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В бифуркации Андронова — Хопфа Вы говорите, что вместе полярного радиуса будете использовать его квадрат, но затем в выводе уравнения на$\rho$ у Вас $(x^2+x^2)^2$ заменяется на $\rho$ , но исходя из сказанного выше следует, что оно должно быть равно $\rho^2$ . Я проверил Фазовые портреты, в них радиус предельного цикла всё-таки равен $\sqrt{-\epsilon /c}$ . Я исправил все места. $\rho_*=\sqrt{-\epsilon /c}$