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

1. [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 $|x| < \sqrt{\frac{28}{3}}$

2. $1^{2} - 1 \cdot 3 + 3^{2} = k$

$k = 7$

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

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

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

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

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

3. 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.