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First order ODE

  1. Jul 26, 2006 #1
    Hi, can someone please help me with the following ODE? I need to find the general solution.

    [tex]
    y = xy' + \frac{1}{{y'}}
    [/tex]

    Rearranging I get a quadratic in dy/dx.

    [tex]
    x\left( {\frac{{dy}}{{dx}}} \right)^2 - y\left( {\frac{{dy}}{{dx}}} \right) + 1 = 0
    [/tex]

    [tex]
    \frac{{dy}}{{dx}} = \frac{{y \pm \sqrt {y^2 - 4x} }}{{2x}}
    [/tex]

    I don't know what to do from this point nor am I sure if I've started the right way. Any help would be good thanks.
     
  2. jcsd
  3. Jul 26, 2006 #2

    J77

    User Avatar

    That's a first order equation of degree two - you have to look at the discriminant you've found to find where solutions exist...

    edit: I see sid deleted his reply - lucky I didn't quote it:wink:
     
  4. Jul 26, 2006 #3

    siddharth

    User Avatar
    Homework Helper
    Gold Member

  5. Jul 26, 2006 #4
    Now that you mention the name of the DE, I remember doing a question on it last year. It's too bad that I've put that booklet containing the problem and the books which I did questions in away in storage. Anyway thanks for the help. It looks like I need to diff both sides wrtx.
     
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