I'm reading that in the confluent hypergeometric equation $$z\frac{d^2w}{dz^2}+(c-z)\frac{dw}{dz}-aw = 0$$ the change of variables $x = bz$ transforms this into
$$\frac{x}{b}\frac{b-x}{b}(b^2w'')+\left(c-(a+b+1)\frac{x}{b}\right)(bw')-abw = 0$$
This makes little sense to me--I see of course how $z$ transforms into $\frac{x}{b}$ but where did $\frac{b-x}{b}$ come from? I can identify this as $(1-z)$ but that was not a coefficient of the second-order term before the transformation.
And I would think that, if $w(z)=v(zb)$, then
$$\frac{d}{dz}w(z) = \frac{d}{dz}v(zb) = v'(zb)\cdot b$$
and
$$\frac{d^2}{dz^2}w(z) = \frac{d}{dz}v'(zb)\cdot b = b^2v''(zb)$$
and from here we could evaluate the DE in terms of $v$. But it seems to me something very different is going on. Perhaps $b$ here is not a constant as I've been assuming, but treating it as a variable would not result in the transformation described above, so I'm pretty thoroughly confused.