Question

x(1-y^2)dx+y(1-x^2)dy=0

Answer 1

ДУ с разделяющимися переменными:

$\displaystyle x(1-y^2)dx+y(1-x^2)dy=0\\x(1-y^2)dx=-y(1-x^2)dy|*\frac{1}{(1-x^2)(1-y^2)}\\\frac{xdx}{1-x^2}=-\frac{ydy}{1-y^2}\\-\int\frac{d(1-x^2)}{1-x^2}=\int\frac{(1-y^2)dy}{1-y^2}\\-ln|1-x^2|=ln|1-y^2|+C\\\frac{1}{1-x^2}=C(1-y^2)\\(1-x^2)(1-y^2)=C;y=^+_-1;x=^+_-1$