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Math Help - Reduction of order, dependent variable missing trouble

  1. #1
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    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 u=y' u' = y'' for the missing dependent variable y, separating variables and integrating I can get the equation in the below form

    -\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

    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
    \frac{1}{u} = \frac{1}{x} + C

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

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

    Thanks for any help.
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  2. #2
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    Re: Reduction of order, dependent variable missing trouble

    The two answers are, in fact, the same. Starting from your \frac{x}{1+ Cx} we can "multiply and divide" by C to get
    \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)
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  3. #3
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    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.
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