Question

Однородное ДУ
x^{2} +y^{2} =2xyy'

Answer 1

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