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Math Help - Differential equations/ separable

  1. #1
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    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
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by CookieC View Post
    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 \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?
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  3. #3
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    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
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  4. #4
    A Plied Mathematician
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    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.
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  5. #5
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    note that we can avoid partial fractions.

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

    we can save time, oh yeah! xD
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