Linearly independent solutions of Cauchy-Euler differential equation

Question

Are $y_1=x^3$ and $y_2=x^2|x|$ ($x\in \mathbb R$) linearly independent solutions of the Cauchy-Euler homogeneous differential equation: $x^2y''-4xy'+6y=0$ on $\mathbb R$?

Clearly, $y_1$ and $y_2$ are linearly independent functions on $\mathbb R$ and on plugging, both satisfy the given differential equation. Hence, it appears that they are LI solutions of the given ODE, but solving this equation gives $x^2$ and $x^3$ as the two independent solutions and both of them are defined for all $x\in \mathbb R$ (which again contradicts the fact that the Cauchy-Euler equation is defined for $x>0$). Now, my confusion is:

$1$. Can a second order ODE have three LI solutions viz. $x^2$, $x^3$ and $x^2|x|$ over $\mathbb R$?

$2$. Though the coefficients of $y'$ and $y$ viz. $-4/x$ and $6/x^2$ are not defined at $x=0$ but its solutions are defined there. So, will it be correct to talk about the behaviour of solutions in the domain $x\in \mathbb R$?

Thanks!

Answer 1

1. Yes, as you showed yourself. Note that the usual statement of the theorem that linear equation of the $k$-th order has $k$ linearly independent solutions assumes that the equation can be written as $y^{(k)}+a_{k-1}(x)y^{(k-1)}+\ldots$.

2. In your problem you do not have coefficients of the form $-4/x$. So no problem with $x=0$. It is a different story that you have a coefficient at the highest derivative that can turn into zero. In this case the uniqueness theorem does not apply, but you still of course can say that some solutions defined on all $\mathbb R$ satisfy the equation,.

Answer 2

A couple of things to note:

• every possible solution goes through the origin.

• if $x>0$, then $x^3$ and $x^2|x|$ are the same thing. (Was $x^2|x|$ what it said in the solution? That's a strange way to write it.)

• Here's where the domain stuff is important: supposed you have initial conditions with positive $x$ values. Say $y(1)=1$ and $y'(1)=2$. This will give you a unique solution for positive values of $x$. But for negative values of $x$ there are still an infinite number of possible solutions, each one equally valid.

If you wanted to define a solution for all real numbers using the initial conditions from above, you could get any piecewise function of the form: $y=x^2$ if $x>0$, $y=c_1 x^2+c_2 x^3$ if $x\leq 0$