3
$\begingroup$

I'm sure it's staring at me, but how does one solve this?

$$ (y')^2 + y = xy' $$

Thanks.

  • 1
    That's [Clairut's equation](http://en.wikipedia.org/wiki/Clairaut%27s_equation). [Wolfram alpha](http://www.wolframalpha.com/input/?i=%28y%27%29^2%2By%3Dxy%27) gives a clean solution step by step in this case, just click the 'show steps' button.2011-12-19
  • 0
    That should have been Clairaut, actually.2011-12-19
  • 0
    One solution is $y = x-1$?2011-12-19
  • 1
    @H.M.Šiljak: I think you meant [this one](http://www.wolframalpha.com/input/?i=%28y%27%29%5E2%2By%3Dxy%27).2011-12-19
  • 0
    Ok, got it. Differentiating both sides wrt x 2y'y'' + y' = y' + xy'' i.e., y''(2y'-x) = 0 which can now just be solved for each case in turn, y'' = 0 and 2y' = x Thanks everyone.2011-12-19
  • 0
    So if you found solution, it is worth to post an answer to your own question.2011-12-19
  • 0
    @Gigili: strange, both links (yours and mine) look exactly the same to me :)2011-12-19
  • 0
    @H.M.Šiljak: Eh, how so? It should be $*(y')^2* + y = xy'$ as OP mentioned.2011-12-19
  • 0
    Got it - weird enough, the ^ in url didn't copy properly. Thank you.2011-12-19

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