# Dimension of the solution of a second order homogenous ODE

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 $N$.

• I don't believe your solution is differentiable at $0$ for what that matters. Mar 30, 2017 at 19:12
• @Mark It definitely is. first derivative $lim \frac{h^2|h|-0}{h} = 0$ and second derivative $lim \frac{3h^2-0}{h}=0$ for RHD and $lim \frac{-3h^2-0}{-h}=0$ for LHD Mar 30, 2017 at 19:16
• My mistake, you're right. Mar 30, 2017 at 19:18
• And BTW, an interval containing 0 is the one causes headaches. For domains lying completely on either side of 0, $x^2|x|$ simply reduces to $x^3$ or $-x^3$ depending on which side you're on and the dimension reduces to 2. Mar 30, 2017 at 19:21

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 $(0,\infty)$