How to solve the differential equation: $yy'^2-2xy'+y=0$ Solve the differential equation: $yy'^2-2xy'+y=0$ (*)
I have 
$(*)\Leftrightarrow (yy')^2-2xyy'+y^2=0\\
\Leftrightarrow z'^2-4xz+4z^2 =0 \ \ (z=y^2)$
I can't know how to continue.
Could you give me some hints. Thanks for helping.
 A: $$yy'^2-2xy'+y=0$$
$z'^2-4xz+4z^2 =0\quad$ is false (two typos in it). The correct equation is :
$$ (yy')^2-2xyy'+y^2=0 \quad\to\quad z'^2-4xz'+4z =0 \quad (z=y^2)$$
The polynomial form of the ODE draw us to try $z=ax+b$
$$a^2-4xa+4(ax+b)=0 \quad\to\quad a^2+4b=0 \quad\to\quad b=-\frac{a^2}{4}$$  $$z(x)=ax-\frac{a^2}{4}$$
The try is successful and the solution contains an arbitrary constant $a$.
$$y(x)=\pm\sqrt{ax-\frac{a^2}{4}}$$
GENERAL SOLUTION :
The above calculus is a short way to obtain some particular solutions. More generally :
$$(yy')^2-2xyy'+y^2=(x-yy')^2-x^2+y^2=0$$
Change of function : $2u=x^2-y^2 \quad\to\quad u'=x-yy'$ :
$$(u')^2-2u=0$$
First solution : $u=0 \quad\to\quad x^2-y^2=0\quad\to\quad y=\pm x$
Second solution, with $u\neq 0$ :
$$u'=\pm\sqrt{2u}\quad\to\quad \frac{u'}{2\sqrt{u}}=\pm \frac{1}{\sqrt{2}} \quad\to\quad \sqrt{u}=\pm \frac{x-c}{\sqrt{2}}$$
$$2u=(x-c)^2=x^2-y^2$$
$$y=\pm\sqrt{x^2-(x-c)^2}$$
$$y=\pm\sqrt{2cx-c^2}=\pm\sqrt{ax-\frac{a^2}{4}}\quad \text{ with } a=2c$$
Of course, this is more complicated than the above trial method. But this makes sure that no solution is forgotten.
