Question

решите
(1+y^2)dx+(1+x^2)dy=0

Answer 1

Разделяем переменные

$(1+y^2)dx+(1+x^2)dy=0 \ \ (1+y^2)dx=-(1+x^2)dy \ \ - frac{dy}{1+y^2} = frac{dx}{1+x^2} \ \ - intlimits {frac{dy}{1+y^2}} , dx = intlimits {frac{dx}{1+x^2}} , dx \ \ -arctgy=arctgx+C \ \ tg(-arctgy)=tg(arctgx+C) \ \ tg(arctgy)=-tg(arctgx+C) \ \ y=-tg(arctgx+C)$