1. ## Differential 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

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
Observe that $\displaystyle \dfrac{1}{y(1-y^2)}=\dfrac{1}{y(1-y)(1+y)}$ (its the same for x as well...). Now apply partial fractions in order to integrate this.

Can you continue?

3. 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

4. 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.

5. 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