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Math Help - [SOLVED] IVP for an exact Diff Eqn

  1. #1
    Senior Member Pinkk's Avatar
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    [SOLVED] IVP for an exact Diff Eqn

    Solve the given initial value problem and determine at least approximately where the solution is valid.

    (2x-y)dx + (2y-x)dy=0 where y(1)=3

    I really have no idea how to go correctly solve this. I eventually got to the primitive x^{2}-xy+y^{2}=k where k is some constant. After that I'm lost because the solution in the back of the book says the answer is:

    y=\frac{x + \sqrt{28-3x^{2}}}{2} with |x| < \sqrt{\frac{28}{3}}
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  2. #2
    Super Member Random Variable's Avatar
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     1^{2} - 1 \cdot 3 + 3^{2} = k

     k = 7


     y^{2} -xy = 7 - x^{2}

     y^{2} - xy +\frac{1}{4}x^{2} = 7 - x^{2} + \frac{1}{4} x^{2}

     \Big(y-\frac{1}{2}x\Big)^{2} = 7 - \frac{3}{4} x^{2} = \frac{28 -3x^{2}}{4}

     y - \frac{1}{2}x = \pm \frac{\sqrt{28-3x^{2}}}{2}

     y = \pm \frac{x + \sqrt{28-3x^{2}}}{2}
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  3. #3
    Senior Member Pinkk's Avatar
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    Ah, thanks, it was relatively simply. I'm new to Differential Equations so I assumed I had to some something extra my professor hadn't explained to us. Thanks.
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