TITLE: Dimension of the solution of a second order homogenous ODE
QUESTION [9 upvotes]: So I have a 2nd order homogenous ODE $$x^2y''-4xy'+6y=0$$ over any interval containing $0$.
This is a standard Cauchy-Euler equation with the roots of the auxiliary equation being 2 and 3. Pretty simple, right?
But here's the twist. Consider the solution of the form $x^2|x|$. It's twice differentiable over $R$ and it satisfies the ODE. What's more, it's linearly independent from $x^2$ and $x^3$. So the basis contains three elements and so the dimension is 3.
What did I miss? I remember my professor teaching us that a linear Nth order ODE has a vector space of dimension $N$.

REPLY [6 votes]: The claim about the solution space being linear or affine with dimension equal to the order of the linear ODE is only valid over intervals where the coefficients of the ODE are defined and continuous where the leading coefficient is $1$. Normalizing to
$$
y''-\frac4xy′+\frac6{x^2}y=0
$$
shows that the coefficients are only defined and continuous on either $(-\infty,0)$ or $(0,\infty)$