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

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


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That's Clairut's equation. Wolfram alpha gives a clean solution step by step in this case, just click the 'show steps' button. – H. M. Šiljak Dec 19 '11 at 16:34
That should have been Clairaut, actually. – H. M. Šiljak Dec 19 '11 at 16:35
One solution is $y = x-1$? – AD. Dec 19 '11 at 16:44
@H.M.Šiljak: I think you meant this one. – Gigili Dec 19 '11 at 16:49
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. – Simon S Dec 19 '11 at 16:50

Differentiating both sides with respect to x:

$$2y^\prime y^{\prime\prime}+y^\prime =y^\prime+xy^{\prime\prime}$$

i.e., $y^{\prime\prime}(2y^\prime −x)=0$ which can now just be solved for each case in turn, $y^{\prime\prime}=0$ and $2y^\prime=x$

Thanks everyone.

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Note that not every solution to $y'' = 0$ and $2y' = x$ automatically works because the differentiation step is not reversible. But it's easy to plug the solutions to those into the original equation and find out which solutions do work. – Zarrax Dec 20 '11 at 1:54

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