Let's say I have a differential equation
$$ \frac{dy(x)}{dx}=f(y,x) \tag{1} $$
Then, I notice that this would be much easier to solve if $x$ is substituted via $z=ax+b$. Is it correct to write it like this:
$$ \frac{dy(\frac{z-b}{a})}{dz/a}=f(y,\frac{z-b}{a}) \tag{2} $$
It feels like I have forgotten the chain rule here. The expression on the left hand side still looks like something I can evaluate using the chain rule:
$$ \frac{dy(\frac{z-b}{a})}{dz/a} = a \left.\frac{dy(g)}{dg}\right|_{g=\frac{z-b}{a}} \frac{dg}{dz} = \left.\frac{dy(g)}{dg}\right|_{g=\frac{z-b}{a}} \tag{3} $$
So that the system becomes
$$ \left.\frac{dy(g)}{dg}\right|_{g=\frac{z-b}{a}} = f(y,\frac{z-b}{a}) \tag{4}$$
or again
$$ \frac{dy(g)}{dg} = f(y,g) \tag{5}$$
Where is the problem in the notation here? I think in (2), the way it is written, I am not supposed to apply the chain rule but rather replace $y$ with some other function which then hides the full dependence on $z$. But can I write this down more cleanly?
Yeah, the notation is a bit confusing. Here's some alternatives.
Remembering that $y$ is implicitly a function, so you have actually been given $\dfrac{\mathrm d y}{\mathrm d x}(x)= f(y(x), x)$ and on substituting $g(z)$ for $x$, then this becomes: $$\begin{align}\dfrac{\mathrm d y}{\mathrm d g}(g (z)) &= f(y(g(z)),g(z))&&= y'(g(z))&=\left.\dfrac{\mathrm d y(x)}{\mathrm d x}\right\vert_{x:=g(z)}\\y'\!\left(\dfrac{z-b}{a}\right)&=f\!\left(y\!\left(\dfrac{z-b}{a}\right), \dfrac{z-b}{a}\right)\end{align}$$
If you want to derive $y(g(z))$ with respect to $z$ you will have:
$$\begin{align} \dfrac{\mathrm d y(g(z))}{\mathrm d z} &=\dfrac{\mathrm d y}{\mathrm d g}(g(z))\cdot \dfrac{\mathrm d g}{\mathrm d z}(z)&&=(y\circ g)'(z) = y'(g(z))\cdot g'(z)\\&=\dfrac 1a~f\!\left(y\!\left(\dfrac{z-b}{a}\right), \dfrac{z-b}{a}\right)&&\text{since }g(z)=\dfrac{z-b}{a} \end{align}$$