# Thread: Reduction of order, dependent variable missing trouble

1. ## Reduction of order, dependent variable missing trouble

Hi I've been having trouble with a reduction of order problem from my textbook. By making the appropriate substitutions of $\displaystyle u=y'$ $\displaystyle u' = y''$ for the missing dependent variable $\displaystyle y$, separating variables and integrating I can get the equation in the below form

$\displaystyle -\frac{1}{x} = -\frac{1}{u} + C$

My main problem from here is I can't seem to solve for u in such a way as to make the next integration come out to the answer given in the book. After struggling with this for a while, I actually found the exact same problem in a Differential Equations Demystified book I purchased. They go in one step from what is above to

$\displaystyle u = \frac{1}{C} \left( 1 - \frac{1}{1+Cx}} \right)$

How did they do this? And is there any intuition behind why you would decide to solve for u in this way as opposed to what I tried to do which was more along the lines of
$\displaystyle \frac{1}{u} = \frac{1}{x} + C$

$\displaystyle \frac{1}{u} = \frac{1+Cx}{x}$

$\displaystyle u = \frac{x}{1+Cx}$

Thanks for any help.

2. ## Re: Reduction of order, dependent variable missing trouble

The two answers are, in fact, the same. Starting from your $\displaystyle \frac{x}{1+ Cx}$ we can "multiply and divide" by C to get
$\displaystyle \frac{1}{C}\frac{Cx}{1+ Cx}= \frac{1}{C}\frac{Cx+1- 1}{Cx+ 1}= \frac{1}{C}\left(\frac{Cx+ 1}{Cx+ 1}- \frac{1}{Cx+ 1}\right)= \frac{1}{C}\left(1- \frac{1}{Cx+1}\right)$

3. ## Re: Reduction of order, dependent variable missing trouble

Makes sense now that I see it done, thanks. I would have never thought to do the +1 -1 thing.