I’m reading a lecture in which this ODE: $$p\frac{\Bbb d^2v}{\Bbb dp^2} + 2(l+1-p)\frac{\Bbb dv}{\Bbb dp}+\big(k-2(l+1)\big)v = 0$$ is rearranged by the identity $x = 2p$ into this: $$x\frac{\Bbb d^2v}{\Bbb dx^2}+\big(2(l+1)-x\big)\frac{\Bbb dv}{\Bbb dx}+\left(\frac{1}{2}k-(l+1)\right)v = 0.$$
To be clear, $v$ is initially a function of $p$ (I’m guessing it would then become a function of $\frac{1}{2}x$ but I’m unsure). Also, $l$ and $k$ are variables.
I understand how the coefficient of $\frac{\Bbb dv}{\Bbb dx}$ is achieved as $2p$ is just substituted for $x$. However the coefficient of $v$ is suddenly halved and the coefficient of $\frac{\Bbb d^2v}{\Bbb dx^2}$ suddenly doubled (or so it appears to me) and I can’t work out why.