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?