Unique prolongation of the solution of a Cauchy's problem to all $\mathbb R$

Question

Given the following $$(PC_{a,b})\begin{cases}3x^2y''-10xy'+4y=0\\y(2)=a\\y'(2)=b\end{cases}\text{ with }a,b\in\mathbb R$$ (A) We can uniquely extend the solution to all $\mathbb R\iff b=2a$.
(B) we can extend the solution to $\mathbb R$ and there is an infinite number of prolongations of the solution of $(PC_{a,b})$ $\iff b=2a$.
(C) We can extend the solution to $\mathbb R\iff b=a=0$.
(D) The solution is always extendible to $\mathbb R$ without imposing any condition on $a$ and $b$.

$\textbf{My attempt:}$ The equation is defined in $\mathbb R\times\mathbb R$.
I considered the linear operator $\mathcal L$ such that $\mathcal L(y)=3x^2y''-10xy'+4y$ and since the equation is homogeneous, the solutions of $(PC_{a,b})$ correspond to the generators of the space $Ker(\mathcal L)$.
Substituting $\tilde y=x^{\alpha}$, I found the solutions $x^4$ and $x^{1/3}$, so $$\tilde y(x)=Hx^4+Kx^{1/3}, H,K\in\mathbb R$$ is a solution for $(PC_{a,b})$. I observed that the constant function $y\equiv0$ is a solution for $(PC_{a,b})\iff$$ y(2)=a=y'(2)=b=0$.
Calculating the value of the solution in $0$, we get $\tilde y= 0$, but $y'(0)=4Hx^3+\dfrac{K}{3x^{2/3}}\Big|_{x=0}\to \infty$, so we can prolong $\mathcal C^1$ and then $\mathcal C^2$ the function $\tilde y$ with the function $y\equiv0\iff K=0$.
I don't see other particular existence conditions for the solution $\tilde y$, since $Hx^4+Kx^{1/3}$ is defined in $\mathbb R$.

$\underline{Important:}$ In general, I can't well distinguish the cases in which we have a unique prolongation of the solution or an infinite number of prolongations. What's the main difference between these two situations?

Answer 1

1. The equation is an ODE only for $x\ne 0$
2. The equation is of the Euler-Cauchy type. The characteristic polynomial is $$ 0=3m(m-1)-10m+4=3m^2-13m+4=(3m-1)(m-4) $$ so indeed $x^{1/3}$ and $x^4$ are the basis solutions. Both can be extended to be functions on all of $\Bbb R$.
3. Note that $x^{1/3}$ has no derivative in $x=0$, so any gluing together of solutions for $x<0$ and $x>0$ can only have the condition of continuity. But then one can glue together arbitrary solutions pairs, with different constants left and right from $x=0$.
4. To go in direction of uniqueness of the combined solution one needs to add a condition at $x=0$. This can be that both parts be differentiable in $x=0$, with a continuous derivative. This requires the coefficient of $x^{1/3}$ to be zero on both sides. Unfortunately, the coefficient on the right does not uniquely determine the coefficient on the left. This does not change if one adds the continuity of the second derivative. But more one should not demand from a second order equation.

In conclusion, to solve the task you need to know what "can extend" means, as a continuous function, or as a continuously differentiable function, or a twice so function,...?