Question

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

Answer 1

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

$y=pmsqrt{x^2+Cx}$