Posted by stevie on Monday, April 13, 2009 at 4:01pm.
how to integrate:
y' = 2xy/(x^2-y^2)
integration math - Sharpay, Monday, April 13, 2009 at 4:12pm
first 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 4:27pm
I 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
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 4:37pm
Reiny is right...sharpay sweetie you really need to know what your talkin about before you answer someones questions...you don't 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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