I've been doing some work, and I am profoundly confused, hopefully the answer will put me at ease. Thing is, for example, I have a differential equation of the form $$ \ddot{w}(z)+p(z)\dot{w}(z)+q(z)w(z)=0 $$ and I'm asked to do the change of variable kz, so we would have to compute $$ \frac{d}{dz}w(kz)\quad\text{ and }\quad \frac{d^2}{dz^2}w(kz) $$ and substite in our original equation. So far so good, my question arises when substituting on $p$ and $q$ (which are polynomials), my first inclination was to do $p(kz)$ and $q(kz)$ however an observation from my teacher was made: it is needed that the substitution on $p$ and $q$ be $\tfrac{1}{k}z$ instead, so it would be $$ p\left(\frac{1}{k}z\right)\quad\text{ and }\quad q\left(\frac{1}{k}z\right). $$ The result I was looking to achieved was achieved doing it this way, but why is the substitution on the polynomials different than on $w$ itself? Moreover, it appears that the inverse should be on the polynomials, I'm at awe at this.
And is this the general case?
Any help, insight or bibliography would be much appreciated, if a geometric point of view can be provided I would be even more grateful.
In the case where $z=kx$ and $k$ is a constant, if you replace $z$ everywhere you get $$ u(x)=w(kx)\\ \dot u(x)=k\dot w(kx)\\ \ddot u(x)=k^2\ddot w(kx) $$ so that inserted you get $$ \frac1{k^2}\ddot u(x)+\frac{p(kx)}{k}\dot u(x)+q(kx)u(x)=0 $$ Now you could also replace $k$ by $k^{-1}$ to get another version of this formula that appears closer to what your teacher wanted to obtain.