Question

Дифференциальное уравнение
2xy*y'=x^2+y^2

Answer 1

$2xyy'=x^2+y^2\\y'=\frac{x^2+y^2}{2xy}\\y'=\frac{t^2x^2+t^2y^2}{2txty}=\frac{t^2(x^2+y^2)}{t^22xy}=\frac{x^2+y^2}{2xy}\\f(x;y)=f(tx;ty)=>y=ux,\:y'=u'x+u\\u'x+u=\frac{x^2+u^2x^2}{2ux^2}\\u'x=\frac{1+u^2}{2u}-u=\frac{1-u^2}{2u}\\\frac{du}{dx}=\frac{1-u^2}{2ux}\\\frac{2u}{1-u^2}du=\frac{dx}{x}\\\int{\frac{2udu}{1-u^2} }=\int{\frac{dx}{x}}\\-\ln(1-u^2)=\ln(x)+\ln(C)\\\frac{1}{1-u^2}=x+C\\u^2=-\frac{1}{x+C}=-\frac{1-x-C}{x+C}=\frac{x+C}{x}\\u=\pm\sqrt{\frac{x+C}{x}}\\\frac{y}{x}=\pm\sqrt{\frac{x+C}{x}}$

$y=\pm\sqrt{x^2+Cx}$