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    The point  (x, y) lies on the curve  C such that

     \displaystyle (x^2-1) \left (\frac{dy}{dx} \right )^2 -2xy\frac{dy}{dx} +y^2=1 .

    Find the possible equations of  C .

    Any ideas?

    Edit : I have put the correct question now. Also it's only first order actually.
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    (Original post by Ano123)
    The point  (x, y) lies on the curve  C such that

     \displaystyle (x^2-1)\frac{d^2y}{dx^2} -2xy\frac{dy}{dx} +y^2=1 .

    Find the possible equations of  C .

    Any ideas?
    Why do you think an (elementary) solution exists?
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    (Original post by Zacken)
    Why do you think an (elementary) solution exists?
    This one does now.
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    (Original post by Ano123)
    This one does now.
    Now it makes sense.

    Think of it as a quadratic in \frac{\mathrm{d}y}{\mathrm{d}x} - this gets you \displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} =\frac{xy \pm \sqrt{x^2 + y^2 - 1}}{x^2 - 1} or something along those lines.

    It's a first order ODE, by the way - not a second order one.
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    (Original post by Ano123)
    The point  (x, y) lies on the curve  C such that

     \displaystyle (x^2-1) \left (\frac{dy}{dx} \right )^2 -2xy\frac{dy}{dx} +y^2=1 .

    Find the possible equations of  C .

    Any ideas?

    Edit : I have put the correct question now. Also it's only first order actually.
    Consider differentiating both sides w.r.t. x, and a lot of stuff cancels (but be careful not to lose any solutions when dividing through by a certain factor later). This will reduce the ODE to a trivial solution and a second order ODE that is "morally" a first order separable one.
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    I already have obtained the answers really. But it can always help to see how others do it.
 
 
 
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