Solve the given initial value problem and determine approximately the interval of existence?

Question

Here's the question:

Solve the given initial value problem and determine approximately the interval of existence. $$ (2x-y)dx+(2y-x)dy=0, \ \ y(1)=3. $$

I solved and got this: $$ x^2-xy+y^2=7. $$

But how do I approximate the interval of existence? Thank you !

Answer 1

Solving for $y$ we have:$$y^2-xy+x^2-7=0$$ for domain of existence we must consider where discriminant is not negative i.e.$$4x^2-4.(x^2-7)\ge 0$$which leads to always true $28\ge 0$. So the existence domain is all $(x,y)\in\Bbb R^2$.