# Partial Fractions in Differential Equations

1. Jul 19, 2014

### checkmatechamp

I'm a little rusty with partial fractions, and I can't seem to find my error once I get up to that point.

1. The problem statement, all variables and given/known data

dy/dx = (y^2 - 1) / x

2. Relevant equations

3. The attempt at a solution

Cross-mutliply

x dy = (y^2 - 1) dx

Divide by the appropriate terms

dy / (y^2 - 1) = dx / x

So I'm integrating 1 / (y^2 - 1) dy and 1/x dy

Obviously, the integral of 1/x is ln|x| + c, but I'm having trouble with integrating the y terms. This is what I did so far.

A / (y + 1) + B/ (y - 1) = 1 / (y^2 - 1)

Ay - A + By + B = 1

Combine the y terms and constant terms

A + B = 0 (Since there's no y term)
-A + B = 1

2B = 1

B = 0.5

So I integrate 0.5 / (y - 1) and get 0.5*ln|y - 1|

So that means 0.5*ln|y - 1| = 1/|x| + c

ln|y - 1| = 2/x + c

y - 1 = e^(2/x + c)

y - 1 = e^(2/x) * e^c (which is just a constant)

y = ce^(2/x) + 1

But the thing is that I checked online to find out how to integrate the y term, and found this: http://calc101.com/partial_1.html

My question is: Why does 1 / (y + 1)(y - 1) get split up into 1/(2y - 2) and 1/(2y + 2)? Also, why are they subtracting those 2 terms rather than adding them?

2. Jul 19, 2014

### verty

You found that B = 1/2, but you forgot to find a value for A. Also notice that ${0.5 \over (y - 1)} = {1 \over (2y - 2)}$.

3. Jul 19, 2014

### checkmatechamp

Wow, I guess I saw that A + B = 0, and somehow thought A was 0.

So if B is 1/2, that means A = -1/2, so I'm integrating -0.5/(y + 1) + 0.5(y - 1), which gets me -0.5*ln|y + 1| + 0.5*ln|y - 1| = ln|x| + c

Multiplying both sides by 2 gets me:

-ln|y + 1| + ln|y - 1| = 2*ln|x| + c

Raise e to both sides.

e^(1/(y + 1)) * e^(y - 1) = e^(2*ln|x| + c)

e^(1/(y + 1)) * e^(y - 1) = e^(2*ln|x|) * e^c

e^(1/(y + 1)) * e^(y - 1) = ce^(2*ln|x|)

Is it possible to get that in explicit form? Or should I divide both sides by e^(2ln|x|) and just leave it in terms of c?

Thank you very much!

4. Jul 20, 2014

### verty

It is possible to get a closed form solution. That said, it is a little tricky.

You seemed quite eager to get rid of the logs as soon as possible, but the log form is actually easier to work with. Actually, you made a mistake in your algebra. Perhaps look over the log/exponent laws to refresh your knowledge.