How Do You Use Long Division to Simplify (1 - u)/(1 + u)?

[solomon684]

Homework Statement

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.

Homework 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)

The Attempt at a Solution

Any help would be greatly, appreciated. Thanks guys

 

[HallsofIvy]

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}

 

[tiny-tim]

hi solomon684!

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.

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)

 

[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

 

[Ray Vickson]

Homework Statement

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.

Homework 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)

The Attempt at a Solution

Any help would be greatly, appreciated. Thanks guys

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