# Long Division to Integrate

1. Feb 15, 2012

### solomon684

1. The problem statement, all variables and given/known data
The problem is about solving the homogenous differential equation (x2 + y2)dx + (x2 - xy)dy = 0 using substitution, in this case y=ux. This is the example they go through in the textbook (A First Course in Differentia Equations with Modeling Applications, 9th Edition, Zill, Example 1 Section 2.5). Anyway, that stuff does not really matter.

You reach this equation and need to integrate it: [(1 - u)/(1+u)]du + dx/x = 0
The next equation is [1 - (2/(1+u))]du + dx/x, and it says they got there using long division. I cannot figure out how to go from [(1 - u)/(1+u)]du to [1 - 2/(1+u)]du using long division, since I never learned this, and was wondering if someone could provide a quick explanation on how to do this.

2. Relevant equations
Go from [(1 - u)/(1+u)]du to [1 - (2/(1+u))]du using long division (the function needs to be integrated, which is the reason for the long division)

3. The attempt at a solution

Any help would be greatly, appreciated. Thanks guys

2. Feb 15, 2012

### HallsofIvy

Staff Emeritus
You are dividing -u+ 1 by u+ 1. Clearly, u divides into -u -1 times. Multiplying u+ 1 by -1 gives -u- 1 and subtracting the remainder is (-u+ 1)-(-u- 1)= 2.

That is, u+ 1 divides into -u+ 1 -1 times with remainder 2.

$$\frac{-u+1}{u+1}= -1+ \frac{2}{u+1}$$

3. Feb 15, 2012

### tiny-tim

hi solomon684!
the best way to explain long division is to actually do it …

eg (1 + u + 3u2)/(1+u)

write the top line first, with the highest powers on the left: 3u2 + u + 1

now look at the first item, it looks like the start of 3u(1 + u)

so subtract 3u(1 + u) from the whole thing …

that leaves -2u + 1

now that looks like the start of -2(1 + u)

so subtract -2(1 + u) from the whole thing …

that leaves 3

so the answer is: (3u2 + u + 1)/(1 + u) = 3u - 2 + 3/(1 + u)

4. Feb 15, 2012

### solomon684

Hm, I kind of get it, but at the same time I'm still pretty lost and feel like I'd never be able to figure it out on my own.

I looked up polynomial long division and tried doing it the way the first example here is done, and got an answer of -1 with remainder 2. When doing it that way, is it safe to say it can always be rewritten as answer + remainder/original denominator? Because that is how this example works out but I am not sure if it is always true

5. Feb 15, 2012

### Ray Vickson

Whether or not you call it long division, it IS the type of thing you must have done in Calculus 101, when you took integration. $$\frac{1-u}{1+u} = \frac{2-(1+u)}{1+u} = \frac{2}{1+u} - 1.$$ So, I don't go from (1-u)/(1+u) to 1 - 2/(1+u), but I do go to 2/(1+u) - 1.

RGV