For example, suppose I have the following ODEs: $$(y')^2-2y'+1=0\tag{1}\label{eq:ODE1}\,;$$ $$(y')^2=0\tag{2}\label{eq:ODE2}\,,$$ where $y$ is dependent on $x$.
Then are they linear?
Immediately I would think not since they are not of the form $a_n(x)y^{(n)}+\cdots+a_1(x)y'=a_0(x)$ (and WolframAlpha seems to agree with this). However, if I rewrite $\eqref{eq:ODE1}$ and $\eqref{eq:ODE2}$ as $$(y'-1)^2=0\,;\quad y'=1\tag{1*}$$ and $$y'=0\,,\tag{2*}$$ now both equations are apparently linear.
What am I misunderstanding here? As far as I can tell it may be one of two things: either the first two equations are linear and there is a mistake with how WolframAlpha obtains this result or how I entered the equations, or my rearrangement was wrong and the pairs of equations are not equivalent.
I can't seem to find any similar questions on here or anywhere else so forgive me if this is a duplicate.
Reducing a nonlinear equation to a linear one can be quite a delicate issue. Yes, in your case, as Vasya observed, they are equivalent in the sense that any solution of the one is a solution of the other, and vice versa. But consider the case of a Bernoulli equation $$ \tag{1} y' = \sqrt{y}, $$ which can be reduced, via substitution $u = \sqrt{y}$, to the (certainly linear) equation $$ u' = 2. \tag{2} $$ $u(x) = 2x$ is a solution of $(2)$ for all $x \in \mathbb{R}$, and it corresponds to $y(x) = 4x^2$, which is a solution of $(1)$ only for $x \in [0, \infty)$. On the other hand, $y \equiv 0$ is a solution of $(1)$ which does not correspond to any solution of $(2)$. And, $$ y(x) = \begin{cases} 0 & \text{ for } x < 0 \\ 4x^2 & \text{ for } x \ge 0 \end{cases} $$ is yet another solution of $(1)$.