I asked something related to this and this was a secondary question, after some answers. So I prefer to ask this in a new post because this question is extensive. Sorry for ask this simple things. )=
If I have an ODE, let's say of second order, and has the form $ \frac{d^2 y}{dx^2} = f \Big( {\frac{dy}{dx},y,x} \Big), $ where clearly $y$ is a function that depends on $x$.
There are two important kinds of change of variable. The second complains me, but I put i) also in case someone notes an error. In fact if someone know if this can be writted in a optional hide, but I don't know how to do it.
i) If I use some change of the form $s=g(x)$, then I have to compute $\frac{d^2 y}{dx^2}, \frac{dy}{dx}$ with the new variable. I know how to do it. $ \frac{dy}{dx} = \frac{dy}{ds}\frac{ds}{dx}. $ Then for the second, we have $ \frac{d^2 y}{dx^2} = \frac{d}{dx} \Big( \frac{dy}{dx} \Big) = \frac{d}{dx} \left( \frac{dy}{ds} \frac{ds}{dx} \right) = \frac{ds}{dx} \cdot \frac{d}{dx} \Big( \frac{dy}{ds} \Big) + \frac{dy}{ds} \cdot \frac{d}{dx} \Big( \frac{ds}{dx} \Big) .$ For the first term we have $ \frac{ds}{dx} \cdot \left( \frac{d}{dx} \Big( \frac{dy}{ds} \Big) \right) = \frac{ds}{dx} \cdot \left( \frac{d}{ds} \Big( \frac{dy}{ds} \Big) \cdot \frac{ds}{dx} \right) = \frac{d^2 y}{ds^2} \Big( \frac{ds}{dx} \Big)^2 , $ and the second is obviously equal to $ \frac{dy}{ds} \cdot \frac{d^2 s}{dx^2}. $ So we have that $ \frac{d^2 y}{dx^2} = \frac{d^2 y}{ds^2} \Big( \frac{ds}{dx} \Big)^2 + \frac{dy}{ds} \cdot \frac{d^2 s}{dx^2} $
ii) Here is my problem, a change of variable of the form $s= g(y)$. Here the technique is different because, in the other I start with $\frac{dy}{dx}$ and changed with the chain rule, but here is different because in a way it clearing in an indirect way. I start differentiating the equality with respect to $x$, I have $ \begin{align*} s &= g(y) \\ \frac{ds}{dx} &= \frac{dg}{dy} \cdot \frac{dy}{dx} \end{align*}$ But for the second, what can I do?