How to show that a particular solution $y = \sqrt{|x|}(-2)$ is unique We have a differential equation: $2xy’=y$ and $y(-4)=-4$. Prove that the solution is unique.
I have solved the DE:
$y = \sqrt{|x|}D$, where $D \in \mathbb{R}$
How to show that the particular solution for $y(-4) =-4$ is unique.
I got the the particular solution, but how to show that it is unique:
$D = -2 \implies y(-4) = -4$, $y = \sqrt{|x|}(-2)$
 A: Using directly the Picard–Lindelöf theorem. Consider the Cauchy problem $\begin{cases}y'=f(x,y),\\y(x_{0})=y_{0} \end{cases}$ where $f:(x,y)\longmapsto \frac{y}{2x}$ is a real-valued function defined on rectangle $\Delta\subseteq \mathbb{R}^{2}$ which contains the point $(x_{0}=-4,y_{0}=-4)$. Notice that $f$ and $\frac{\partial f}{\partial x}$ they are continuous over $\left]-\infty,0\right[\times \mathbb{R}$ (and of course over $\left]0,+\infty\right[\times \mathbb{R}$ but since in the problem the data is $x_{0}<0$ so we will not work over this domain. Moreover $y/x$ is continuous over whole of $\mathbb{R}^{2}$ except for $x=0$ where the function blows up) so over that domain Picard-Lindelöf give the existence and uniqueness for the Cauchy problem.
Moreover we can find the solution over that domain using variable separation with $\frac{{\rm d}y}{{\rm d}x}=\frac{y}{2x}$ give $2\ln|y|=\ln|x|+c$ then $|y|^{2}=c|x|$ then $\sqrt{|y|^{2}}=c\sqrt{|x|}$ so $|y|=c\sqrt{|x|}$ and setting $y(x_{0})=y_{0}$ we get $y=\pm 2\sqrt{|x|}$ but $y_{0}<0$ then the only solution over that domain is given by $\boxed{y=-2\sqrt{|x|}}$.
