When studying about ODE's I've stumbled upon some types that are solvable by using parameterization (as explained by my book which unfortunately isn't written in English), for example:
$$t = f(y, y')$$ $$y = ty' + \psi(y')$$ $$y = t\phi(y') + \psi(y')$$
Second and third being Clairaut's and D'Alembert's equation, respectively. In my book, it is explained that those three types of equations can be solved by using parameter $p = y'$, substituting into original equation, calculating total differential and solving equation where $t$ is an unknown function of $p$. Then, plugging back we get solution where $y$ and $t$ as functions of parameter $p$. For example, let's look at this Clairaut's equation done in my book:
$$y = ty' + \sqrt{y'^2+1}$$ Let $p = y'$, then $dy = pdt$ (also, why is this allowed?)
Substituting, we have:
$$y = tp + \sqrt{p^2+1}$$ Taking differential on both sides, we get: $$dy = tdp + pdt + \frac{p}{\sqrt{p^2+1}}dp$$ $$(t + \frac{p}{\sqrt{p^2+1}})dp = 0$$ From which we conclude that:
$1)$ $dp = 0$ and we get the solution $p = c$, and plugging back we get that $y = tc + \sqrt{c^2+1}$ or
$2)$ $t + \frac{p}{\sqrt{p^2+1}} = 0$, now plugging in we get the solution:
\begin{cases} t = -\frac{p}{\sqrt{p^2+1}}\\ y = -\frac{p}{\sqrt{p^2+1}} + \sqrt{p^2+1}\ \end{cases}
My question is, why is this method correct, I couldn't find any proof and it seems weird to me that we can parametrize our starting equation just like that.