<s>[INST] How to solve $y=(xy'+2y)^2$? What kind of differential equation is this thing and how to solve it?
$$y=(xy'+2y)^2$$
$$y=x^2y'^2+4xyy'+4y^2$$
 [/INST] the differential equation $y = (x\frac{dy}{dx} + 2y)^2$ is singular at $x = 0.$ one has to find the solution away from $x = 0$ and patch it up around 
$x = 0$ and $y = 0$ or $y + \frac{1}{4}.$ there is a boundary layer at $x = 0.$
now we will find the outer solution that is valid away from $x = 0.$
$y = (x\frac{dy}{dx} + 2y)^2$ implies $y \ge 0$ so we can make a change of variable $y = u^2, u \ge 0.$ this transforms the differential equations to 
$$2xu \frac{du}{dx} + 2u^2 = \pm u , \frac{d(xu)}{dx} = \pm \frac{1}{2}$$ this has the general solution $$ u = \pm \frac{1}{2} + \frac{C}{x}, y = \left( \pm \frac{1}{2} + \frac{C}{x} \right)^2 $$
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