Solving H.S. Bear's Diff Eq Problem: (1-y^{2}) dx - xy dy = 0

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bryanosaurus
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I am reviewing differential equations, going through H.S. Bear's book Diff Eq: Concise Course.
The problem set for the variables separate section were pretty easy and straightforward except for this one, which I can't see how to arrive at the answer given in the book. I'm probably just missing something silly, so maybe another pair of eyes looking at it will clear it up.

(1-y[tex]^{2}[/tex]) dx - xy dy = 0

the solution given is (x[tex]^{2}[/tex])(1-y[tex]^{2}[/tex]) = c
 
Hi bryanosaurus! :smile:
bryanosaurus said:
(1-y[tex]^{2}[/tex]) dx - xy dy = 0

erm … dx/x = ydy/(1 - y²) … ? :smile:
 
Okay I got it now, I was fudging a sign when getting rid of the natural logs.
Thanks :)