xy'^2 + yy' = 0 where y' = dy/dx(adsbygoogle = window.adsbygoogle || []).push({});

The answer is C1 = y and C2 = xy but I get this:

y'(xy' + y) = 0 where y' = 0 and thus y = C1

For the other solution:

xy' + y = 0

y' = -y/x

y = -y ln x + C2

C2 = y + y ln x

Full question is here: http://www.cramster.com/solution/solution/640396

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# How to solve this simple polynomial?

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