Question

Диффиринциальное Уравнение
xy(1+x^2)y'=1+y^2

Answer 1

$xy(1+x^2)y'=1+y^2\\\\\frac{dy}{dx}=\frac{1+y^2}{xy(1+x^2)}\\\\\int \frac{y\, dy}{1+y^2}=\int \frac{dx}{x(1+x^2)}\\\\\frac{1}{2}\int \frac{2y\, dy}{1+y^2}=\int (\frac{A}{x}+\frac{Bx+C}{1+x^2})dx\\\\\frac{1}{x(1+x^2)}=\frac{A}{x}+\frac{Bx+C}{1+x^2}=\frac{A(1+x^2)+x(Bx+C)}{x(1+x^2)}\\\\1=A+Ax^2+Bx^2+Cx\\\\x^2|\, A+B=0\\\\x|\, C=0\\\\x^0|\, A=1\; \to \; B=-A=-1$

$\frac{1}{2}\int \frac{d(1+y^2)}{1+y^2}=\int (\frac{1}{x}-\frac{x}{1+x^2})dx\\\\\frac{1}{2}ln|1+y^2|=\ln|x|-\frac{1}{2}ln|1+x^2|+ln|C|\\\\\sqrt{1+y^2}=\frac{Cx}{\sqrt{1+x^2}}\\\\\sqrt{(1+y^2)(1+x^2)}=Cx$