how to integrate:
y' = 2xy/(x^2-y^2)
3 answers
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.!.
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
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.
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.
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!!
0 answers