Question

(1+y^2)dx+xdy=0 полное решение

Answer 1

$(1+y^2)dx+xdy=0|*frac{1}{x(1+y^2)}\frac{dx}{x}+frac{dy}{1+y^2}=0\frac{dx}{x}=-frac{dy}{1+y^2}\intfrac{dx}{x}=-intfrac{dy}{1+y^2}\ln|x|=-arctg(y)+C\ln|x|+arctg(y)=C$