Classifying a strange ODE I have the given ODE:
\begin{equation}
y'= \frac{y^3}{x^3} + \frac{y}{x} + 1
\end{equation}
which turns out to give a strange  solution:

with the following solution curves:

One of the curves, where $C=100$ has the following shape, which is really strange, nearly linear like $y=x$ over the origin, then it grows slowly like a quadratic equation.

I am looking for a classification of this ODE. To me this appears as a nonlinear nonhomogeneous differential equation. Am I missing some more details?
 A: It is quite a homogeneous equation (not in the linear sense), in that you can set $y=xu$ to get
$$
xu'+u=u^3+u+1\\
xu'=u^3+1=(u+1)(u^2-u+1)
$$
The terms you get are quite typical for the anti-derivative of a partial fraction decomposition, here with terms
$$
\frac{u'}{u+1},~~~\frac{(2u-1)u'}{u^2-u+1},~~\text{and}~~\frac{u'}{(u-\frac12)^2+\frac34}.
$$
As long as $u$ is relatively small, the $1$ will dominate in $1+u^3$, so around points where $u=0$ you get approximately $u'=1$, $u=x+c$, $y=x^2+cx$. If $c$ is large enough, the linear term will dominate for some time.
A: As @Lutz Lehmann answered, after $y=x u$, the equation becomes $$xu'=(u+1)(u^2-u+1)$$ Now, swithching variables
$$\frac {x}{x'}=(u+1)(u^2-u+1)\quad \implies \quad  \frac {x'}{x}=\frac 1{(u+1)(u^2-u+1)}$$ Then, partial fraction decomposition
$$\frac {x'}{x}=\frac{2-u}{3 \left(u^2-u+1\right)}+\frac{1}{3 (u+1)}$$
Integration
$$\log(x)+C=\frac{1}{\sqrt{3}}\tan ^{-1}\left(\frac{2 u-1}{\sqrt{3}}\right)+\frac 16 \log \left(\left|\frac{(u+1)^2}{u^2-u+1}\right|\right)$$
