Basis of solutions of $x^{2}y''-4xy'+6y=0$?

Question

What is the basis of solution space spanned by the solutions of the homogeneous ODE given by $$x^{2}y''-4xy'+6y=0$$ My work: The solution is of the type $x^{m}$.

By solving the auxiliary equation for this ODE, which is $m(m-1)-4m+6=0$ implies $m=2,m=3$. Hence, the two solutions are $x^{2}$ and $x^{3}$ and thus the basis is {$x^{2}$, $x^{3}$}.

But according to the solution I have the basis is {$x^{2}$, $x^{3}$, $x^{2}|x|$}, but I actually want to know why $x^{2}|x|$ is explicitly specified as a solution?

Answer 1

The equation is singular at $x=0$. Thus you get a solution space in the strict, basic sense of the existence theorems for $x>0$ and one for $x<0$, both have basis $x^2,x^3$. Now you can make up a combined function space over $\Bbb R$ made up of piecewise defined functions which are in the respective solution spaces, connected by the value zero at $x=0$.

One can make up some generalized solution idea by demanding that these functions be twice continuously differentiable also in $x=0$. Now you can check that this is still a linear space, that the continuation of $x^2$ has to be $x^2$, but that the continuation of $x^3$ can by any multiple of $x^3$. Any such continuations of the cubic term can be represented as linear combination of $x^3$ and $|x|x^2$. Or $\max(0,x)^3$ and $\min(0,x)^3$.

To my taste this is not a solution space of an ODE, but a function space associated to the solutions of the ODE. But this is just my point-of-view.

Answer 2

The equation is an Euler equation with the auxillary poynomial $r(r-1)-4r+6=(r-2)(r-3)$. In this case the two linearly independent solutions which form a basis of the space are $|x|^2$ and $|x|^3$ You have found a pair of linearly independent solutions. So the general solution is $y=ax^{2}+b|x|^{3}$. It is clear that the dimension of the space of solutions cannot be greater than $2$, since is a second order linear equation.