Anyone out there who has the patience to list that out- please?> I appreciate your help & can use the info provided on my homework, but I have just got to know how to do that. I'm sure it's probably something simple I'm not seeing.

Thanks a million!!

Thread closed to prevent further deletion of questions.