# Solve the following DE: $2y(y'+2)=x(y')^2$

I'm stuck trying to understand how to solve this differential equation: $$2y(y'+2)=x(y')^2$$ The main problem is to understand what type it is. I have never come across anything like this before. Could anyone give me a hint?

At first, I thought it is a Lagrange equation, so that was my attempt: $$y=\frac{x(y')^2}{2y'+4}$$ $$y'=p\Rightarrow y'=p=\frac{(p^2+2xpp')(2p+4)-2xp^2}{(2p+4)^2}$$ And then I got stuck trying to solve this equation for $$x$$ in terms of $$p$$.

Start using $$y=x \,z(x)$$ to make $$-x^2 z'(x)^2+z(x)^2+4 z(x)=0$$ Now, switch variables to make $$\frac{x'}x=\pm \frac 1{\sqrt{4z+z^2}}$$ which seems to be simple.
$$2y(y'+2)=x(y')^2$$ This is D'alembert's differential equation: $$y=x \left (\dfrac {y'^2}{2(y'+2)}\right)$$ Is of the form: $$y=x f(y')+g(y')$$
You made a little mistake here : $$p=\frac{(p^2+2xpp')(2p+4)-2xp^2}{(2p+4)^2}$$ It shoul be: $$p=\frac{(p^2+2xpp')(2p+4)-2xp^2\color {red}{p'}}{(2p+4)^2}$$ Then it factorize nicely into: $$p(p+4)(p+2-xp')=0$$
$$\begin{cases} p=0 \\ p+4=0 \\ p+2-xp'=0 \end{cases}$$ And $$y=0$$ is also a solution.