I'm stuck on a question in my book, dy/dx = [y(1-y^2)]/[x(1-x^2)]

I know its separable but when i separate it, it comes to:

{1/[y(1-y^2)]} dy = {1/[x(1-x^2)]} dx and i dont know how to integrate both sides could someone please help

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- Jul 28th 2010, 07:15 PMCookieCDifferential equations/ separable
I'm stuck on a question in my book, dy/dx = [y(1-y^2)]/[x(1-x^2)]

I know its separable but when i separate it, it comes to:

{1/[y(1-y^2)]} dy = {1/[x(1-x^2)]} dx and i dont know how to integrate both sides could someone please help - Jul 28th 2010, 07:26 PMChris L T521
- Jul 28th 2010, 07:48 PMCookieC
I used partial fractions and ended up with A = 1 B =1/2 C = -1/2 then put this back into the fractions and integrated to get

lny -1/2ln(1-y) -1/2ln(1+y) = lnx - 1/2ln(1-x) - 1/2ln(1+x) + c but the answer in my book is x^2(1-y^2) = ky^2(1-x^2) where K is a constant but i have no idea how they ended up with that :S - Jul 29th 2010, 01:22 AMAckbeet
Your answer is correct... and the book's answer is correct. To get the book's answer from yours, take your equation and exponentiate both sides. Use exponent rules to simplify, and I fancy you'll need to square both sides. That should get you the book's answer, after a re-definition of the constant of integration.

- Jul 29th 2010, 06:18 PMKrizalid
note that we can avoid partial fractions.

we consider $\displaystyle \displaystyle\int\frac{dy}{y(1-y^2)}$ now if we put $\displaystyle y=\dfrac1t$ the integral becomes $\displaystyle \displaystyle\int\frac t{1-t^2}\,dt$ thus the original integral equals $\displaystyle \displaystyle\frac{1}{2}\ln \left( \frac{y^{2}}{y^{2}-1} \right)+k.$

we can save time, oh yeah! xD