How to solve ode of the form $ a_n(y')^n + a_{n-1}(y')^{n-1} + \cdots + a_1 y' + a_0 =0 $ $
a_n(y')^n + a_{n-1}(y')^{n-1} + \cdots + a_1 y'+a_0 =0
$
I am also unclear on how to describe this as it is not nth order. The polynomial being in the derivative is not something that I think I have seen before.
 A: Well. As pointed out in the comments by Seth.
Let us consider an example
\begin{align}
(y')^2+2y'+1=0
\end{align}
then it follows $y' = -1$ so $y = -t+C$. 
So, in general, you will get $y'=$ const. Then $y=\text{const}. t+C$.
A: 
TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.

Let $y:[0,T)\rightarrow\mathbb{R}$ where $T\leq\infty$ be a continuously differentiable function satisfying the ODE
$$
a_{n}(y^{\prime}(t))^{n}+\cdots+a_{1}y^{\prime}(t)+a_{0}=0.
$$
By the fundamental theorem of algebra, if $a_{n}\neq0$, the corresponding polynomial
$$
a_{n}r^{n}+\cdots+a_{1}r+a_{0}
$$
has $n$ complex roots, call them $z_{1},\ldots,z_{n}$. We can re-express the ODE as
$$
\left(y^{\prime}(t)-z_{1}\right)\cdots\left(y^{\prime}(t)-z_{n}\right)=0.
$$
By the above, $y^{\prime}(t)=z_{k(t)}$ where $k(t)\in\{1,\ldots,n\}$ for all $t$.
In fact, $t\mapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.
