<s>[INST] How to solve following differential equation $y = \frac{xy'}{2} + \frac{y'^2}{x^2}$? Given the following differential equation:
$$y = \frac{xy'}{2} + \frac{y'^2}{x^2}$$
I tried to solve it downgrading and as a quadratic equation for unknown $y'$ but it did not bring any results
UPD Possible solution:
$$y = \frac{xy'}{2} + \frac{y'^2}{x^2}$$
Let $y' = p$:
$$y = \frac{xp}{2} + \frac{p^2}{x^2}$$
Now multiply both side by 2:
$$2y = xp + 2\frac{p^2}{x^2}$$
$$d(2y) = d(xp) + d(2\frac{p^2}{x^2})$$
$$2dy = xdp+pdx + 2\frac{x^2d(p^2)-p^2d(x^2)}{x^4}$$
$$2pdx = xdp+pdx + 2 \frac{x(x2pdp - p^22dx)}{x^4}$$
Then multiply both side by $x^3$:
$$x^3pdx = x^4dp + 4xpdp - 4p^2dx$$
$$(x^3p + 4p^2)dx = (x^4 + 4xp)dp$$
$$p(x^3 + 4p)dx  = x(x^3 + 4p)dp$$
$$(x^3 + 4p)(pdx - xdp) = 0$$
Now solve $(x^3 + 4p) = 0$ and $(pdx -xdp) =0 $ we got $p = \frac{-x^3}{4}$ and $p = xConst$
Solution is: 
$y = -\frac{x^4}{16}$ and $y = \frac{x^2Const}{2} + Const^2$
 [/INST] Hint. If we differentiate the equation
$$
y = \frac{xy'}{2} + \frac{y'^2}{x^2}
$$
we obtain
$$
y'=\frac{y'}{2}+\frac{xy''}{2}+\frac{2y'y''}{x^2}-\frac{2y'^2}{x^3}
$$
or
$$
-\frac{y'}{2}+\frac{xy''}{2}+\frac{2y'y''}{x^2}-\frac{2y'^2}{x^3}=0.
$$
or
$$
(xy''-y')\left(\frac{1}{2}+\frac{2y'}{x^3}\right)=0.
$$
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