# How to obtain dimension of solution space of ODE?

We are given the equation,

$$x^2y''-4xy'+6y=0$$

And we have to get the dimension of solution space in $(-1,1)$.

My Attempt: I tried substituting $$x=e^z$$ and I get from that, the following,

$$y=c_1x^2+c_2x^3$$

and concluded that dimension is $2$. But, I realize that at $x=0$, my ODE in standard form will face issues, and this solution may not work. How do I go about solving this?

This will depend on precisely what you mean by a "solution" when the differential equation is singular. Every solution is of the form $y = c_1 x^2 + c_2 x^3$ on $(0,1)$ and on $(-1,0)$, but do the coefficients $c_1$ and $c_2$ for $x > 0$ and for $x < 0$ have to be the same? It is reasonable to require $y$ to be twice differentiable, since the differential equation involves $y''$. Now the second derivative at $x=0$ is $2 c_1$, so $c_1$ needs to be the same on both sides, but there's no reason for $c_2$ to be the same. Thus we have solutions of the form
$$y(x) = \cases {c_1 x^2 + c_2 x^3 & for x \le 0\cr c_1 x^2 + c_3 x^3 & for x > 0\cr}$$ forming a vector space of dimension $3$.