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$2xy+y^2-2x^2\dfrac{dy}{dx}=0$; $y=2$ when $x=1$.

My reference gives the solution $y=\dfrac{2x}{1-\log x}$, but is it really the solution ?

My Attempt $$ 2xy+y^2-2x^2\dfrac{dy}{dx}=0\implies \dfrac{dy}{dx}=\dfrac{2xy+y^2}{2x^2}=\frac{y}{x}+\frac{1}{2}\frac{y^2}{x^2}\\ \text{Put }v=\frac{y}{x}\implies y=vx\\ \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}=v+\frac{1}{2}v^2\implies x\dfrac{dv}{dx}=\frac{1}{2}v^2\\ 2\int v^{-2}dv=\int\frac{dx}{x}\implies -\frac{2}{v}=\log|x|+C\\ \boxed{\frac{-2x}{y}=\log|x|+C} $$ $y=2$ 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 ?

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Obviously, the differential equation is not defined for $x=0$. The initial condition tells us that the maximal interval for the solution is $(0,\infty)$. On that interval, $\log|x|=\log x$.

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  • $\begingroup$ thanx. i think thats whr my doubt really is, how does the initial condition tells us the maximal interval for the solution is $(0,\infty)$ ? $\endgroup$ – ss1729 Jun 23 '18 at 10:23
  • $\begingroup$ The plane $x=0$ in $(x,y,y')$ phase/state space is a singularity for the differential equation, it divides the domain of the ODE in explicit form $y''=f(x,y,y')$ into two halves. The initial point is in the half-space $x>0$, thus any solution through this initial point will have to stay in the same half-space. $\endgroup$ – Lutz Lehmann Jun 23 '18 at 11:59
  • $\begingroup$ ok well. actually i am not familiar with singularity and all. could u bit simplify ur explanation if i have only basic knowledge abt differential equations. $\endgroup$ – ss1729 Jun 23 '18 at 14:17
  • $\begingroup$ That is just a more general way to say that you would have to divide by zero, or that you get a vertical slope, which is not defined and thus these points can not be inside the domain of a differential equation. $\endgroup$ – Lutz Lehmann Jun 23 '18 at 14:28
  • $\begingroup$ I think I only understand, $y=\dfrac{2x}{1-\log|x|}\implies y=\begin{cases}\dfrac{2x}{1-\log x} \text{ for } x>0\\\dfrac{2x}{1-\log(-x)} \text{ for } x<0\end{cases}$. So the condition "$y=2$ when $x=1$" does not include in the second case. Does this has something to do with my doubt ? $\endgroup$ – ss1729 Jun 23 '18 at 15:31

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