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

(y')^2 + y = xy'

Thanks.

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    Got it - weird enough, the ^ in url didn't copy properly. Thank you.2011-12-19

1 Answers 1

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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.2011-12-20