How to solve $y=(xy'+2y)^2$? What kind of differential equation is this thing and how to solve it?
$$y=(xy'+2y)^2$$
$$y=x^2y'^2+4xyy'+4y^2$$
 A: Lets try a substitution $y = x^{\alpha}u$
this results in
$$
x^{\alpha}u = x^2\left(\alpha x^{\alpha-1}u + x^{\alpha}u'\right)^2 + 4x\left(\alpha x^{\alpha-1}u + x^{\alpha}u'\right)x^{\alpha}u + 4x^{2\alpha}u^2 
$$
dividing by $x^{\alpha}$ and collecting like terms we find
$$
u = \left(\alpha^2+4\alpha + 4\right)u^2 + (2\alpha + 4)x^{\alpha+1}uu' + x^{\alpha+2}u'^2
$$
if we set $\alpha = -2$
then we have
$$
\alpha^2+4\alpha + 4 = 4 -8 + 4 = 0\\
2\alpha + 4 = -4 + 4 = 0
$$
thus we have
$$
u = u'^2
$$
this results in
$$
u' = \pm\sqrt{u} \implies 2\sqrt{u} = \pm x + C
$$
integrating leads to
$$
2u^{1/2} = \pm x + C
$$
replacing the subs, we obtain
$$
y = x^{-2}\left(\pm x + C\right)^2 = \left(\frac{C}{x}\pm 1\right)^2
$$
lets try the solution (always do this)
$$
xy' + 2y = -2\left(\frac{C}{x}+1\right)\frac{C}{x^2}x + 2\left(\frac{C}{x}+ 1\right)^2\\
(xy'+2y)^2 = \left(\frac{C}{x}+1\right)^2\left[-\frac{2C}{x} + 2\left(\frac{C}{x}+ 1\right)\right]^2 = \left(\frac{C}{x}+1\right)^2 = y
$$
A: given is the equation $$y=(xy'+2y)^2$$ from here we get $$\pm\sqrt{y}=xy'+2y$$
clearly is $$y(x)=0$$ one solution in the other case we have
$$\frac{dx}{x}=\frac{dy}{\pm\sqrt{y}-2y}$$
A: 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 $$ u = \pm \frac{1}{2} + \frac{C}{x}, y = \left( \pm \frac{1}{2} + \frac{C}{x} \right)^2 $$
A: Notice that $xy' + 2y$ is just a factor of $x$ away from equalling $(x^2 y)'$, 
so that 
$$y = \left(xy' + 2y\right)^2 = \left(\frac{x^2y' + 2xy}{x}\right)^2 = \frac{(x^2y)'^2}{x^2}$$
or 
$$x^2 y = (x^2 y)'^2.$$
Now, what function is equal to the square of its own derivative? 
If we can solve $v = v'^2$ then we'll be done upon letting $y = v/x^2$.
Unfortunately, $v = v'^2$ does not have a unique one-parameter family of solutions.
Certainly, $v = 0$ is one solution. Supposing that $v\ne 0$ on some open interval, we can separate variables: 
$$dx = \pm v^{-1/2}~dv \\ x = \pm 2 v^{1/2} + c \\ v = \frac{1}{4}(x-c)^2.$$
The interval of existence can only be extended until $v$ vanishes, 
at which point there is no reason to believe $v$ continues to "look like the way it did". 
We can patch two solutions together so long as differentiability is maintained; 
thus,
the full solution to $v = v'^2$ is 
$$v(x) = 0;$$
$$v(x) = \frac{1}{4}(x-c)^2$$
for $c$ in $\mathbb{R}$;
$$v(x) = \begin{cases}0 & x \le b \\ \frac{1}{4}(x-b)^2 & b \le x \end{cases}$$
for $b$ in $\mathbb{R}$;
$$v(x) = \begin{cases}\frac{1}{4}(x-a)^2 & x \le a \\ 0 & a \le x \end{cases}$$
for $a$ in $\mathbb{R}$;
and
$$v(x) = \begin{cases}\frac{1}{4}(x-a)^2 & x \le a \\ 0 & a \le x \le b \\ \frac{1}{4}(x-b)^2 & b \le x\end{cases}$$
for $a < b$ in $\mathbb{R}$.
The first four solutions can be seen as "limiting cases" of the fifth solution.
