Global existence of ODE $-\frac{1}{2}y+\frac32xy'+yy'=0$ In my paper, the ODE with $i\in \mathbb{N}$
$$-\frac{1}{2i}y+\frac{2i+1}{2i}xy'+yy'=0; y(1)=a$$
does not have global solution if $a>0$ or $a<-1.$
At first, I tried to prove this in the case of $i=1:$
$$-\frac12y+\frac32xy'+yy'=0; y(1)=1$$
but I don't know how to start to prove that above ode does not have global solution.
Please give me some help.
And I wonder if we can verify the global existence of solution when $-1<a<0$?
 A: The Initial Value Problem (IVP) is given by
$$\text{(E)}: \begin{cases}\displaystyle-\frac{1}{2k}y+\left(\frac{2k+1}{2k}\right)xy'+yy'=0,\\y(1)=a, \end{cases}$$where $k\in \mathbb{N}$ and $y=y(x)$ and $a$ one constant such that which allows to study the behaviour of the global solution for $(E)$. Notice the equation is a homogeneous equation and we can find the solution as follows:
\begin{align*}
\frac{1}{2k}y+\left(\frac{2k+1}{2k}\right)xy'+yy'=0 &\overset{y=xv}{\implies} x\left( \frac{2k+1}{2k}\right)\left(x\frac{{\rm d}v}{{\rm d}x}+v \right)-\frac{x}{2k}v+x\left(x\frac{{\rm d}v}{{\rm d}x}+v \right)v=0,\\
&\implies \frac{{\rm d}v}{{\rm d}x}=-\frac{2k(v^{2}+v)}{x(2k+2k+v+1)},\\
&\implies \int \frac{2k+2kv+1}{v(v+1)}\, {\rm d}v=\int -\frac{2k}{x} \, {\rm d}x,\\
&\implies -\log\left| v+1 \right|+(2k+1)\log|v|=-2k\log|x|+c,  \quad c\in \mathbb{R},\\
&\implies -\log\left|1+\frac{y}{x}\right|+(2k+1)\log\left|\frac{y}{x}\right|=-2k\log|x|+c
\end{align*}
Setting $y(1)=a$ we get the particular solution for $(E)$:
$$\log|a+1|+(2k+1)\log\left|\frac{y}{x}\right|+2k\log|x|=2k\log|a|+\log|a|+\log\left|1+\frac{y}{x}\right|,$$where $y=y(x)$. The result about the globality of the solution and the conditions on the constant $a$ as follows.
