$("x" - "y"^2"x") "dx" - ("y" + "x"^2"y")"dy" = 0,$ when x = 2, y = 0
x (1 - y2) dx - y (1 + x2)dy = 0
$"x"/(1 + "x"^2) "dx" - "y"/(1 - "y"^2)"dy"$ = 0
$"2x"/(1 + "x"^2) - "2y"/(1 + "y"^2)"dy" = 0$
Integrating both sides, we get
$int "2x"/(1 + "x"^2) "dx" + int(-2"y")/(1 - "y"^2) "dy" = "c"_1$
Each of these integrals is of the type
$int ("f"'("x"))/("f"("x")) "dx" = log |"f"("x")| + "c"$
the general solution is
log |1 + x2| + log |1 - y2| = log c, where c1 = log c
log |(1 + x2)(1 - y2)| = log c
(1 + x2)(1 - y2) = c
When x = 2, y = 0, we have
(1 + 4)(1 - 0) = c
c = 5
the particular solution is (1 + x2)(1 - y2) = 5.
