$2xy+y^2-2x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$; $y=2$ when $x=1$.
My reference gives the solution $y=\dfrac{2x}{1-\log x}$, but is it really the solution ?
My Attempt $$ 2xy+y^2-2x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\implies\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2xy+y^2}{2x^2}=\frac{y}{x}+\frac{1}{2}\frac{y^2}{x^2}\\ \text{Put }v=\frac{y}{x}\implies y=vx\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}=v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}=v+\frac12v^2\implies x\dfrac{\mathrm{d}v}{\mathrm{d}x}=\frac{1}{2}v^2\\ 2\int v^{-2}dv=\int\frac{\mathrm{d}x}{x}\implies -\frac{2}{v}=\log|x|+C\\ \boxed{\frac{-2x}{y}=\log|x|+C} $$ $y=2$ when $x=1$$\implies C=-1$, $$ \frac{-2x}{y}=\log|x|-1\implies \color{red}{y=\dfrac{2x}{1-\log|x|}} $$ How do I justify going from $y=\dfrac{2x}{1-\log\color{red}{|}x\color{red}{|}}$ to $y=\dfrac{2x}{1-\log x}$ ?
What I understand
I only have basic knowledge on differential equations, thus not familiar with the ideas like singularity and all.
I think I only understand a hint, $$ y=\dfrac{2x}{1-\log|x|}\implies y=\begin{cases}\dfrac{2x}{1-\log x} \text{ for } x>0\\\dfrac{2x}{1-\log(-x)} \text{ for } x<0\end{cases} $$ So the condition "$y=2$ when $x=1$" does not include in the second case. Does this has something to do with my doubt ?
Obviously, the differential equation is not defined for $x=0$. The equation in its original form exists, but at $x=0$ it does not contain any derivatives. Thus the domain in the $(x,y)$ space consists of two disjoint components, one with $x>0$ and one with $x<0$. Solutions of the ODE can have their graph in only one of the components
The initial condition tells us that the maximal interval for the solution has to be inside $(0,\infty)$. On that interval, $\log|x|=\log x$.
The solution formula tells us that any solution in both components can be continuously extended to the point $(x,y)=(0,0)$, so that any pair of solutions, one from each component, can be joined to form a function that could be said to be a solution in some extended sense.