Однородное ДУ x^{2} +y^{2} =2xyy'

Решите задачу:

$x^2+y^2=2xyy'\; |:x^2\\\\1+(\frac{y}{x})^2= 2\cdot \frac{y}{x}\cdot y'\\\\u=\frac{y}{x}\; ,\; y=ux\; ,\; y'=u'x+u\\\\1+u^2=2u\cdot (u'x+u)\\\\2uu'x=1+u^2-2u^2\\\\2uu'x=1-u^2\; ,\; \; u'=\frac{1-u^2}{2ux}\; ,\; \; \frac{du}{dx}=\frac{1-u^2}{2ux}\\\\\int \frac{2u\, du}{1-u^2}=\int \frac{dx}{x}\\\\-\int \frac{d(1-u^2)}{1-u^2}=\int \frac{dx}{x}\\\\-ln|1-u^2|=ln|x|-lnC\\\\ln|x|+ln|1-u^2|=lnC\\\\x\cdot (1-u^2)=C\\\\x\cdot (1-\frac{y^2}{x^2})=C\\\\\frac{x^2-y^2}{x}=C\; ,\; \; x^2-y^2=Cx\\\\y^2=x\cdot (x-C)\\\\y=\pm \sqrt{x\cdot (x-C)}$