When could these be equal? Differentials given $ax^2y''+bxy'+cy=0$ a,b, and c are real. x is positive. I want to show that 
$$a \frac{d^2y}{dv^2} +(b-a)\frac{dy}{dv} +cy =0.$$
This is a problem in a elementary text, I have found a more rigours method online, but I'm trying to make sense of the last step in my own less rigours idea. 
Using the coefficients $a$, $b$, and $c$, I observe that if this is true then
$$xy'=\frac{dy}{dv}\quad\text{and}\quad
x^2y''= \frac{d^2y}{dv^2} - \frac{dy}{dv},$$
so if I show the two statements above are true I will have justified the subsitution.
Let $\ln x = v$
$$\frac{1}{x} = \frac{dv}{dx}$$
$\ln$ is one to one so it has an inverse and 
$$\begin{align*}
x &= \frac{dx}{dv}\\
xy'&= \frac{dx}{dv} \frac{dy}{dx}\&&\text{by the chain rule}\\
xy'&= \frac{dy}{dv}.
\end{align*}$$
Great one down one to go. Continuing with the above...
$$x \frac{dy}{dx} = \frac{dy}{dv}.$$
I  take the derivative with respect to $x$ of both sides, using the product rule on the right.
$$x \frac{d^2y}{dx^2} + \frac{dy}{dx} = \frac{d}{dx}  \left( \frac{dy}{dv} \right)$$
I dont feel so good about the right side, but I  keep going anyway.
$$x \frac{d^2y}{dx^2} = \frac{d}{dx}  \left( \frac{dy}{dv} \right) - \frac{dy}{dx}$$
since $x$ is $e^v$, $x$ is its own derivative with respect to $v$. I multiply through by $x$:
$$x^2 \frac{d^2y}{dx^2} = x \frac{d}{dx}  \left( \frac{dy}{dv} \right) - \frac{dy}{dx} \frac{dv}{dx}$$
chain rule on the far right.
$$x^2 \frac{d^2y}{dx^2} = x \frac{d}{dx}  \left( \frac{dy}{dv} \right) - \frac{dy}{dv}$$
I'm so close but I'm stuck! all I want to say is:
$$x^2y''= \frac{d^2y}{dv^2} - \frac{dy}{dv}$$
and it is very sugestive to write:
$$x^2 \frac{d^2y}{dx^2} = \frac{dx}{dv} \frac{d}{dx}  \left( \frac{dy}{dv} \right) - \frac{dy}{dv}$$
but what could it mean to cross out the $dx/dx$? that is in my way? as I  was told before here such operations are "dubious" though I'm still trying to grasp why.
But I  know this is true so it must be the case that for these functions
$$\frac{dx}{dv} \frac{d}{dx}  \left( \frac{dy}{dv} \right) = \frac{d^2y}{dv^2}$$
perhaps I  can say that the kind of functions that work in the above are just the kind I'm working with and then I  would be done? 
 A: Start with $$a{x^2}y'' + bxy' + cy = 0$$
which is an Euler-Cauchy DE of second degree. Now make the change of variables, 
$$x = e^z$$
From here you get that
$$dx = e^z dz$$
So now you have
$$\frac{{dy}}{{dx}} = \frac{{dy}}{{dz}}\frac{{dz}}{{dx}}=\frac{{dy}}{{dz}}\frac{1}{x}$$
Thus
$$bx\frac{{dy}}{{dx}} = b\frac{{dy}}{{dz}}$$
Which is what you correctly derived.
For the second equality go this way
$$\eqalign{
  & x = {e^z}  \cr 
  & \frac{{dy}}{{dx}} = {e^{ - z}}\frac{{dy}}{{dz}}  \cr 
  & \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {{e^{ - z}}\frac{{dy}}{{dz}}} \right)  \cr 
  & \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dz}}\left( {{e^{ - z}}\frac{{dy}}{{dz}}} \right)\frac{{dz}}{{dx}}  \cr 
  & \frac{{{d^2}y}}{{d{x^2}}} = \left( {{e^{ - z}}\frac{{{d^2}y}}{{d{z^2}}} - {e^{ - z}}\frac{{dy}}{{dz}}} \right)\frac{{dz}}{{dx}}  \cr 
  & \frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 2z}}\left( {\frac{{{d^2}y}}{{d{z^2}}} - \frac{{dy}}{{dz}}} \right)  \cr 
  & {x^2}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{d^2}y}}{{d{z^2}}} - \frac{{dy}}{{dz}} \cr} $$
Notice that what is used is the chain rule, in a rather "algebraic" way. 
You can prove by induction that
$${x^n}{D^n}y = \mathcal{D}\left( {\mathcal{D} - 1} \right) \cdots \left( {\mathcal{D} - n + 1} \right)y$$
Where $\mathcal{D}=\dfrac{d}{dz}$ and $D = \dfrac{d}{dx}$
A: With respect to $\frac{dx}{dx}$, the reason such operations are "dubious" is because there is a subtle difference between differential division and applying the differential operator ($\frac{d}{dx}$), so even though they are written the same way, you could argue that it should be written $\frac{dx}{dx}$ and $\frac{d}{dx}x$. The distinction is clear in higher order differentials: $\frac{dy^2}{dx^2}$ vs $\frac{d^2y}{dx^2}$($=\left(\frac{d}{dx}\right)^2y=\frac{d^2}{dx^2}y$).
