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Math Help - y' = (x^2 + 3y^2) / (2xy)

  1. #1
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    y' = (x^2 + 3y^2) / (2xy)

    In a DiffEQ video, the problem y' = (x^2 + 3y^2) / (2xy) is solved as a homogeneous differential equation. The steps are all fine and the solution is found to be x^2 + y^2 - Cx^3 = 0.

    If the answer is correct, then the derivative of the solution should equal the RHS of the question. Am I understanding DiffEQs correctly?

    If yes, then what's wrong here? I take the derivative of the solution and solve for y'
    2x + 2y y' -3Cx^2 = 0
    y' = (3Cx^2 - 2x) / (2y).

    But I don't see how to make that look like the question. My derivative looks correct, so what am I missing?
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  2. #2
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    Re: y' = (x^2 + 3y^2) / (2xy)

    Quote Originally Posted by mathDad View Post
    In a DiffEQ video, the problem y' = (x^2 + 3y^2) / (2xy) is solved as a homogeneous differential equation. The steps are all fine and the solution is found to be x^2 + y^2 - Cx^3 = 0.

    If the answer is correct, then the derivative of the solution should equal the RHS of the question. Am I understanding DiffEQs correctly?

    If yes, then what's wrong here? I take the derivative of the solution and solve for y'
    2x + 2y y' -3Cx^2 = 0
    y' = (3Cx^2 - 2x) / (2y).

    But I don't see how to make that look like the question. My derivative looks correct, so what am I missing?
    2x+2y y' - 3Cx^2=0 ; multiply both sides by x

    2x^2 +2x y y' = 3C x^3

    2x y y'=3Cx^3 - 2x^2 ; substitute in implicit function definition

    2x y y'=3(x^2 + y^2) - 2x^2

    2x y y'=x^2+3y^2

    y' = \frac{x^2+3y^2}{2x y}
    Last edited by romsek; February 13th 2014 at 07:31 PM.
    Thanks from mathDad
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  3. #3
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    Re: y' = (x^2 + 3y^2) / (2xy)

    y' = (x^2 + 3y^2) / (2xy)-14-feb-14.png
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