Question

Решить диф.уравн. xy(1+x^2)*y'=1+y^2

Answer 1

$xy(1+x^2)\cdot y'=1+y^2\\\\xy(1+x^2)\cdot \frac{dy}{dx}=1+y^2\\\\ \int \frac{y\, dy}{1+y^2} = \int \frac{dx}{x(1+x^2)} \\\\.\quad \frac{1}{x(1+x^2)} = \frac{A}{x} + \frac{Bx+C}{1+x^2} = \frac{A(1+x^2)+x(Bx+C)}{y} \\\\x^2\; |\; A+B=0\\\\x\; |\; C=0\\\\x^0\; |\; A=1\quad \Rightarrow \quad B=-A=-1\\\\\int \frac{dx}{x(1+x^2)} =\int \frac{dx}{x}-\int \frac{x\, dx}{1+x^2}=ln|x|-\frac{1}{2}\int \frac{2x\, dx}{1+x^2} =$

$=ln|x|-\frac{1}{2}ln|1+x^2|+C_1$

$\int \frac{y\, dy}{1+y^2}=\frac{1}{2}ln|1+y^2|+C_2$

$\frac{1}{2}ln(1+y^2)=lnx-\frac{1}{2}ln(1+x^2)-\frac{1}{2}lnC\\\\\frac{1}{2}(ln(1+y^2)+ln(1+x^2)+lnC)=lnx\\\\2lnx=ln(C(1+x^2)(1+y^2)\\\\x^2=C(1+x^2)(1+y^2)$