Solve the given initial value problem and determine at least approximately where the solution is valid $(2x - y) dx + (2y- x) dy = 0, y(1) = 3 $.
My Solution: Let $M=2x-y \Rightarrow M_y=-1$ and let $N=2y-x\Rightarrow N_x=-1$, therefore the given DE is exact. Let $\Psi_x=M=2x-y\Rightarrow \Psi = x^2-xy+h(y)\Rightarrow \Psi_y=-x+h'(y)\Rightarrow h'(y)=\Psi_y+x=2y-x+x=2y \Rightarrow \Psi=x^2-xy+2y=C$, initial value condition given is $y(1)=3 \Rightarrow C=(1)^2-(1)(3)+2(3)=4 \Rightarrow x^2-xy+2y=4\Rightarrow y=\frac{4-x^2}{2-x}=2+x$, but the given answer is $y = [x+\sqrt{28-3x^2}]/2, |x|<\sqrt{28/3} $