Solve $y''+(y')^2y=0$ My idea is to set $v=y'$, then I get that
$$
y'' = v \frac{dv}{dy}.
$$
Hence the equation reads
$$
v\frac{dv}{dy}+v^2y=0\Leftrightarrow v\left(\frac{dv}{dy}+vy\right)=0
$$
and hence
$$
v = 0\text{ or }\frac{dv}{dy}+vy=0.
$$
For the second option, I get separate variables to get the equation
$$
\int \frac{1}{v}\, dv = -\int y\, dy
$$
which gives
$$
v = C e^{-\frac{y^2}{2}}
$$
Is this correct and how can I proceed?
I am just wondering how to get the solution(s) which are given here (see Example 2) which is
$$
\begin{cases}y=C,\\\frac{y^2}{3}-2Cy=2x+C^*\end{cases}
$$
Of course, from $v=0$, I also get that $y=C$ for some constant $C$ is a solution. But what about the other solution?

Solve instead
$$x'=c_1\,e^{\frac{y^2}{2}}$$ which will give
$$x=\frac{\sqrt{\frac{\pi }{2}} \text{erfi}\left(\frac{y}{\sqrt{2}}\right)}{c_1}+c_2$$ Do not try to make it explicit in $y$.
