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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 $N$.

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  • $\begingroup$ I don't believe your solution is differentiable at $0$ for what that matters. $\endgroup$ Mar 30, 2017 at 19:12
  • $\begingroup$ @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 $\endgroup$
    – Akshit
    Mar 30, 2017 at 19:16
  • $\begingroup$ My mistake, you're right. $\endgroup$ Mar 30, 2017 at 19:18
  • $\begingroup$ 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. $\endgroup$
    – Akshit
    Mar 30, 2017 at 19:21

1 Answer 1

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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 $(0,\infty)$

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  • $\begingroup$ So there is no bound on the dimension of vector space when the coefficients are not continuous? $\endgroup$
    – Akshit
    Mar 31, 2017 at 2:20
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    $\begingroup$ Also, is there a proof somewhere regarding the dimension? I cannot find it and it keeps bugging me that continuity of the coefficients can affect the dimensions $\endgroup$
    – Akshit
    Mar 31, 2017 at 2:21
  • $\begingroup$ If the coefficients are not continuous, you leave the domain of ODE that is covered by the classical existence theorems. So you have to define what in that generalized context a solution is and if/how they exist. The dimension of the solution space is then a secondary concern. $\endgroup$ Mar 31, 2017 at 6:37

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