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April 18, 2015

April 18, 2015

Posted by **stevie** on Monday, April 13, 2009 at 4:01pm.

y' = 2xy/(x^2-y^2)

- integration math -
**Sharpay**, Monday, April 13, 2009 at 4:12pmfirst of all, the first y doesn't need a 1 exponent, so when you fix that error, you can ask that algebraic equation of a question.!.

- integration math -
**Reiny**, Monday, April 13, 2009 at 4:27pmI will assume that by y' you mean dy/dx

so it looks like your derivative is the result of an implicit derivative.

so le't work it backwards

y'(x^2 - y^2) = 2xy

y'x^2 - y'y^2 - 2xy = 0

looks like it could have been**(x^2)(y) - (1/3)y^3 = 0**

let's differentiate:

x^2(y') + y(2x) - y^2y' = 0

sure enough!! it works

I must admit there was no real method to what I did, just some observation of patterns and lucky guessing.

- integration math -
**Miss Logan:)**, Monday, April 13, 2009 at 4:37pmReiny is right...sharpay sweetie you really need to know what your talkin about before you answer someones questions...you dont want people thinking of you as a you know...ditz as i like to say...good luck in life sharpay and keep it up REINY!!

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