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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 $p$.

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Hint

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.

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    $\begingroup$ How did you find the first expression (after the substitution)? $\endgroup$
    – Bonrey
    Jun 20 '20 at 9:41
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$$2y(y'+2)=x(y')^2$$ 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.

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