Uniqueness of solution of Cauchy-Euler ODE, with conditions 
Suppose we have two "similar" non-homogeneous ODE:

*

*$x^2y''-6y=12$

*$x^2y''+2xy'-6y=12$
Find a solution that satisfies $y(1)=3$.

I solved the equations as usual and got that the solution families are:

*

*${y(x) = -2 +\frac{c_1}{x^{2}} + c_2x^{3} }$

*${y(x) = -2 +\frac{c_1}{x^{3}} + c_2x^{2} }$
Now my question is, how do I determine that the solution that satisfies $y(1)=3$ is unique AND is defined for each ${x \in \mathbb{R} }$?
I checked the Wronskian for both cases and it turned out that the first case has $W=5$ and the seconds has $W=\frac{5}{x^{2}}$,  what does it necessarily mean that the Wronskian is not defined at one point? ($x=0$)
 A: Sorry for the extended banter.  So anyway, here is the correct answer, summarized.
The theorem that you want to use is the following (it can be found in pretty much every undergraduate DiffEq textbook):  Let $I$ be an open interval, and let $p,q,f$ be continuous functions on $I$.  Consider the equations:
\begin{equation}\tag{NH}y''+p(x)y'+q(x)y=f(x)\end{equation}
and
\begin{equation}\tag{H}y''+p(x)y'+q(x)y=0\end{equation}
Suppose $y_1$ and $y_2$ are linearly independent solutions to (H) on $I$, and $y_p$ is any solution to (NH) on $I$.  Then every solution to (NH) on $I$ has the form
$$y(x)=y_p(x)+c_1y_1(x)+c_2y_2(x)$$
where $c_1,c_2\in\mathbb{R}$.
Now, rewrite your equation (2) as
$$y''+2x^{-1}y'-6x^{-2}y=12x^{-2}$$
Note that this is (NH) with $p(x)=2x^{-1}$, $q(x)=-6x^{-2}$, and $f(x)=12x^{-2}$, all continuous on $(0,\infty)$.  Note that $y_1(x)=x^{-3}$ and $y_2(x)=x^2$ are LI solutions to (H) in this case, and that $y_p(x)=-2$ is a solution to (NH).  Hence, every solution to (NH) on $(0,\infty)$ has the form
$$y_+(x)=-2+c_1x^{-3}+c_2x^2.$$
I'm calling it $y_+$ instead of just $y$ because this is only true for solutions defined on $(0,\infty)$.  Similarly, every solution to (NH) on $(-\infty,0)$ has the form
$$y_-(x)=-2+d_1x^{-3}+d_2x^2.$$
If $y$ is a solution on $\mathbb{R}$, then $y$ must agree with $y_-$ and $y_+$ on their domains. Note that $c_1=d_1=0$, since  otherwise we would have $\lim_{x\to 0}y(x)$ either infinite or nonexistent, violating differentiability of $y$.  Thus, $y(x)=-2+c_1x^2$ on $(0,\infty)$ and $=-2+d_1x^2$ on $(-\infty,0)$.  Since we require $y(1)=3$, this means $c_1=5$.  Note that $y'(x)=10x$ if $x>0$ and $y'(x)=2d_1x$ if $x<0$. Since $y''$ exists, this means $d_1=5$ as well. For continuity we must include $y(0)=-2$ and now we get
$$y(x)=-2+5x^2$$
as the unique solution to (2) on $\mathbb{R}$ with $y(1)=3$.
Case (1) is similar, but with one crucial difference. Similar to (2), we find that
$$y(x)=\left\{\begin{array}{ll}-2+d_2x^3,&x<0\\-2+5x^3&x\geq 0\end{array}\right.$$
But now $d_2$ can be any value at all, because we get
$$y'(x)=\left\{\begin{array}{ll}3d_2x^2,&x<0\\15x^2&x\geq 0\end{array}\right.$$
and hence
$$y''(x)=\left\{\begin{array}{ll}6d_2x,&x<0\\30x&x\geq 0\end{array}\right.$$
so that (1) is satisfied on $\mathbb{R}$ with $y(1)=3$.
However, note that if we require that $y$ be three times differentiable then $d_2=5$ so the solution is unique in that special case.
